pages.uwc.edupages.uwc.edu/shubhangi.stalder/mat 110/college...page iii about the authors dr....
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COLLEGE ALGEBRA
Stalder & Martin
Page i
Dedication
This book is dedicated to
All students who want to learn
And
All teachers who want to teach.
Page ii
College Algebra
Dr. Shubhangi Stalder
University of Wisconsin Waukesha
&
Dr. Paul Martin
University of Wisconsin Marathon
All material in this book is copyrighted to the authors and can only be used for non-profit educational
purposes only. Please contact the authors for any other use.
First Edition, 2016
Page iii
About the authors Dr. Shubhangi Stalder is a full professor of mathematics at the University of Wisconsin
Waukesha. She has received her doctoral degree in mathematics from the University of
Wisconsin Milwaukee in 1993. She has decades of teaching experience and her focus has
always been to reach out to those who are struggling in
mathematics. Her main belief is that everyone can learn basic
mathematics if they tried. The key is to understand the βWhyβ
and the βHow, and to be able to see the patterns across different
mathematical processes. She believes that in the long run rote
memorization does not work to learning mathematics. She uses
yoga and meditation techniques with her students who
experience math and test anxiety and continues to include
mindfulness practice in her teaching of mathematics. She has received the UW System Board of
Regents Teaching in Excellence Award (the state of Wisconsinβs highest teaching award), the
UW Colleges Chancellorβs Excellence in Teaching Award, and the UW Colleges Kaplan Teacher
Award.
Dr. Paul Martin is a full professor of mathematics at the University of Wisconsin Marathon. He received his doctoral degree in mathematics from the University of Wisconsin Madison in 1994. He has decades of teaching experience and his focus has always been to help his students see how mathematics connects to the real world. He does this through building 3-dimensional models, modeling physical processes such as heat loss through the attic of a house, to connecting mathematics to his studentsβ other classes from chemistry to music. As a teacher, Martin stresses the importance of reasoning over memorization. He has received several teaching awards over his career including the prestigious UW Colleges Chancellorβs Excellence in Teaching Award.
Page iv
Acknowledgements This book is the first edition. The material in this book is continuation of the developmental and
Intermediate Algebra text by the same authors.
The authors would like to thank their families who gave their professional opinions, their
constant support and put up with long hours on this project (without their support it would
have been difficult to put this book together in a such a short time).
We also would like thank Arman Banimahd, Amanuel Teweldemedhin, and Adam Figarsky for
their willingness to discuss the content and try it out in the classroom.
We also want to thank Kent Kromarek who was our sounding board and input into some of the
History of Mathematics presented in the book and for reviewing and editing the book. In
addition, we want to thank all the countless brave souls (students and teachers) who were
willing to give us their time in trying this material, and giving us their feedback to make this
project better. We hope their work and ours has made it possible to get a product that we hope
will be useful to other students and teachers.
Table of Contents
CHAPTER 1: ALL ABOUT FUNCTIONS AND RELATIONS ............................................................ 1
1.1 INTRODUCTION AND DOMAIN AND RANGE OF FUNCTIONS AND RELATIONS ....................................................1
WORKSHEET SEC 1.1 ....................................................................................................................................... 16
EXERCISES 1.1 ................................................................................................................................................ 17
1.2 INVERSE FUNCTIONS AND A BRIEF LIBRARY OF FUNCTION TYPES .............................................................. 23
OTHER TYPES OF FUNCTIONS ............................................................................................................................. 27
WORKSHEET SEC 1.2 ....................................................................................................................................... 36
EXERCISES 1.2 ................................................................................................................................................ 37
1.3 EXPONENTIAL AND LOGARITHMIC FUNCTIONS .................................................................................... 50
COMPOUND INTEREST AND THE NUMBER π .......................................................................................................... 52
EVALUATING LOGARITHMIC FUNCTIONS .............................................................................................................. 56
PROPERTIES OF LOGARITHMS ............................................................................................................................ 59
WORKSHEET SEC 1.3A ..................................................................................................................................... 64
EXERCISES 1.3A .............................................................................................................................................. 65
ANOTHER CLASS OF FUNCTIONS: SEQUENCES ...................................................................................................... 69
FIBONACCI SEQUENCE AND THE GOLDEN RATIO .................................................................................................... 68
WORKSHEET SEC 1.3B...................................................................................................................................... 73
EXERCISES 1.3B ............................................................................................................................................... 75
EVEN AND ODD FUNCTIONS .............................................................................................................................. 77
WORKSHEET SEC 1.3C...................................................................................................................................... 80
EXERCISES 1.3C ............................................................................................................................................... 81
1.4 ARITHMETIC AND COMPOSITION OF FUNCTIONS ..................................................................................... 83
WORKSHEET SECTION 1.4A ............................................................................................................................... 87
EXERCISES 1.4A ............................................................................................................................................... 88
COMPOSITION OF FUNCTIONS ........................................................................................................................... 89
WORKSHEET SECTION 1.4B ............................................................................................................................... 94
EXERCISES 1.4B .............................................................................................................................................. 95
CHAPTER 2: GRAPHING FUNCTIONS AND RELATIONS ...................................................................... 101
2.1 LIBRARY OF FUNCTIONS .................................................................................................................. 101
HORIZONTAL AND VERTICAL SHIFT OF FUNCTION GRAPHS .................................................................................... 106
HORIZONTAL AND VERTICAL STRETCH/COMPRESSION AND REFLECTIONS ................................................................ 110
COMBINATION OF TRANSLATIONS, STRETCHES AND REFLECTIONS .......................................................................... 117
WORKSHEET SEC 2.1 ..................................................................................................................................... 121
EXERCISES 2.1 .............................................................................................................................................. 122
2.2 GRAPHING CONIC SECTIONS ............................................................................................................ 129
2.2A PARABOLAS .......................................................................................................................................... 130
WORKSHEET SEC 2.2A ................................................................................................................................... 138
EXERCISES 2.2A ............................................................................................................................................. 139
2.2B CIRCLES AND ELLIPSES ............................................................................................................................. 140
CIRCLES........................................................................................................................................................ 144
WORKSHEET SEC 2.2B.................................................................................................................................... 146
EXERCISES 2.2B ............................................................................................................................................. 147
2.2C HYPERBOLAS .............................................................................................................................. 148
SUMMARY AND TIPS ON IDENTIFYING CONIC SECTIONS ........................................................................................ 151
WORKSHEET SEC 2.2C.................................................................................................................................... 154
EXERCISES 2.2C ............................................................................................................................................. 155
2.3 GRAPHING POLYNOMIAL FUNCTIONS ......................................................................................................... 159
WORKSHEET SEC 2.3 ..................................................................................................................................... 182
EXERCISES 2.3 .............................................................................................................................................. 183
2.4 GRAPHING RATIONAL FUNCTIONS ..................................................................................................... 190
LINEAR ASYMPTOTES OF RATIONAL FUNCTIONS .................................................................................................. 190
RECIPROCAL OF A FUNCTION ........................................................................................................................... 197
ZEROS OF THE NUMERATOR AND DENOMINATOR ............................................................................................... 201
END BEHAVIOR OF A GRAPH OF A RATIONAL FUNCTION ....................................................................................... 206
WORKSHEET SEC 2.4 ................................................................................................................................... 214
EXERCISES 2.4 .............................................................................................................................................. 215
CHAPTER 3 SOLVING EQUATIONS AND INEQUALITIES ..................................................................... 222
3.1 QUADRATIC EQUATIONS ................................................................................................................. 222
REVIEW OF COMPLEX NUMBERS ...................................................................................................................... 224
QUADRATIC EQUATIONS ................................................................................................................................. 227
COMPLETING THE SQUARE .............................................................................................................................. 228
QUADRATIC FORMULA ................................................................................................................................... 230
WORKSHEET SEC 3.1 ..................................................................................................................................... 234
EXERCISES 3.1 .............................................................................................................................................. 235
3.2 POLYNOMIAL EQUATIONS OF DEGREE THREE OR HIGHER ........................................................................ 241
RATIONAL ZEROS THEOREM: ........................................................................................................................... 241
DIVISION AND FACTOR THEOREMS ................................................................................................................... 242
COMPLEX ZEROS OF POLYNOMIALS WITH REAL COEFFICIENTS .............................................................................. 242
USING DIVISION TO FIND ZEROS....................................................................................................................... 245
WORKSHEET SEC 3.2 ..................................................................................................................................... 249
EXERCISES 3.2 .............................................................................................................................................. 250
3.3 EXPONENTIAL AND LOGARITHMIC EQUATIONS ..................................................................................... 258
LOGARITHMIC EQUATIONS AND INEQUALITIES .................................................................................................... 258
EXPONENTIAL EQUATIONS AND INEQUALITIES .................................................................................................... 260
WORKSHEET SEC 3.3 ..................................................................................................................................... 266
EXERCISES 3.3 .............................................................................................................................................. 267
3.4 SYSTEMS OF EQUATIONS ................................................................................................................ 276
WORKSHEET SEC 3.4 ..................................................................................................................................... 281
EXERCISES 3.4 .............................................................................................................................................. 282
3.5 INTRODUCTION TO MATRICES AND GAUSS ELIMINATION METHOD ........................................................... 284
ELEMENTARY ROW OPERATIONS ON A MATRIX OF NUMBERS. .............................................................................. 285
WORKSHEET SEC 3.5 ..................................................................................................................................... 289
EXERCISES 3.5 .............................................................................................................................................. 290
Functions and Relations Page 1
Chapter 1: All about Functions and Relations
1.1 Introduction and Domain and Range of Functions and Relations Lecture
Functions and Relations Part 1 (13.15 min) https://www.youtube.com/watch?v=2HCxvI_S-WQ
We very often consider how different things are related in many situations in every-day life. Some
examples include: each person is related to their biological mother; the price of gas in dollars per gallon
is related to time; temperature in Fahrenheit degrees is related to temperature in Celsius degrees; your
height in inches is related to your age in years; the profit of an automaker is related to the annual car
sales; the number of movie ticket sales at a theater is related to the ticket price. Many of these
relationships can be quantified using numerical measures of each related quantity, e.g., the height of a
person in inches on his birthday is related to his age in years, or the relationship between Fahrenheit
and Celsius temperatures is given by the equation β =9
5β+ 32 for any value of β.
The connections between different quantities can be referred to as relations.
Oftentimes we think of a direction in the relation in that when the value of one quantity (called an
input) is known, this determines a value of the second quantity (called an output).
A function is a directed relation where every individual input has a unique output.
We will focus on relations between two items or characteristics that are connected to each other. For
example, think of the students in a class and their height in inches. We could represent these data as a
set of ordered pairs or as a table.
A={(Robert, 72 ), (Sarah, 61 ), (Matt, 70 ), (Robin, 61 ), (Sarah, 65)}
Name of Student Height in Inches
Robert 72
Sarah 61
Matt 70
Robin 61
Sarah 65
In the data pairs above, two people could have different names and have the same height. Also two
people can have the same name but different heights, or two could have the same height and the same
name with a larger class. The way this data is presented suggests that the name of the person is the
βinputβ, while the height of that person is the βoutputβ.
Clearly the relation above is not a function, since the single input βSarahβ has two different outputs 61ββ
and 65ββ. Note that if the input column included middle initial,
Functions and Relations Page 2
e.g., B={(Robert A, 72 ),(Sarah K, 61 ), (Matt L, 70 ), (Robin P, 61 ),(Sarah M, 65)}, then each input has a
single output and π΅ would be a functional relation of the form (ππππ, βπππβπ‘). However the relation
π΅ in the other direction from height to name would not be a function since in that direction, the input of
61β yields two different names!
Consider the sister relationship where any person with one or more sisters is the input and that personβs
sister(s) is the output. Since there are people who have more than one sister, this is not a function
relationship. Next, consider the biological mother relationship where any person is an input and the
output is that personβs biological mother. Explain why this is a function. Note if we reverse the direction
of this mother relation, it is not a function, i.e., starting with an input of a mother with more than one
child, that mother would give rise to at least two different outputs (her children).
Another way to express some relations and functions is through an equation with two variables. For
example, π¦ = 3π₯ + 1, where the input π₯ is any real number, and π¦ is the corresponding output number.
This is a relation between π₯ and π¦ that is a function since for any value of the input π₯, there is only one
output, namely 3π₯ + 1. Weβd say this equation represents a function from π₯ to π¦. In the relation π₯ =
π¦2, where input π₯ is any nonnegative real number, π¦ is not a function of π₯, since the input of π₯ = 4 will
yield two different outputs π¦ = 2, π¦ = β2.
In order to distinguish two different functions like π¦ = 3π₯ + 1 and π¦ = π₯ β 5, we need a notation that
clearly describes the two βπ¦βs as being different. We can use the notation π(π₯) = 3π₯ + 1 and g(π₯) =
π₯ β 5 where we are giving names of π and π to these two different functional relations. We would read
that as βf of π₯ is 3π₯ + 1β and βg of π₯ is π₯ β 5β. We could also write this as e.g., π¦ = π(π₯) = 3π₯ + 1
which identifies the function equation π¦ = 3π₯ + 1 with the function name π. It is very important to
make sure you learn to read and process this notation correctly. Many students who make mistakes in
working with functions confuse the notation π(π₯) with π times π₯. Remember that the notation π(π₯) is
NOT read as π times π₯. You can think of it as π( ____ ), and the blank can be filled in with many different
inputs.
In the notation π(π₯) = 3π₯ + 1 , π is the name of the function, and π₯ is called the input or argument,
π(π₯) refers to the output. This statement simply says that the function π computes an output by
multiplying an input by 3 and then adding 1.
The collection of all the input π₯ values we can have in a function so that the output is well
defined is called the domain of the function. The collection of all the output values of a function
is called the range of the function.
Working with Domain and Range of a Function or a Relation
Playing
Remember there cannot be a function or a relation without having a domain and range. In mathematics we are always trying to use our imagination to see what all possibilities are in every new scenario.
Some questions to ask are
Functions and Relations Page 3
1. What kinds of sets of objects we can be used for domains and ranges. Try to come up with functions for each of the domains: {All the houses in the city you live in}, {All the employees at the school you are attending}, {All the days in time since you were born}.
2. Do functions have to have one input variable and one output variable, or could we have multiple inputs and one unique output? For example can you think of a function perhaps given by an equation where the output depends on two separate real numbers as inputs?
Before you continue reading, spend some time just thinking and trying to answer these questions. There are many different answers. Try being creative and see what you can come up with.
A function of one variable means the input can be described by just one variable, e.g., π(π₯) = 3π₯ + 1. When a function has multiple inputs and one output we call it a function of several variables. For example, the volume of a box is a function of three variables, its length π, its width π€ and its height β and we might write this as π = π(π, π€, β) = π β π€ β β. The domain of a function can be anything you want it to be from a finite set of objects to infinite sets. We will mostly work with functions of a single variable in this book.
Next we develop some notation and conventions on how work with functions efficiently. Any letter really can represent the single input. For example, we may use letters like π₯, ππ π‘ to represent an input to a function. Similarly we use a different letter to represent the corresponding single output, e.g., π¦, or π§ might represent the output of a function. The functional relationship itself will often be denoted by the letter π or some other appropriate letter. A statement like π¦ = π(π₯) simply says that π¦ represents an output, π₯ an input and they are related by the function called π. (Remember that this is not saying to multiply π by π₯!) To define what this relationship is, weβd have to elaborate on the meaning of π¦ and π₯ and how π¦ and π₯ are related perhaps by some equation. We can also look at the function notation as follows where the object π₯ is transformed by the application of the function π to become a new object called π(π₯) (read as π of π₯)
π₯πβ π(π₯)
This notation is credited to the mathematician Leonard Euler and was developed around 1734.
To denote what kind of function we are working with mathematicians may use the notation shown
below. Writing π: RβR to means we have a real number as an input and also a real number as the
output. Such a function is called a real valued function of one variable. Writing π: RΓRβ R to mean the function π is taking an ordered pair of real numbers and transforming them into a real number. Such a function is called a real valued function of two variables. Functions can have as many input variables as
needed. Writing π: RnβR to means we have an n-tuple as an input and output is a real number.
We can represent functions in many ways. Below are some of the ways.
a) Using function notation as described above. 1 π(π₯) = π₯2
Here the domain is set of all real numbers and range is set of all non-negative real numbers.
2 π(π₯, π¦) = π₯2 + π¦2 Here the domain is the set of ordered pairs of the type (π₯, π¦) with π₯ and π¦ both real numbers and the range is the set of all non-negative real numbers.
3 π(π‘) = 2π‘ β 5 Here the domain is set of all real numbers and range is the set of all real numbers.
Functions and Relations Page 4
b) As a set of objects. For example, 1. π: {(2,β1), (4,2), (3,2), (0,4)}.
This means the input 2 to π produces the output β1. The input 0 produces the output 4 and so on. Here the domain is the set {2,4,3,0} and the range is the set {β1,2,4}. A set is just a group of objects and we donβt need to list the β2β twice.
2. π: {(π₯, π¦)|π¦ = π₯2, π₯ β R}. Read this as π is the function that is a collection of all ordered pairs of the type (π₯, π¦) in which π¦ = π₯2 and π₯ is a real number. Here the domain is the set of all real numbers and range is set of all non-negative real numbers.
c) As formulas or equations in two or more variables. For example,
1. π¦ = π₯2, Here the domain is the set of all real numbers and range is set of all non-negative real numbers.
2. π§ = π₯2 + π¦2, Here the domain is the set of ordered pairs of the type (π₯, π¦) with π₯ and π¦ both real numbers and the range is the set of all non-negative real numbers.
d) As English sentences modeling scenarios. For example, 1. Lindaβs parents loaned her $12,480 interest free for her college tuition and books. Linda
promised to pay her parents back $80 a week until the loan is completely paid off.
Write a function equation that tell Linda how much money she still owes π₯ weeks after
she began paying on the loan.
The equation would look like: Amount owed by Linda = π¦ = 12480 β 80 β π₯ The
domain is all whole numbers less the number of weeks it takes to pay off the loan, i.e., 12480
80= 156. The range is all whole numbers counting down by 80βs from 12480 to 0.
2. The period of a pendulum is proportional to the square root of its length. Here domain is set of all positive real numbers and range is also set of positive real numbers.
e) As graphs.
1. Input is the π₯ coordinate of the point and output is the π¦ coordinate of the point. For example input ofβ2 will give you output of 5
since (β2,5) is a point plotted above. Here the domain is {β5,β3,β2,0,1,2,3,4,7} and range is {β1,0,1,2,3,4,5}.
2. Input is the π₯ coordinate of a point on the graph of the function π(π₯) =π₯2, then output is the π¦ coordinate of that point. For
example input ofβ2 will give you output of 4 since (β2,4) is a point on the graph above. Here the domain is the set of all real numbers and range is set of all non-negative real numbers.
Functions and Relations Page 5
Below are more examples of functions.
1. F= π(πΆ) =9
5πΆ + 32 . This statement is a description of the degrees centigrade to Fahrenheit
conversion function. The input is the temperature in centigrade denoted as πΆ, the output is the temperature in Fahrenheit πΉ, and the name of the function is π. In this course our functions usually will be defined in this way, i.e., π¦ = π(π₯) =some algebraic formula that involves π₯.
Also note that a function is the relation between input and output and that π(π₯) =9
5π₯ + 32 is
the same function as π(πΆ) =9
5πΆ + 32 since they both have exactly the same set of
(input, output) ordered pairs.
2. π =4
3ππ3 is the formula for volume V of a sphere where the radius= π is known. This is a
polynomial function. We might call this function simply π in which case weβd have π = π(π) =4
3ππ3 where we use the same variable to denote an output and the name of the relationship.
Likewise, the area of a circle function can be expressed as π΄ = π΄(π) = ππ2 where π =radius of the circle.
3. The height of a baseball at a given time might be given by π¦ = β(π‘) = 2 + 160π‘ β 16π‘2 where π¦ is the height above the ground in feet and π‘ is the number of seconds after the ball is hit by the bat and β is the name of this relationship between time and height. This is also an example of a polynomial function. This notation allows us to easily describe the height at several different times, e.g.,β(2) = 258 says that when the input is 2 seconds, the height is 258 feet. Other questions can also be stated easily using this notation, e.g., βdetermine when β(π‘) =100β says that the output of the height function is 100 and we are to find the input π‘ when this happens. You could try guessing and checking to find out that this happens at π‘ β0.66π ππ, and π‘ β 9.34 π ππ. You can use the quadratic formula to find exact answers to 100 =2 + 160π‘ β 16π‘2. We will review this in a later section.
4. Another type of function that is often useful is called an exponential function where the input is actually in the exponent. For example π(π‘) = 7.4(1.012)π‘ might be used as a model of the worldβs population π‘ years after 2016 where the output is in billions of people. Thus π(10) =7.4(1.012)10 β 8.34 billion. This number 8.34 billion is the projection of the world population in 2026.
Functions and Relations Page 6
Playing with function notation
If we say π(π₯) = 3π₯ + 1, then that means that we really have
π( ______) = 3 Γ (______) + 1
That means if you are asked to evaluate the following you must replace the blank above with whatever
takes its place in the notation
π(2) = 3 Γ (2) + 1 = 6 + 1 = 7
π(100) = 3 Γ (100) + 1 = 300 + 1 = 301
π(π) = 3(π) + 1 = 3π + 1
π(π + β) = 3(π + β) + 1 = 3π + 3β + 1
See if you can use these examples to work on the following practice problems.
Practice Problems
1. Let π(π₯) = 7π₯2 + 1 and π(π‘) = βπ‘ + 7. Find the values of
a. π(4)
b. π(π + β)
c. π(β2)
d. π(π + 1)
e. π(4)
f. π(3) + π(π)
2. Let π΄ππ πππ’π‘π(π‘) = |π‘|. Find the values of
a. π΄ππ πππ’π‘π(β5.1)
b. π΄ππ πππ’π‘π(10)
c. π΄ππ πππ’π‘π(β1200)
d. π΄ππ πππ’π‘π(1400)
π( ______) = 3 Γ (______) + 1
π(πΆππΏπ·) = 3 Γ (πΆππΏπ·) + 1
Functions and Relations Page 7
Solutions
1. Let π(π₯) = 7π₯2 + 1 and π(π‘) = βπ‘ + 7. Find the values of
a. π(4) = 7 β 42 + 1 = 7 β 16 + 1 = 113
b. π(π + β) = 7 β (π + β)2 + 1
c. π(β2) = 7 β (β2)2 + 1 = 7 β 4 + 1 = 29
d. π(π + 1) = βπ + 1 + 7
e. π(4) = β4 + 7 = 9
f. π(3) + π(π) = 7 β (3)2 + 1 + βπ + 7 = 63 + 1 + βπ + 7 = 71 + βπ
2. Let π΄ππ πππ’π‘π(π‘) = |π‘|. Find the values of
a. π΄ππ πππ’π‘π(β5.1) = |β5.1| = 5.1
b. π΄ππ πππ’π‘π(10) = |10| = 10
c. π΄ππ πππ’π‘π(β1200) = |β1200| = 1200
d. π΄ππ πππ’π‘π(1400) = |1400| = 1400
Vertical Line Test:
The graph of a relation is a function from π₯ to π¦ if and only if every vertical line intersects the
graph at no more than one point.
Examples
1. Below π¦ is a function of π₯ as it passes the vertical line test.
2. Below π¦ is not a function of π₯ as it does not pass the vertical line test.
3. Below π¦ is a function of π₯ as it passes the vertical line test.
As you can see from the graph above, any vertical line will intersects the graph at only one point.
As you can see from the graph above, any vertical line will intersects the graph at only one point for the domain π₯ β₯ 0.
As you can see on the graph above, the vertical line intersects the graph in two points.
Functions and Relations Page 8
Playing
If you look at examples 1 and 3 above, a natural question that we can ask is what information would a
horizontal line test provide? We can see that if a horizontal line is drawn in example 1, it intersects the
graph at two points. This means that a single π¦-value where the horizontal line is at is related to two
points or two π₯-values. Thus one π¦ β (π‘π€π π₯ β π£πππ’ππ ) and π₯ is not a function of π¦. As an equation
the relation π¦ = π₯2, has two π₯-solutions for each π¦ and thus π₯ is not a function of π¦.
However, in example 3, all horizontal lines hit the graph at most one time and thus π₯ is a function of π¦
for graph 3. The relation in 3. has equation π¦ = βπ₯ which can also be converted to π¦2 = π₯, with π¦ β₯
0. Thus each π¦ gives rise to a single π₯. Note that we have the same set of ordered pairs here, but the
function directions are in opposite directions thus π¦ = βπ₯ starts with π₯ to produce the square root of
π₯ and this is the square root function. However, π₯ = π¦2 takes as input a π¦-value and the output is the
square of this input, thus in this direction from π¦ β π₯ we have the squaring function.
A function that satisfies the horizontal line test is called a one-one-to function.
One-to-One Function: A function in which each output comes from only one input is called a
one-to-one function.
Horizontal Line Test:
A function from π₯ to π¦ is one-to-one if and only if every horizontal line intersects the graph at
no more than one point.
To check algebraically whether a function is one-to-one, we need to make sure that the outputs of a
function π are equal, i.e., π(π) = π(π) , only when π = π.
Functions and Relations Page 9
Examples
Determine if the functions below given by a graph or an equation are one-to-one.
1. Below π¦ is not a one-to-one function of π₯ as it does not pass the horizontal line test.
2. Below π¦ is not a one-to-one function of π₯ as it does not pass the horizontal (or vertical) line test.
3. Below π¦ is a one-to-one function of π₯ as it passes the horizontal line test.
As you can see from the graph above any horizontal line intersects the graph in only one point.
As you can see from the graph above the horizontal line will intersect the graph in two points.
As you can see from the graph above the horizontal line intersects the graph in two points.
4. π¦ = 3π₯ β 1 We could graph the function and use horizontal line test as above or check algebraically whether the function is one-to-one. If π₯ = π, and π₯ = π are two real values, then the outputs π(π) = 3π β1 = 3π β 1 = π(π) , solving the equations for π we get 3π = 3π, or π = π. That is the function output values are only equal if π = π implying the function is one-to-one.
5. π¦3 = π₯ Checking algebraically whether the function is one-to-one: For two real values π¦ = π, and π¦ = π If π(π) = π3 = π3 = π(π), solving the equations for π we get π = π Thus the function values are only equal if π = π implying the function is one-to-one.
6. π¦ + 3π₯2 = 4 or π¦ = 4 β 3π₯2 Checking algebraically whether the function is one-to-one: For two real values π₯ = π, and π₯ = π, π(π) = 4 β3π2 = 4 β 3 = π(π), we get π2 = π2 or that π = Β±π. That implies the function is not one-to-one. For example π₯ =2 = π and π₯ = β2 = π both give the same output π¦ = 4 β12 = β8.
Functions and Relations Page 10
Practice Problems
1. Determine the following
i. Is the relation a function or not.
ii. Is the relation a one-to-one function
iii. Domain and Range
A. Domain Range
E Bus F
G
H
o Function o One-to-One
o Not a Function o Not One-to-One
Domain: Range:
B. Domain Range
0 π β10 π 4 π 34 π
o Function o One-to-One
o Not a Function o Not One-to-One
Domain: Range:
C. {(π, π), (π, 2), (π, 2), (π, 4)}
o Function o One-to-One
o Not a Function o Not One-to-One
Domain: Range:
D. {(π, β1), (π, 2), (π, 2), (π, 4)}
o Function o One-to-One
o Not a Function o Not One-to-One
Domain: Range:
E. An Olympic size swimming pool has a capacity of 2,500,000 liters of water. The pool currently holds 100,000 liters of water, and water is being pumped at 400,000 liters/hour into the pool. Write a relation that represents the amount of water π (liters) in the pool after π‘ hours.
o Function o One-to-One
o Not a Function o Not One-to-One
Domain: Range:
Functions and Relations Page 11
F.
o Function o One-to-One
o Not a Function o Not One-to-One
Domain: Range:
G.
o Function o One-to-One
o Not a Function o Not One-to-One
Domain: Range:
H. π(π₯) = |π₯|
o Function o One-to-One
o Not a Function o Not One-to-One
Domain: Range:
I. π₯2 + π¦2 = 25
o Function o One-to-One
o Not a Function o Not One-to-One
Domain: Range:
J. ππππ(π‘) =5
9(π‘ β 32)
o Function o One-to-One
o Not a Function o Not One-to-One
Domain: Range:
K. π¦3 = π₯
o Function o One-to-One
o Not a Function o Not One-to-One
Domain: Range:
Functions and Relations Page 12
L.
o Function o One-to-One
o Not a Function o Not One-to-One
Domain: Range:
M.
o Function o One-to-One
o Not a Function o Not One-to-One
Domain: Range:
N.
o Function o One-to-One
o Not a Function o Not One-to-One
Domain: Range:
O.
o Function o One-to-One
o Not a Function o Not One-to-One
Domain: Range:
Functions and Relations Page 13
Practice Problems Solutions
1. Determine the following
i. Is the relation a function or not.
ii. Is the relation a one-to-one function
iii. Domain and Range
A. Domain Range
E Bus F
G
H
β Function o One-to-One
o Not a Function β Not One-to-One
Domain: {E,F,G,H} Range: {Bus}
B. Domain Range
0 π β10 π 4 π 34 π
o Function o One-to-One
β Not a Function o Not One-to-One
Domain:{0,β10,4,34} Range: {π, π, π, π}
C. {(π, π), (π, 2), (π, 2), (π, 4)}
o Function o One-to-One
β Not a Function o Not One-to-One
Domain: {π, π, π} Range: {π, 2,4}
D. {(π, β1), (π, 2), (π, 2), (π, 4)}
β Function o One-to-One
o Not a Function β Not One-to-One
Domain: {π, π, π, π} Range: {β1,2,4}
E. An Olympic size swimming pool holds 2,500,000 liters of water. The pool currently holds 100,000 liters of water, and water is being pumped at 400,000 liters/hour into the pool. Write a function that represents the amount of water π (liters) in the pool at π‘ hours. Find the domain and range of this function.
We will represent the amount of water time π‘ hours as π΄(π‘). Since there is already 100,000 liters of water in the pool and then we are adding more water per hour we have π = π΄(π‘) = 100,000 + 400,000π‘. To find domain and range we have to think about what π‘ values we can have. Note that the water in the pool cannot exceed the pool size of 2,500,000 liters. So we have 100000 β€ 100000 + 400000π‘ β€ 2500000 solving the inequality we will get that 0 β€400000π‘ β€ 2500000 β 100000 0 β€ 400000π‘ β€ 2400000 or
0 β€ π‘ β€2400000
400000 or 0 β€ π‘ β€ 6, also note that when time π‘ = 6 hours the pool will be at full
capacity of 2,500,000π and at time π‘ = 0 the pool would have 100,000 liters. Domain = [0,6] , Range [100000,2500000] The function is one-to-one as the graph is a line with slope π = 100,000 and thus passes the horizontal (and vertical) line tests.
Functions and Relations Page 14
F.
π¦ is a function of π₯ and vice versa.
β Function β One-to-One
o Not a Function o Not One-to-One
Domain: (ββ,β) Range: (ββ,β)
G.
Only π₯ is a function of π¦
o Function o One-to-One
β Not a Function o Not One-to-One
Domain: [0,β) Range: (ββ,β)
H. π(π₯) = |π₯| For every real number π₯ = π, we have π¦ =|π| which results in a unique real number.
β Function o One-to-One
o Not a Function β Not One-to-One
Domain: (ββ,β) Range: [0,β)
I. π₯2 + π¦2 = 25 For every real number π₯ = π, we have π¦ =
Β±β25 β π2 which are two real numbers for any π that satisfies β5 β€ π β€ 5.
o Function o One-to-One
β Not a Function o Not One-to-One
Domain: [β5,5] Range: [β5,5]
J. ππππ(π‘) =5
9(π‘ β 32)
For every real number π‘ we will get a unique
real number 5
9(π‘ β 32).
For any two values of = π , and π‘ = π we
have 5
9(π β 32) =
5
9(π β 32) or π β 32 =
π β 32, or π = π which means the function is one-to-one.
β Function β One-to-One
o Not a Function o Not One-to-One
Domain: (ββ,β) Range: (ββ,β)
K. π¦3 = π₯ π¦ is a function of π₯ and vice versa. For every real number π₯ = π we have one
and only one output π¦ = βπ3
. For any two values of = π , and π¦ = π we
βπ3
= βπ3
or that π = π which means the function is one-to-one.
β Function β One-to-One
o Not a Function o Not One-to-One
Domain:(ββ,β) Range: (ββ,β)
Functions and Relations Page 15
L.
Domain: [β6,β1] βͺ [2,7] Range: {π, π}
Each π₯ coordinate has a unique π¦ coordinate, so π¦ is a function of π₯. But many π₯ coordinates give the same output, e.g., π¦ = 2, so π¦ is not a one-to-one function of π₯.
β Function o One-to-One
o Not a Function β Not One-to-One
M.
Domain:[β6,2] Range: {π, π} Each π₯ coordinate between β3 and β1 has two π¦ coordinates associated to it on the graph making π¦ not a function of π₯.
o Function o One-to-One
β Not a Function o Not One-to-One
N.
Each π₯ coordinate has a unique π¦ coordinate, so π¦ is a function of π₯. But many π₯ coordinates give the single π¦-coordinate π¦ = 5, so π¦ is not a one-to-one function of π₯.
β Function o One-to-One
o Not a Function
β Not One-to-One
Domain:[βπ, π] Range: {π, π}
O.
Each π₯ coordinate has a unique π¦ coordinate, so π¦ is a function of π₯. But several π₯ coordinates have the same output, so π¦ is not a one-to-one function of π₯.
β Function o One-to-One
o Not a Function β Not One-to-One
Domain:[βπ, π] Range: [βπ, π]
From above examples you can see that sometimes even when the domain is an interval, the range can
be discrete points and other times it is an interval of π¦-values. So there is no fixed pattern, just pay
careful attention to the function and remember the π¦-coordinates are the range values and the π₯-
coordinates are the domain values.
Range
Domain
Range
Domain
Range
Domain
Domain
Range
Functions and Relations Page 16
Section1.1 Worksheet Date:______________ Name:__________________________
Concept Meaning in words and/or examples as required
1. Relation
2. Function
3. Domain of a function
4. Range of a function
5. All the different ways to represent a function (give examples you make up not the ones that appear in the book)
6. One-to-one function
7. Give an example of a relation that is not a function
8. Given an example of a function that is not one-to-one.
Difficulties encountered in the section:
Functions and Relations Page 17
Exercises 1.1
1. Determine the following. (For graphs, the direction of the relation is from π₯ to π¦.) i. Is the relation a function or not.
ii. If the relation is a function, is it one-to-one? iii. Domain and Range of the relation.
A. Domain Range
d Math F
I
j
o Function o One-to-One
o Not a Function o Not One-to-One
Domain: Range:
B. Domain Range
5 π ππ β20 ππ’ππππ 13 πππ€ 7 πππ
o Function o One-to-One
o Not a Function o Not One-to-One
Domain: Range:
C. {(β1, π), (2,8), (5, β3), (2, β3)}
o Function o One-to-One
o Not a Function o Not One-to-One
Domain: Range:
D. {(2,β1), (2,1), (π, 2), (π, 4)}
o Function o One-to-One
o Not a Function o Not One-to-One
Domain: Range:
E.
o Function o One-to-One
o Not a Function o Not One-to-One
Domain: Range:
F.
o Function o One-to-One
o Not a Function o Not One-to-One
Domain: Range:
Functions and Relations Page 18
G. π(π₯) = 3βπ₯ β 1
o Function o One-to-One
o Not a Function o Not One-to-One
Domain: Range:
H. π₯2 + π¦2 = 9 (Imagine the graph!)
o Function o One-to-One
o Not a Function o Not One-to-One
Domain: Range:
I. ππππ(π‘) =5
9(π‘ β 32)
o Function o One-to-One
o Not a Function o Not One-to-One
Domain: Range:
J. π¦5 = π₯
o Function o One-to-One
o Not a Function o Not One-to-One
Domain: Range:
K.
o Function o One-to-One
o Not a Function o Not One-to-One
Domain: Range:
L.
o Function o One-to-One
o Not a Function o Not One-to-One
Domain: Range:
Functions and Relations Page 19
M.
o Function o One-to-One
o Not a Function o Not One-to-One
Domain: Range:
N.
o Function o One-to-One
o Not a Function o Not One-to-One
Domain: Range:
2. Draw the graph of π¦ = π(π₯) where the domain is β3 β€ π₯ β€ 5 and the Range is β1 β€ π¦ β€ 3 and π(0) = 2 and the function is one-to-one.
3. Draw the graph of π¦ = π(π₯) where the domain is β3 β€ π₯ β€ β1 βͺ 1 β€ π₯ β€ 3 and the Range is π¦ β {β2, 3} and π(2) = β2. Is your function one-to-one?
6. For each if the following equations, determine whether:
In column 1, is π¦ a function of π₯? In column 2, is π₯ a function of π¦?
A. π¦2 = 2π₯ Function Not a function
B. 7π₯ β 5π¦ = 10 Function Not a function
C. 2π₯ β 7π¦ = 9 Function Not a function
D. π₯ = 3π¦2 Function Not a function
E. 6π₯ + |π¦| = 3 Function Not a function
F. π₯3 = π¦ Function Not a function
G. π₯2 + 2 = π¦ Function Not a function
H. π₯2π¦2 = 9 Function Not a function
I. π₯2 + 2π¦2 = 8 Function Not a function
J. π₯π¦ = 5 Function Not a function
Functions and Relations Page 20
7. The function β is defined by the following rule: β(π₯) = 4π₯ + 5. Complete the following table.
π₯ β(π₯) β4
β2
2
4
5
8. The functions π and π are defined as follows:
π(π₯) = β3π₯ + 2, π(π₯) = 3π₯3 + 5. Find π(3) and π(β3). Simplify your answers as much as possible. π(3) = π(β3) =
9. The function π is defined as follows: π(π₯) =4π₯
3π₯β15.
Find π(4). Simplify your answer as much as possible.
10. The function β is defined as follows:
β(π₯) =π₯2β3π₯β10
π₯2β14π₯+45.
Find β(6). Simplify your answer as much as possible.
11. Fill the table using the function rule
π(π₯) = βπ₯ + 6. Simplify your answers as much as possible.
π₯ π(π₯) β9
0
1/4
4
25
12. The function β is defined as follows:
β(π₯) = βπ₯ β 13
. Evaluate each output value and simplify your answers as much as possible. β(9) = β(126) =
β(β26) =
Functions and Relations Page 21
13. Simplify the expressions below for π(π₯) =2π₯2 + 3π₯ β 5. π(β6) =
π(π) = π(π + β) =
14. With π(π₯) and πΉ(π₯) being the mother and father of person π₯, try to evaluate: π(π¦ππ’ππ πππ) =
πΉ(π(π¦ππ’ππ πππ)) =
π(πΉ(π¦ππ’ππ πππ) =
π(πΉ(π(π¦ππ’ππ πππ))) =
15. Describe one or more functions and what the relations mean through using each of four methods: by a formula; by a table; by a graph; and through a sentence or two of words. State the domain and range of your function. (You could do this all for a single function or you can come up with different functions for each method.)
16. Create a function π¦ = π(π₯) that satisfies the following criteria. a. Is One-to-one b. Has domain all real numbers c. π(2) = β1
Questions about your function- i) What is the domain of π(π₯)? ii) What is the range of π(π₯)?
Functions and Relations Page 22
17. Create a function π¦ = π(π₯) that satisfies the following criteria. a. Has domain [β2, 1] b. For at least two π₯βs in the domain of π, π(π₯) = 4
Questions about your function- i) What is the domain of π(π₯)? ii) What is the range of π(π₯)? iii) Is your function one-to-one?
18. Create a function π¦ = π(π₯) that satisfies the following criteria. a. Is One-to-one b. Has domain [β2, 1] βͺ [3,10] c. π(β2) = 1 d. π(3) = 0 e. π(0) = 5 f. For at least one π₯ in the domain of π, π(π₯) = β4
Questions about your function- i) What is the domain of π(π₯)?
ii) What is the range of π(π₯)?
Functions and Relations Page 23
1.2 Inverse Functions and a Brief Library of Function Types
Functions and Relations Part 2 (10:21 min) https://www.youtube.com/watch?v=RZwGOfAeqp4
Functions and Relations Part 3 (10:41 min)https://www.youtube.com/watch?v=gPBE_QqVwbk
In section one we developed the basic concept of a function and the domain and range sets as well as
what it means for a function to be one-to-one. In this section weβll start by switching the direction of
functional relations to obtain what is called the inverse relation. Sometimes these inverse relations will
also be functions and it turns out that will be the case exactly when a function is one-to one. Weβll also
get familiar with a collection of basic types of functions that come up frequently in modeling in the
natural and social sciences. Even some very simple functions can lead to very beautiful images. As an
example, the simple function π(π₯) = π₯2 + π can be used to create Fractal Images called βJulia Setsβ Julia
Set video. The kinds of functions you can create is limited only by your imagination.
Recall in the previous section our definition of one-to-one functions.
One-to-One Function: A function in which each output π¦ comes from only one input π₯ is called
a one-to-one function.
We define the inverse of a function to be the same pairing as the original function, but with the direction reversed.
Inverse Function: If π¦ = π(π₯) is a one-to-one function, then the function denoted by πβ1 is
called the inverse of π function and for every π¦ in the range of π, πβ1 takes this π¦ as an input
and gives the one π₯-value in the domain of π as its output. Thus when π¦ = π(π₯), we also have
π₯ = πβ1(π¦). In other words the inverse function simply maps π¦ back to π₯. Also π being one-to-
one is exactly what is required for πβ1 to be a function, i.e., only one π₯-value to go back to.
Note: The notation πβ1 is the name of the inverse function and the β1 is not an exponent.
πβ1 reverses the direction of the π relation. The roles of the domain and range for πβ1 are reversed from that for π. Thus the range of π is the domain of πβ1 and the domain of π is the range of πβ1.
Functions and Relations Page 24
The examples below illustrate these properties of π and πβ1 and the one-to-one necessity of π for πβ1 to be a function.
1.
π₯πβ π¦
Domain Range
β1 1
2
0 1 1 2 2 4 3 8
π₯πβ1
β π¦
Range Domain
β1 1
2
0 1 1 2 2 4 3 8
π₯πβ π¦
Domain Range
β1 1
0 1 0
2 2
β2
π₯πβ1
β π¦
Range Domain
β1 1
0 1 0
2 2
-2
As you can see the function π is a one-to-one function. You can also see that if we reverse the direction it remains a function. This is the concept of the inverse function. You can see in the inverse function the roles that π₯ and π¦ play are reversed. Domain of π = {β1,0,1,2,3} = Range of πβ1
Range of π = {1
2, 1,2,4,8} = Domain of πβ1
As you can see the relation πβ1 is not a function. For example π¦ = 1 gets mapped back to two π₯-values. We can still look at the inverse relation but πβ1 is not a function. We will concentrate on one-to-one functions when working with the concept of inverse functions.
2. If π: {(1, π), (β2, π‘), (4, π), (8, π)} (which is a one-to-one function), then the function where the
domain values become the range values and vice versa for each of the pairs is the inverse function. So πβ1: {(π, 1), (π‘, β2), (π, 4), (π, 8)} this new relation is also a function. Domain of π = {1,β2,4,8} = Range of πβ1
Range of π = {π, π‘, π, π} = Domain of πβ1 3. If π¦ = π(π₯) is the graph shown below in red, then the purple graph would represent its inverse
function. Since all points (π₯, π¦) on the original graph would become (π¦, π₯) in the inverse function graph, you can see that all points where π¦ = π₯ will remain the same. So essentially it means if we graph the function π¦ = π(π₯) its inverse function is the same graph reflected across the line π¦ = π₯.
Functions and Relations Page 25
Steps to graph inverse functions: To sketch the graph of π¦ = πβ1(π₯), switch the π₯,π¦-roles of all the points on the graph of π¦ = π(π₯). If(π₯ = π, π¦ = π(π) = π), or (π, π), is on the graph of π, then (π₯ = π, π¦ = π) is on the graph of π¦ = πβ1(π₯). You may notice that each of the point pairs (π, π) and (π, π) lie directly across the diagonal line π¦ = π₯ from each other. In fact the complete graphs of π¦ = π(π₯) and π¦ = πβ1(π₯) are reflections of each over the line π¦ = π₯. Finding formulas for the inverse function when given a formula for the function:
Consider the function π¦ = π(π₯) = π₯Β³. The relationship from π₯ to π¦ is given by π¦ = π₯Β³. When
we want the inverse function, we want to start with a π¦-value from the output of π and
somehow determine what input π₯ the function π uses to get to that π¦. To do this we take the
equation for π, i.e., π¦ = π₯Β³ and solve it for π₯ to see how to get π₯ when π¦ is known.
Details: Solve the equation π¦ = π(π₯) = π₯Β³ for π₯ to get π₯ = βπ¦3 .
Thus π₯ = πβ»ΒΉ(π¦) = βπ¦3 is the formula that tells what πβ1 does to the input π¦ to get π₯.
We usually (but not always) use π₯ to indicate the input and π¦ for the output of a function. Doing
this for the function πβ»ΒΉ, means we switch the π₯ and π¦ roles to:
π¦ = πβ»ΒΉ(π₯) = βπ₯3
.
Typically we switch the π₯, π¦ roles first as π₯ = π(π¦) and then solve for π¦ to obtain π¦ = πβ»ΒΉ(π₯).
This way, we'd start with π₯ = π(π¦) ππ π₯ = π¦Β³ β βπ₯3
= π¦ β π¦ = πβ»ΒΉ(π₯) = βπ₯3
.
Steps to finding a formula for π = πβπ(π) when given a formula for π = π(π) : Step 1: Write the original one-to-one function as π¦ = π(π₯) Step 2: Switch the roles of π₯ and π¦ so we have π₯ = π(π¦) Step 3: Solving for π¦ will give us the inverse function π¦ = πβ1(π₯).
Examples: Find the inverse function to each of the functions below.
1. π(π₯) = 2π₯ β 3 Step 1: π¦ = 2π₯ β 3 Step 2: π₯ = 2π¦ β 3 Step 3: π₯ + 3 = 2π¦
π₯+3
2= π¦
Solution: πβ1(π₯) =π₯+3
2
Since both the function and its inverses are lines Domain of π = (ββ,β) = Range of πβ1 Range of π = (ββ,β) = Domain of πβ1
2. π(π₯) = π₯2, πππ π₯ β₯ 0 Step 1: π¦ = π₯2 πππ π₯ β₯ 0. Keeping the domain at π₯ β₯ 0 is makes the function one-to-one. Step 2: π₯ = π¦2 πππ π¦ β₯ 0 Here π₯ β₯ 0 too, so taking square roots will give a real π¦-value.
Step 3: βπ₯ = π¦ πππ π₯ β₯ 0
Solution: πβ1(π₯) = βπ₯, π₯ β₯ 0 Domain of π = [0,β) = Range of πβ1 Range of π = [0,β) = Domain of πβ1
Functions and Relations Page 26
3. Sketch the graph of the inverse function for each function shown below. Also state the domain
and range of the original function and of its inverse.
a. π¦ = π(π₯).
Domain of π = {β2,2,4,5} Range of π = {β1,2,3,0}
Domain of πβ1 = {β1,2,3,0}
Range of πβ1 = {β2,2,4,5}
b. π¦ = π(π₯)
Domain of π = [β5,0] Range of π = [β2,2]
Domain πβ1 = [β2,2] Range of πβ1 = [β5,0]
π¦ = π(π₯) π¦ = πβ1(π₯)
(β5,β2)
(β2.5,0)
(0,2)
(β2,β5)
(0, β2.5)
(2, 0)
π¦ = πβ1(π₯)
π¦ = π(π₯)
Functions and Relations Page 27
Other Types of Functions
1. Square Root Function: A function defined as π(π₯) = βπ₯, for all real numbers π₯ β₯ 0.
β’ Graph
β’ One-to-One: Yes β’ Inverse function πβ1(π₯) = π₯2, πππ π₯ β₯
0
Step 1: π¦ = βπ₯, πππ π₯ β₯ 0
Step 2: π₯ = βπ¦, πππ π¦ β₯ 0
Step 3: π₯2 = π¦ πππ π¦ β₯ 0, and therefore π₯ β₯0 β’ Domain of π = [0,β) = Range of πβ1 β’ Range of π = [0,β) = Domain of πβ1
Examples: For π(π₯) = βπ₯ find the values of
a. π (2
3)
b. π(100) c. π(3456)
Solution:
a. π (2
3) = β
2
3β0.816
b. π(100) = β100 = 10
c. π(3456) = β3456 β 58.79
2. Polynomial function: A function defined as π(π₯) = π0 + π1π₯ + π2π₯2 + π3π₯
3 +β―+ πππ₯π,
where π0, π1, π2, π3,β¦ππ are all real numbers and nβ₯ 0 is a whole number. The domain of these functions is all real numbers. The range is all real numbers when π is odd, but is more difficult to determine when π is even.
Example
i) Constant Function: A function defined as π(π₯) = π, where π is any real number. A special case of the polynomial function of degree zero.
β’ Generic Graph
β’ Domain: All real numbers β’ Range: {π} β’ One-to-One: NO
π(π₯) = βπ₯
πβ1(π₯) = π₯2, π₯ β₯ 0
(4,2)
(2,4)
π¦ = π
Functions and Relations Page 28
Example If π(π₯) = 5, then find
a. π(3) b. π(β2) c. π(β3456) d. π(π + β) e. Sketch the graph
Solution: a. π(3) = 5 b. π(β2) = 5 c. π(β3456) = 5 d. π(π + β) = 5 e. As you can see from the parts a,b,and c, no matter what π₯-coordinate you plot the π¦-coordinate is always 5 so it is a horizontal line as shown above.
ii) Linear Function: A function defined as (π₯) = ππ₯ + π . This is a polynomial function with degree one. Remember π =slope of the line, π = π¦-intercept.
Example 1. If π(π₯) = 2π₯ β 3, then
a. Find π(β1) b. Find π(0) c. Find π(4) d. Find the inverse function and its graph. e. Sketch the graph of π¦ = π(π₯)
Solution: a. π(β1) = 2(β1) β 3 = β2 β 3 = β5 b. π(0) = β3 c. π(4) = 2(4) β 3 = 8 β 3 = 5 d. Inverse function
π¦ = 2π₯ β 3 π₯ = 2π¦ β 3 π₯ + 3 = 2π¦ π₯ + 3
2= π¦
Inverse function is πβ1(π₯) =π₯+3
2, slope is
1
2 and π¦-intercept is at
3
2.
e. Graph of the function π¦ = π(π₯) is to the right with slope of 2 and π¦-intercept of β3.
2. An Olympic size swimming pool holds 2,500,000 liters of water. If the pool currently holds 100,000 liters of water, and water is being pumped at 400,000 liters/hour into the pool.
a. Write a function that represents the number of liters π of water in the pool at π‘ hours. b. Find the domain and range of this function. c. Sketch the graph of this function. d. Find the inverse of this function, but donβt switch the variable names.
Functions and Relations Page 29
Solution: a. π = π΄(π‘) = 100000 + 400000π‘ Liters,
for 0 β€ π‘ β€ 6 The domain starts at π‘ = 0 and after 6 hours the pool will be full since it holds a maximum of 2,500,000 liters of water.
b. Domain of π΄ = [0,6] and Range of π΄ =[100000,2500000]
c. Graph is to the right. The scale is each tick mark represents 1 million liters on the π¦-axis and 1 hour on the π‘-axis.
d. We solve the original equation for π‘ to get
π‘ =πβ100,000
400,000= π΄β1(π) ,
Domain is 100,000 β€ π β€ 2,500,000 Range is 0 β€ π‘ β€ 6
iii) Square Function: π(π₯) = π₯2 This is a special case of a general second degree polynomial π(π₯) = ππ₯2 + ππ₯ + π .
With π(π₯) = π₯2:
a. Find π (2
3)
b. Findπ(β2) c. Find π(2) d. Find π(π + β) e. Sketch the graph the function π¦ = π₯2 f. Is the function one-to-one?
Solution:
a. π (2
3) = (
2
3)2=
4
9
b. π(β2) = (β2)2 = 4 c. π(2) = (2)2 = 4 d. π(π + β) = (π + β)2 = π2 + 2πβ + β2 e. Plot a few points to sketch the graph of the
function. See to the left f. No. The function is not one-to-one as it does
not pass the horizontal line test.
π₯ π¦ = π₯2 β2 (β2)2 = 4 β1 (β1)2 = 1 0 (0)2 = 0 1 (1)2 = 1 2 (2)2 = 4
iv) Cube Function: π(π₯) = π₯3 This is a special case of a general third degree polynomial function π(π₯) = ππ₯3 + ππ₯2 + ππ₯ + π
With π(π₯) = π₯3:
a. Find π (2
3)
b. Findπ(β2) c. Find π(2) d. Sketch the graph the function π¦ = π₯3 e. Is the function one-to-one? f. Find the inverse function if it is one-to-one.
π₯ π¦ = π₯3 β2 (β2)3 = β8 β1 (β1)3 = β1 0 (0)3 = 0 1 (1)3 = 1 2 (2)3 = 8
Mill
ion
s o
f Li
ters
Number of Hours
Functions and Relations Page 30
Solution:
a. π (2
3) = (
2
3)3=
8
27
b. π(β2) = (β2)3 = β8 c. π(2) = (2)3 = 8 d. Plot a few points to sketch the graph of the
function. See to the left e. Yes. The function is one-to-one as it does pass
the horizontal line test. f. To find the inverse function to π¦ = π(π₯) = π₯3,
switch π₯, π¦ roles so: = π¦3 . Then solving for π¦
we get that inverse function is πβ1(π₯) = βπ₯3
4. Piecewise Defined Function: A piecewise defined function is exactly what its name suggests. It is defined in pieces by two or more equations for different parts of the domain.
Examples:
I. For π(π₯) = {1, for 0 β€ π₯ < 10, for β 1 β€ π₯ < 0β1, for β 2 β€ π₯ < β1
a) Graph the function π b) State the domain of π c) State the range of π d) Evaluate π(β0.5) e) Evaluate π(0.5) f) Evaluate π(β1.5)
Solution: a) Our function can be thought of as being made up
of three separate functions. So to plot the graph we have plot each of the three separate functions in one graph. You can do so by plotting some points until you know how it looks like. In this case all three pieces are constant functions in the given interval so the graph would be as shown below.
π(π₯) = {
1, πππ 0 β€ π₯ < 10, πππ β 1 β€ π₯ < 0β1, πππ β 2 β€ π₯ < β1
b) Domain of π = [β2, 1) c) Range of π = {β1,0,1} d) π(β0.5) = 0 since π₯ = β0.5 falls in
the interval β1 β€ π₯ < 0 e) π(0.5) = 1 since π₯ = 0.5 falls in the
interval 0 β€ π₯ < 1 f) π(β1.5) = β1 since π₯ = β1.5 falls in
the interval β2 β€ π₯ < β1
Range
Functions and Relations Page 31
II. For π(π₯) = {
β1
3π₯ β 1, πππ β 3 β€ π₯ < 0
β2, πππ 0 β€ π₯ < 22π₯ β 4, πππ 3 β€ π₯
a) State the domain of π b) State the range of π c) Graph the function π d) Evaluate π(β3)
e) Evaluate π (β1
2)
f) Evaluate π(1.5) g) Evaluate π(5)
Solution: c) Our function can be thought of as being made
up of three separate functions. So to plot the graph we have plot each of the three separate functions in one graph. The first and the last piece are lines so plotting the two end points of the line segments it is made up of will suffice, and the middle is a constant function like we have seen before.
π₯ π¦ = β
1
3π₯ β 1
π₯ π¦ = 2π₯ β 4
β3 β1
3(β3) β 1
= 1 β 1 = 0
3 2(3) β 4= 6 β 4 = 2
0 β1
3(0) β 1
= 0 β 1 = β1
5 2(5) β 4= 10 β 4 = 6
Solution:
π(π₯) = {β1
3π₯ β 1, πππ β 3 β€ π₯ < 0
β2, πππ 0 β€ π₯ < 22π₯ β 4, πππ 3 β€ π₯
a) Domain of π is [β3,0) βͺ [0,2) βͺ [3,β) We see that this is [β3,2) βͺ [3,β) which also be seen by observing the totality of the π₯-cordinates on the graph.
b) The range of π is {β2} βͺ (β1,0] βͺ[2,β). Look at the graph to see what π¦-cordinates are involved in the graph.
d) π(β3) = β1
3(β3) β 1 = 1 β 1 = 0
since π₯ = β3 falls in the interval β3 β€π₯ < 0
e) π (β1
2) β
1
3(β
1
2) β 1 =
1
6β 1 = β
5
6
since π₯ = β1
2 falls in the interval 3 β€
π₯ < 0 f) π(1.5) = β2 since π₯ = β1.5 falls in
the interval 0 β€ π₯ < 2 g) π(5) = 2(5) β 4 = 10 β 4 = 6
Before we get into the next few types of functions we need to get familiar with a few definitions.
Range
Functions and Relations Page 32
5. Rational Function: A rational function is defined as the ratio of two polynomial functions denoted as
π (π₯) =π(π₯)
π(π₯), for polynomial functions π(π₯), π(π₯). The domain of π (π₯) = is all π₯ such that π(π₯) β 0
Examples
I. π (π₯) =1
π₯
a. Find π (2) b. Domain of π (π₯) c. Sketch the graph of π (π₯) d. Is π (π₯) one-to-one? e. What is the inverse
function of π (π₯) Solution:
a. π (2) =1
2
b. Domain of π (π₯) = all real numbers π₯ β 0
= (ββ, 0) βͺ(0,β)
c. We canβt have π₯ = 0, but we see the π¦β²π get very large as π₯ β 0. As you can see from the graph, as π₯ approaches zero from the positive side the π¦ coordinate shoots to infinity, and as π₯ approaches zero from the negative side the π¦ coordinate shoots to negative infinity. Also when π₯ β +β, the π¦ coordinate approaches zero from the positive side and as π₯ β ββ, the π¦ coordinate approaches zero from the negative side. The lines π₯ = 0 and π¦ = 0 are called horizontal and vertical asymptotes respectively. We will study these in-depth later.
d. Yes. π (π₯) passes the horizontal line test.
e. Inverse function of π¦ =1
π₯ π₯ =
1
π¦ solving for π¦ we get
π¦ =1
π₯ or π β1(π₯) =
1
π₯
For now, we will plot a bunch of points to get a sense of the shape of the graph. We will deal with rational functions in more detail later. We must avoid π₯ = 0 as the denominator cannot be zero.
π₯ π¦ =
1
π₯
π₯ π¦ =
1
π₯
β0.1 1
β0.1= β10
β10 1
β10= β0.1
β0.01 1
β0.01= β100
β100 1
β100= β0.01
β0.001 1
β0.001= β1000
β1000 1
β1000= β0. .001
0.1 1
0.1= 10
10 1
10= 0.1
0.01 1
0.01= 100
100 1
100= 0.01
0.001 1
0.001= 1000
1000 1
1000= 0. .001
From the chart the closer to zero the π₯ values come from left or the right of zero, the π¦ values either shoot to β, or ββ respectively. In other words, if we were to zoom in closer and closer to π₯ = 0, the graph resembles the vertical line π₯ = 0. Such a line is then denoted by the dotted line you see and is called a horizontal asymptote. Similarly if we zoom out so that π₯ values shoot to either β or ββ then the π¦ coordinate seems to get closer to zero from either above or below respectively. In other words, we were to zoom out a lot, the graph would resemble the line π¦ = 0. Such a line is called the horizontal asymptote.
Functions and Relations Page 33
II. π (π₯) =2π₯β1
π₯2β4π₯+3=
2π₯β1
(π₯β3)(π₯β1)
a. Find π (5) b. Domain of π (π₯)
Solution:
a. π (5) =2(5)β1
(5β3)(5β1)=
10β1
(2)(4)=
9
8
b. Domain of π (π₯) = All real numbers for which denominator is non-zero, i.e., (π₯ β 3)(π₯ β 1) β 0 = All real numbers so that π₯ β 3 or π₯ β 1 = (ββ, 1) βͺ (1,3) βͺ (3,β)
6. Exponential Function: An exponential function is defined as π(π₯) = ππ₯, where π is a positive real number not equal to 1.
Properties of an exponential function β’ Domain: All real
numbers β’ Range: (0,β) β’ One-to-One: Yes! β’ Points on the
graph to plot are
π π = ππ
β2 πβ2 =
1
π2
β1 πβ1 =
1
π
0 π0 = 1
1 π1 = π
2 π2
Graph of π(π₯) = ππ₯, for π > 1
Graph of π(π₯) = ππ₯, for π < 1
For exponential functions we can see that for all π > 0 the negative powers of
π gives us values of the type 1
ππ where π > 0. The value of
1
ππ will keep on
getting smaller and smaller taking the π¦-coordinate to zero either as π₯ goes to β, or ββ, depending on whether π > 1 or 0 < π < 1 respectively. Therefore the line π¦ = 0 is called the horizontal asymptote. In other words, the graph resembles the line π¦ = 0 when we zoom out sufficiently.
Horizontal Asymptote π¦ = 0 Horizontal Asymptote π¦ = 0
Functions and Relations Page 34
Example I. If π(π₯) = 2π₯, then
a. Find π(β1) b. Find π(3) c. Find π(β3) d. Find π(π + β) e. Sketch the graph of the
function. Solution:
a. π(β1) = (2)β1 =1
2
b. π(3) = 23 = 8
c. π(β3) = 2β3 =1
8
d. π(π + β) = 2π+β e. Sketch the graph of the function.
See graph to the right. You can see from the chart that as π₯ approaches ββ the π¦-coordinate approaches zero. This makes π¦ = 0 the horizontal asymptote.
Graph
π π = ππ
β2 2β2 =
1
22=1
4
β1 2β1 =
1
21=1
2
0 20 = 1
1 21 = 2
2 22 = 4
Example
II. If π(π₯) = (1
3)π₯
, then
a. Find π(β1) b. Find π(2) c. Find π(β2) d. Sketch the graph of the
function. Solution:
a. π(β1) = (1
3)β1
= 3
b. π(2) = (1
3)2=
1
9
c. π(β2) = (1
3)β2
= 32 = 9
d. Sketch the graph of the function. See graph to the right. You can see from the chart that as π₯ approaches β the π¦-coordinate approaches zero. This makes π¦ = 0 the horizontal asymptote.
Graph
π π = (
1
3)π₯
β2 (1
3)β2
= 32 = 9
β1 (1
3)β1
= 3
0 (1
3)0
= 1
1 (1
3)1
=1
3
2 (1
3)2
=1
9
Horizontal Asymptote π¦ = 0
Horizontal Asymptote π¦ = 0
Functions and Relations Page 35
Playing
As you can see all the exponential functions discussed here are one-to-one and therefore have inverses.
If we follow our steps to find an inverse function of say π(π₯) = 2π₯ we get the following
π¦ = 2π₯ and then π₯ = 2π¦. At this step we try to solve the equation for π¦ but with π¦ in the exponent, we
are unable to isolate π¦ using basic tools for solving equations. To get around this problem,
mathematicians created a new function to represent this inverse function and called it a logarithmic
function. The inverse function to π(π₯) = 2π₯ must solve π₯ = 2π¦for π¦. Thus we seek the exponent π¦ that
the base 2 must be raised to so that this will equal π₯. We call this inverse function the base-2 logarithm
and denote it as πβ1(π₯) = πππ2(π₯) read as log base 2 of π₯. What does it mean to evaluate values of this
inverse function?
First note that π₯ = 2π¦ and π¦ = πππ2(π₯) are equivalent equations. So if we wanted to find the value of
π¦ = πππ2(8) then we need to find the exponent π¦ for which 8 = 2π¦. This means π¦ = 3 or that
log2(8) = 3. Similarly you can play and see that πππ2 (1
2) = β1 since
1
2= 2β1.
Recall that π¦ = 0 is the horizontal asymptote to the function π¦ = 2π₯. Also since π¦ = log2 π₯ is the
inverse of π¦ = 2π₯ , and the π₯, π¦ roles get reversed, then the line π₯ = 0 will be the vertical asymptote to
the function π¦ = πππ2(π₯). Also, with inverse functionsβ reversal of the domain and range rolls, The
Range of π¦ = 2π₯ being all positive numbers is the domain of π¦ = log2 π₯. And the range of π¦ = log2 π₯ is
the domain of π¦ = 2π₯ which is all real numbers.
7. Logarithmic Function: A function denoted as π¦ = π(π₯) = πππππ₯, where π > 0, π β 1. This means π¦ is the exponent such that ππ¦ = π₯.
Properties of the logarithmic function: β’ To compute the value of
π(π₯) = logπ π₯ means answering: π? = π₯
β’ Graph: β’ Domain:(0,β) β’ Range: All real numbers β’ One-to-One: Yes! β’ Inverse function πβ1(π₯) = ππ₯
Graph of π(π₯) = πππππ₯, for π > 1
Graph of π(π₯) = πππππ₯, for 0 <π < 1
Vertical Asymptote π₯ = 0 Vertical Asymptote π₯ = 0
Functions and Relations Page 36
Section1.2 Worksheet Date:______________ Name:__________________________
Concept Meaning in words and/or examples as required
1. One-to-One function
2. Inverse function
3. Relationship of domain and range of inverse functions to the original function
4. Polynomial functions
5. Rational functions
6. Exponential functions
7. Logarithmic function
8. Describe the difference between the exponential and logarithmic functions and the relationship between them.
Difficulties encountered in the section:
Functions and Relations Page 37
Exercises 1.2
1. The function π is defined by the following rule: π(π₯) = 8π₯. Find π(π₯) for each π₯-value in the table.
π₯ π(π₯) = 8π₯ β2
β1
0
1
2
2. The tables below give exponential functions in the form π¦ = ππ₯. Write the equation for each of the functions.
π₯ π(π₯) π₯ π(π₯)
β2 1
4
β2 1
π2
β1 1
2
β1 1
π
0 1
0 1
1 2
1 π
2 4
2 π2
π(π₯) =________ π(π₯) =_________
3. Evaluate each logarithmic function below. a. log2 16 =
b. log5 25 =
c. log21
8=
d. log3 β3 =
e. log6 1 =
f. log9 3 =
g. log10 0.00001 =
4. Convert each logarithmic expression to its exponential format and then evaluate the unknown value π₯.
a. log2 32 = π₯
b. log16 π₯ = 1.5
c. log5 π₯ = β3
d. log6 π₯ = 3
e. logx 16 = 2
f. logx 25 = 1/2
Functions and Relations Page 38
5. The function π is defined as follows:
π(π₯) = {
3
4π₯ + 2 ππ π₯ β β1
2 ππ π₯ = β1
Find the following. π(β3) = π(β1) = π(β2) =
6. The function π is defined by
π(π₯) =3π₯β4
π₯+5 .
Find the following: π(β3) = π(π₯ + 5) = The domain of π.
7. π(π₯) = βπ₯ β 2 a. Find π(6) b. Find π(2) c. Find domain and range of the function. d. Is π one-to-one? e. If you answered yes to part d, find the inverse function.
Functions and Relations Page 39
8. Find the domain and range of all the relations below either in interval notation or set notation as appropriate.
a) π(π₯) =1
π₯β4
Domain: Range:
b) π(π₯) = βπ₯ β 4 Domain: Range:
c) β(π₯) = β4 β π₯ Domain: Range:
d) π(π₯) = βπ₯ β 4 + 2 Domain: Range:
e)
Domain of π: Range of π:
f)
Domain of β: Range of β:
g)
Domain of π: Range of π:
h)
Domain of π: Range of π: Evaluate π(0) =___________ Find the values of π₯, for which
π(π₯) = 0 _____ , ______
i)
π(π₯) = {
2 ππ π₯ = 1β2 ππ π₯ < 13 ππ π₯ > 1
Domain of π: Range of π:
j)
π(π₯) = {3βπ₯ ππ π₯ > 4
2π₯ β 1 ππ π₯ < β4
Domain of π: Range of π:
Functions and Relations Page 40
9. A species of bacteria doubles every 30 minutes at room temperature. If you initially started with 30000 bacteria at π‘ = 0: a. Find the number of bacteria after 2 hours.
b. Find the number of bacteria at 4 hours.
c. See if you can come up with a formula for an exponential function that describes this where π‘
is the input in minutes and π¦ is the output being the number of bacteria present at time π‘.
10. The radioactive substance Uranium-240 has a half-life of 14 hours. The amount π΄(π‘) of a sample of Uranium-240 remaining in grams after π‘ hours is given by the exponential function
π΄(π‘) = 3000(1
2)
π‘
14.
a. Find the initial amount in the sample.
b. Find the amount remaining after 30 hours. Round your answer to the nearest gram. c. Determine how many grams remain after two years.
11. Suppose Rahul places $2000 in an account that pays 5% interest compounded each year. Assume that no withdrawals are made from the account. Round to the nearest cent. a. Find the amount in the account at the end of 1 year.
b. Find the amount in the account at the end of 2 years.
c. Try to find a formula for the account value after π‘ years. π΄(π‘) =
Functions and Relations Page 41
12. A car is purchased for $26,500. After each year, the resale value decreases by 15%. a. What will the resale value be after 5 years? Round your answer to the nearest dollar.
b. Try to find a formula for the value after π‘ years. π(π‘) =
13. Devon deposited $4000 into an account with 4.4% interest compounded quarterly. a. Assuming the no withdrawals are made, how much will she have in the account after 10
years? Round your answer to the nearest dollar.
b. Try to find a formula for the value after π‘ years.
π(π‘) =
14. At the beginning of a population study, a city had 360,000 people. Each year since, the population has grown by 2.6%. Let π‘ be the number of years since the start of the study. Let π(π‘) represent the cityβs population at time π‘. a. Predict the population 5 years after the start of the study.
b. Try to write an exponential function for the population π(π‘), π‘ years after the study starts.
15. Compare the function π(π₯) = 3π₯ and π(π₯) = 50π₯2 by completing parts a. and b. a. Fill in the table below. Note that the table is already filled in for π₯ = 6
π₯ π(π₯) = 3π₯ π(π₯) = 50π₯2
6 36 = 729 50(62) = 1800
7
8
9
10
b. For all π₯, π₯ β₯ 8 please check which of the following are true statements π(π₯) β₯ π(π₯) π(π₯) = π(π₯) π(π₯) β€ π(π₯)
c. Explain your answer in part b.
Functions and Relations Page 42
16. An organic farmer raises free-range chickens and will deliver eggs to your home for a fee. The farmer sells a one-year contract for delivery of 25 to 100 dozen eggs to your home. The annual cost in dollars for the service is given by π¦ = πΆ(π) = 25 + 4π, where π is the number of dozen eggs you will use over the year.
Domain: a. State the domain variable and what it
stands for.
b. What is the domain here, assuming eggs are available only in full dozen quantities?
c. Determine the value π¦ = πΆ(50) and what it means.
Range a. State the Range variable and what it
stands for.
b. What is the Range here, assuming eggs are available only in full dozen quantities?
c. Determine the value of π if your contract payment was $325.
17. A construction crew needs to pave a road that is 204 miles long. The crew paves 9 miles of the road each day. The length, πΏ (in miles), that remains to be paved after π days is given by the following function. πΏ(π) = 204 β 9π.
Answer the following questions. a. How many miles of the road the crew have left to pave after 13 days?
b. After how many days will there be only 114 miles left to pave?
18. Find the difference quotient π(π₯+β)βπ(π₯)
β
where β β 0 for the function π(π₯) =5π₯2 β6. The graph of this function is given as well. Simplify your answer as much as possible. π(π₯+β)βπ(π₯)
β =
Try to explain what the quotient represents.
Functions and Relations Page 43
19. Sketch the graph of the functions and relations below. Explain clearly how you decided the graph was the shape you drew. Can you determine the domain and range of the functions and relations that you graphed based on the graphs.
a. π¦ = 2π₯
π₯ π¦
b. π¦ = βπ₯
π₯ π¦
c. π¦ = π₯3
π₯ π¦
d. π¦ = π₯4 π₯ π¦
Functions and Relations Page 44
e. π¦ = 4π₯
π₯ π¦
f. π¦ = (1
4)π₯
π₯ π¦
g. π¦ =3
π₯
π₯ π¦
h. π¦ = log4(π₯) π₯ π¦
4
16
1/4
1
1/16
Functions and Relations Page 45
i. π¦ = log10 (π₯)
π₯ π¦
j. π¦ = πππ3(π₯)
π₯ π¦
3
9
1/3
1/9
1
Functions and Relations Page 46
k. π¦ = {
2, ππ π₯ > 1β2, ππ π₯ < 1 0, ππ π₯ = 1
π₯ π¦
l. π¦ = {π₯ + 2, ππ π₯ > 2
π₯2, ππ π₯ β€ 2
π₯ π¦
20. Evaluate the following for the given one-to-one functions below.
a. π(β5) b. πβ1(β5)
c. π(4)
d. πβ1(4)
a. π(2) b. πβ1(0)
c. π(β5)
d. πβ1(β2)
Functions and Relations Page 47
21. Find the inverses of the following one-to-one functions. Then find the domains and ranges of the functions and their inverses.
a. π(π₯) =7π₯+1
2π₯β1
b. π(π₯) = β2π₯ β 1 for π₯ β₯1
2
Domain of π
Range of πβ1
Domain of π
Range of πβ1
Domain of πβ1
Range of π
Domain of πβ1
Range of π
c. β(π₯) = 4π₯ d. π(π₯) = πππ4π₯
Domain of β
Range of ββ1
Domain of π
Range of πβ1
Domain of ββ1
Range of β
Domain of πβ1
Range of π
22. Given that πβ1(π₯) = 3π₯3 + 5 find the formula for π¦ = π(π₯).
Functions and Relations Page 48
23. Obtain the piece-wise formula for the functions whose graphs are given below. If the functions below are one-to-one, please also find the formulas for their inverse functions and sketch their graphs.
a.
b.
c.
d.
Functions and Relations Page 49
24. Find the original function whose inverse function is given below.
a. πβ1(π₯) =7π₯+1
2π₯β1
b. πβ1(π₯) = β1 β π₯, for π₯ β€ 1
c.
d.
25. Create functions with the given properties below and then answer the questions.
A. Create a function π¦ = π(π₯) that satisfies the following criteria. a. Is One-to-one b. Has domain [β2, 1] βͺ [3,10] c. π(β2) = 1 d. π(3) = 0 e. π(0) = 5 f. For at least one π₯ in the domain of π, π(π₯) = β4
Questions about your function- iii) What is the domain of π(π₯)? iv) What is the range of π(π₯)? v) What are the π₯-intercepts of π¦ = π(π₯)? vi) What is the π¦-intercepts of π¦ = π(π₯)? vii) What is πβ1(π₯)?
Functions and Relations Page 50
1.3 Exponential and Logarithmic Functions In section two, we introduced the basic concept of the exponential and logarithmic functions as inverses
of each other.
Review
a. Exponential Function: An exponential function is defined by π(π₯) = ππ₯, where π is a positive real number not equal to 1.
Properties of an exponential function β’ Domain: All real
numbers β’ Range: (0,β) β’ One-to-One: Yes! β’ Points on the
graph to plot are
π π = ππ
β2 πβ2 =
1
π2
β1 πβ1 =
1
π
0 π0 = 1
1 π1 = π
2 π2
Graph of π(π₯) = ππ₯, for π > 1
Graph of π(π₯) = ππ₯, for π < 1
b. Logarithmic Function: A logarithmic function is denoted by π¦ = π(π₯) = πππππ₯, where π > 0, π β 1. This means π¦ is the exponent such that ππ¦ = π₯.
Properties of the logarithmic function: β’ To compute the value of π(π₯)
means answering: π? = π₯ β’ Graph: β’ Domain:(0,β) β’ Range: All real numbers β’ One-to-One: Yes! β’ Inverse function πβ1(π₯) = ππ₯
Graph of π(π₯) = πππππ₯, for π > 1
Graph of π(π₯) = πππππ₯, for 0 <π < 1
We will now explore these functions more fully and learn their properties so we can use them in many
different applications.
Horizontal Asymptote π¦ = 0 Horizontal Asymptote π¦ = 0
Vertical Asymptote π₯ = 0 Vertical Asymptote π₯ = 0
Functions and Relations Page 51
Remember the fact that the logarithmic function is the inverse function of the exponential function. This
fact means the equations below are equivalent.
Exponential Logarithmic Equation Equation
π = ππ ππππ π = π
There are two logarithm bases that are commonly used and show up as functions on scientific
calculators. One of these is logarithm base 10 which we write as πππ10π₯ = log π₯ where the base
subscript is left out. Check out your calculators, you should see a πππ button on it. This represents the
logarithm base ten also called the βCommon Logartihmβ. Check that it works by using the calculator to
take log 1000 and see that the answer is 3.
So: π¦ = log π₯ π₯ = 10π¦ , (A logarithm is always the value of an exponent!)
This notation for common logarithms originated ~1675 about 60 years after they were invented to provide an efficient method for multiplying and dividing numbers and taking powers and roots of numbers. Another motivation for working with logarithms is that it allows us to more easily express quantities that vary over many orders of magnitude. Below are examples where in real life logarithms base ten are used. 1. In chemistry pH of a substance is measured by the formula ππ» = βπππ[π»+] where [π»+] =
hydrogen ions concentration in units of moles/liter. A substance having pH of less than 7 is characterized as being acidic and above 7 to be basic or alkaline. Water has a pH of 7. That means 7 = βπππ[π»+] or that 1 Γ 10β7 moles/liter of hydrogen ions are present in water. The pH of a orange juice is between 3 and 4, while sweet corn has a pH of about 8. Nutrition experts talk about the link of alkaline diets [Foods for which the ash that remains after combustion produces a pH greater than 7 when dissolved in water.] to a reduced risk of cancer. Fruits and vegetables are typically alkaline foods.
2. Seismologistβs measure earthquakes using the Richter scale. The formula for this is given by π =
πππ (πΌ
πΌ0) where πΌ = Intensity or amplitude of a seismograph reading in mm as measured by a
seismograph usually located within 100 kilometers from the epicenter, and πΌ0 = a baseline reading of 0.001 mm. To get a sense of this scale, a Richter scale reading of a 3 is similar to the vibrations felt when a large truck is driving by. A Richter reading of 5, however is 100 times more intense. Also, the energy released by a magnitude 5 earthquake is equivalent to a 500 ton TNT explosion whereas a magnitude 9 earthquake that hit Japan in 2011 was equivalent to a 500,000,000 ton of TNT explosion. This gives you a sense of how logarithms are useful for relating quantities that range over many powers of ten in size. That 2011 earthquake that hit Japan was so powerful it nudged the earthβs axis so that the days on earth shortened by about 1.8 microseconds.
3. Physicists use a similar formula to calculate the decibel level of sound as ππ΅ = 10 πππ (πΌ
πΌ0) where
πΌ = Intensity of the sound measured in πππ‘π‘π /πππ‘ππ2, and πΌ0 = lowest threshold for sound that can be heard and is about 10β12 πππ‘π‘π /πππ‘ππ2 (like a whisper). A 60 decibel sound is like when you are in a restaurant and you hear the noises around you. A jet taking off can expose airport
Functions and Relations Page 52
workers to sounds of 130-150 decibels. (This kind of high intensity noise exposure can lead to hearing loss.)
The other common base for logarithms is the constant "π" where π β 2.718β¦. This is sometimes to
referred to as Eulerβs number (or Napierβs constant) and therefore the letter π is used to denote it. This
special number was first discovered in the mid 1600sβ and was first noticed by Jacob Bernoulli in 1683
when working with compound interest. Letβs play with compound interest formulas to see how Bernoulli
came to this important mathematical constant π.
Compound interest and the number π
Suppose that you invested a 1000$ at an interest rate of 5%/yr. Compound interest means that the
each time interest is paid it gets added to the principal. There are many options for how often the
interest earnings are computed and added to the account. If we compute interest π times a year, then
annual compounding is when π = 1, Quarterly compounding for π = 4 , Monthly for π = 12, and
Weekly compounding for π = 52, and so on β¦
With annual compounding, the principal remains at $1000 until the end of the year when the 5%
interest is added to the principal.
For Annual Compounding:
After 1 year we will have our principal + interest
= 1000 + 1000(0.05) = 1000(1 + 0.05) = 1000(1.05) = 1050$. This will then be the
principal for the interest that is earned during the second year.
After 2 years we will have our principal of 1050 plus 5% interest = 1050 + 1050(0.05)
= 1050(1 + 0.05) = 1050(1.05) = 1000(1.05)(1.05) = 1000(1.05)2 = 1102.50$
We see each 5% interest addition corresponds to multiplying by (1.05) and the value after π‘ years is
given by π(π‘) = 1000(1.05)π‘ dollars.
Annual Compounding Formula π½(π) = π·(π + π)π
For Quarterly Compounding:
When interest is paid quarterly, only ΒΌ of the 5%/yr is added after each ΒΌ year.
After the first quarter, the value will be: 1000 + 1000(0.05
4) = 1000(1 +
0.05
4)1= 1012.50 .
After two quarters, the value becomes: 1000 (1 +0.05
4)2= 1025.16 .
At one year, after 4 interest payments, the value will be: 1000(1 +0.05
4)4= 1050.95 . This is slightly
more money than with annual compounding where the one-year value was $1050.
At two years with 8 interest additions, the value becomes: 1000 (1 +0.05
4)4Γ2
= 1104.49$.
This idea extends to interest being added π times per year as:
Functions and Relations Page 53
Compounding Formula with π interest additions per year π½(π) = π·(π +π
π)ππ
Using this formula the chart below will follow what happens to 1000 dollars when we change the
number of pay periods, and the number of years.
π pay periods in a year π‘ years later
1 2 3 10 15
Yearly 1 1050 1102.50 1157.63 1628.89 2078.93
Semi annually 2 1050.63 1103.81 1159.69 1638.62 2097.57
Quarterly 4 1050.95 1104.49 1160.75 1643.62 2107.18
Monthly 12 1051.16 1104.94 1161.47 1647.01 2113.70
Weekly 52 1051.25 1105.12 1161.75 1648.32 2116.24
Per day 365 1051.26 1105.16 1161.82 1648.66 2116.89
Per hour π = (365 Γ 24) 8760 1051.26 1105.16 1161.82 1648.66 2116.89
Per minute π =(365 Γ 24 Γ 60)
525600 1051.27 1105.17 1161.83 1648.72 2117.00
Per second π =(365 Γ 24 Γ 60 Γ 60)
31536000 1051.27 1105.17 1161.83 1648.72 2117.00
As a mathematician this table should make you ponder a bit as we would have expected our values to
keep getting higher and higher when the number of pay periods increase. We wonder what happens as
π goes to infinity or π β β. We see that the benefit of adding the interest more and more frequently
seems to level off as π gets large. This would suggest that the numbers probably will start to accumulate
closer and closer to the same number no matter how many times interest is paid in a year.
This is the question that Bernoulli pondered. I.e., is there a nice way to describe what happens as the
frequency of interest additions goes to infinity?
Looking at π(π‘) = 1000(1 +0.05
π)ππ‘ as π β β
Bernoulli was a bit of a genius, and manipulated this formula as:
π(π‘) = 1000 (1 +0.05
π)ππ‘
= 1000(1 +1
(π
0.05))
π0.05
β 0.05π‘
= 1000 [(1 +1
π)π
]
0.05π‘
, π =π
0.05
In this way, the question comes down to what happens to (1 +1
π)π
as π β β.
Use your calculator and evaluate this expression for π = 10, 100, 1000, 1000000.
Once you have done so without looking it up on your own make a prediction of what you think this
number is below (1 +1
π)π πβββ ___________________ (Do this before continuing down the page.)
Functions and Relations Page 54
What you should have seen is that the number (1 +1
π)π , as π gets very large seems to settle down at
about (1 +1
π)πββ 2.7182.. In fact the limiting value is an irrational number mathematicians called π.
The first six digits of π are 2.71828β¦. See some of the digits for particular π values below.
π (1 +
1
π)π
100 2.704813829
1000 2.716923932
10000 2.718145927
100000 2.718268237
1,000,000 2.718280469
10,000,000 2.718281692
This number is used in many applications and it is a special kind of irrational number called a
transcendental number.
A transcendental number is a real or complex number that is not a solution of a polynomial equation
with rational coefficients. Numbers that are solutions of such polynomials are called βAlgebraicβ.
You already know another transcendental number π. The complex number π however is an algebraic
number, since π is a solution to the equation π₯2 + 1 = 0. The number β2 is also an algebraic number.
Can you figure out why?
If we let the number of pay periods in the compound interest problem above go to infinity we see that
in the following formula 1000(1 +0.05
π)ππ‘= 1000 [(1 +
1
π)π]0.05π‘
, πβββ 1000π0.05π‘.
Continuous Compounding Formula π½(π) = π·πππ .
In general, the formula π΄ = ππΒ±ππ‘ can be used to describe exponential growth and decay problems
where π is the per unit time rate of increase or decrease of some quantity. The sign for the exponent is
negative in case of decay. We will work in these applications in more detail later.
The βnatural logarithmβ denoted as πππ₯ = πππππ₯ is the inverse of the base-π exponential function
π¦ = ππ₯. Often this is read as βell en of xβ or βthe natural log of xβ.
Review of equivalent logarithmic and exponential equations
Exponential Logarithmic Equation Equation
π = ππ πππππ = π
π = πππ ππππ = π
π = ππ πππ = π
Functions and Relations Page 55
Practice Problems
1. Convert the exponential equations below into their equivalent logarithmic forms.
Exponential Equation Equivalent Logarithmic Equation
a. 8 = 23
b. 1
8= 2β3
c. 10 = 2π₯
d. 5π₯ = 25
e. (1
5)π₯= 125
f. 32π₯β4 = 81
g. 0.4π₯ = 5
h. ππ₯ = 6
i. 10π₯ = 0.0001
j. ππ₯+1 = 0.3
2. Convert the logarithmic equations below into their equivalent exponential forms.
Logarithmic Equation Equivalent Exponential Equation
a. πππ3π₯ = β2
b. πππ₯ = 5
c. log(π₯ + 1) = 5
d. β3 = ππππ₯
e. 1.5 = πππ₯
f. 5 = πππ2π₯
g. 4 = πππ1
2
(π₯)
h. πππ525 = 2
i. πππ3 (1
81) = β4
Functions and Relations Page 56
Practice Problems Solutions
It is helpful when doing these problems to identify the base in the problem. In the exponential version,
the base and exponent are easily identified and the exponent is the input. Also since logarithms are
inverses to exponential functions, the output of the logarithm is the input of the exponential and is the
exponent on the base in the exponential form of an equation.
In general then, π©ππππ¬πππππππ = ππ’πππ‘ππ‘π¦ π¬πππππππ = ππππ©πππ(ππ’πππ‘ππ‘π¦) and
1. Convert the exponential equations below into their equivalent logarithmic forms.
Exponential Equation Equivalent Logarithmic Equation
a. 8 = 23
Base is π
Exponent is π
a. πππ28 = π
b. 1
8= 2β3
Base is π
Exponent is βπ
b. πππ2 (1
8) = β3
c. 10 = 2π₯
Base is π
Exponent is ππ
c. πππ2 10 = π₯
d. 5π₯ = 25
Base is π
Exponent is π
d. π₯ = πππ5(25)
e. (1
5)π₯= 125
Base is π
π
Exponent is π
e. π₯ = πππ1
5
(125)
f. 32π₯β4 = 81 Base is π
Exponent ππ β π
f. 2π₯ β 4 = πππ381
g. 0.4π₯ = 5 Base is π. π
Exponent is π
g. π₯ = πππ0.4(5)
h. ππ₯ = 6 Base is π
Exponent is π
h. π₯ = ln 6
i. 10π₯ = 0.0001 Base is ππ
Exponent is π
i. π₯ = log (0.0001)
j. ππ₯+1 = 0.3 Base is π
Exponent is π + π
j. π₯ + 1 = ln (0.3)
2. Convert the logarithmic equations below into their equivalent exponential forms.
Functions and Relations Page 57
In general here we will use π¬πππππππ = ππππ©πππ(ππ’πππ‘ππ‘π¦) π©ππππ¬πππππππ = ππ’πππ‘ππ‘π¦
Logarithmic Equation Equivalent Exponential Equation
a. πππ3π₯ = β2 Base is π
a. π₯ = πβ2 =1
9
Exponent is βπ b. ln π₯ = 5
Base is π b. π₯ = ππ
Exponent is π c. log(π₯ + 1) = 5
Base is ππ
c. π₯ + 1 = πππ Exponent is π or π₯ = 9999
d. β3 = log π₯ Base is ππ
d. ππβπ = π₯ or π₯ = 0.001 Exponent is βπ
e. 1.5 = ln π₯
Base is π e. ππ.π = π₯
Exponent is π. π f. 5 = πππ2π₯
Base is π f. ππ = π₯ or 32 = π₯
Exponent is π g. 4 = πππ1
2
(π₯)
Base is π
π
g. (π
π)π
= π₯ or 1
16= π₯
Exponent is π h. πππ525 = 2
Base is π h. 25 = ππ
Exponent is π
i. πππ3 (1
81) = β4
Base is π
i. 1
81= πβπ
Exponent is βπ
Given the fact that the output of an exponential function π¦ = ππ₯ always produces a positive real
number, the domain of a logarithmic functions is just positive real numbers.
Practice Problems
1. Find the domain of the functions below.
a. π(π₯) = πππ2(π₯ β 1)
b. π(π₯) = ln (2 β π₯)
c. β(π₯) = log (1 + 2π₯)
Attempt these first before looking on the next page for solutions.
Solutions to Practice Problems
1. Find the domain of the functions below.
a. π(π₯) = πππ2(π₯ β 1)
Since the input has to be positive we must have π₯ β 1 > 0 or π₯ > 1
Domain is (1,β)
Functions and Relations Page 58
b. π(π₯) = ln (2 β π₯)
Since the input has to be positive we must have 2 β π₯ > 0 or 2 > π₯
Domain is (ββ, 2)
c. β(π₯) = log (1 + 2π₯)
Since the input has to be positive we must have 1 + 2π₯ > 0 or 1> β2π₯ or β1
2< π₯
Domain is (β1
2, β)
Evaluating Logarithmic Functions
Evaluating πππππ₯ means asking the question π? = π₯.
Practice Problems
Evaluate the following
1. πππ2(16)
2. πππ1
5
(25)
3. πππ1000
4. ln π3
5. πππ93
Attempt these problems yourself and then look at the answers.
Functions and Relations Page 59
Solutions to Practice Problems
Again, in evaluating logarithms the output is the exponent you are seeking and the base is indicated by
the notation of the logarithm.
Evaluate the following
1. πππ2(16)
Base = 2 and so we are looking for 2? = 16 and we know that 16 = 24 so we have 2? = 24 Since exponential functions are one-to-one the only way this can happen is if we have ?= 4.
Therefore πππ2(16) = 4
2. πππ1
5
(25)
3. Base =1
5 and so we are looking for (
1
5)?= 25 and we know that 25 = 52 = (
1
5)β2
so we have
(1
5)?= (
1
5)β2
Since exponential functions are one-to-one the only way this can happen is if we
have ?= β2. Therefore, πππ1
5
(25) = β2.
4. πππ1000 = 3 since 103 = 1000
5. πππ3 = 3 since π3 = π3
6. πππ93 =1
2 since 9
1
2 = 3
Since the output of logarithm functions are exponents, all the rules of exponents give rise to
corresponding rules of logarithms.
Properties of Logarithms
Playing
Now that we are more comfortable with logarithms and see them as exponents we may wonder how
the laws of exponents give rise to properties of logarithms. Let us play and see what happens.
Let π₯ = πππππ’, π¦ = πππππ£ and π > 0, π β 1. We know from changing logarithmic equations to
exponential equations that we have ππ₯ = π’, and ππ¦ = π£.
1. Now the product property for exponents says π’ β π£ = ππ₯ππ¦ = ππ₯+π¦. Therefore we have
π’π£ = ππ₯+π¦. Now if we put this into its logarithmic form, we get ππππ(π’π£) = π₯ + π¦. Repacing π₯
and π¦, we obtain the product rule for logartihms!
ππππ(π’π£) = ππππ(π’) + ππππ(π£)
2. Now the quotient property for exponents says π’
π£=
ππ₯
ππ¦= ππ₯βπ¦. Therefor we have
π’
π£= ππ₯βπ¦. Now if we put this into its logarithmic form, we get ππππ (
π’
π£) = π₯ β π¦. Repacing π₯
and π¦, we obtain the product rule for logartihms!
ππππ (π’
π£) = ππππ(π’) β ππππ(π£)
Functions and Relations Page 60
3. When we convert the exponential equation π0 = 1, we get 0 = logπ 1 . That is the power that
π must be raised to so that π?=1 is 0 since π0 = 1.
ππππ(1) = 0
4. The power rule for exponents looks like (ππ₯)π = πππ₯ . Putting this into logarithmic form, we
have loga(ππ₯)π = loga π
ππ₯ with ππ₯ = π’ and logπ πππ₯ = ππ₯, we get:
loga(π’)π = ππ₯ and since π₯ = πππππ’
ππππ(π’π) = ππππππ’
5. One final property of logarithms deals with the relationship between the logarithms with
different bases of some number π’. Let π₯ = loga π’ and π§ = logb π’. Is there a relationship
between these two numbers? Can we find the base-π logarithm in terms of the base-
π logarithm? If so, then we can use the base-ten or natural logarithm functions on our
calculators to evaluate logarithms to any base.
We do have that π’ = ππ₯ = ππ§. If we take the base-π logarithm on both sides and use property
(4.) above, we get: loga π’ = loga ππ§ β loga π’ = π§ loga π . Solving for π§ which is logb π’ we
get the useful result π§ = logb π’ =loga π’
loga π.
logb π’ =loga π’
loga π
We can use this to calculate log3 5 by using the common or natural logarithm function on our
calculators. We have: log3 5 =πππ5
πππ3β
0.6990
0.4771β 1.465, ππ log3 5 =
ππ5
ππ3β
1.6094
1.0986β 1.465 . The
individual numerator and denominator will be different values but the ratio is the same. Check
these numbers for yourself.
We can use these properties to decompose products, quotients and powers and radicals that are inputs
to logarithm functions. In the other direction, we can combine the sum and differences of logarithms
that are to the same base.
Functions and Relations Page 61
Practice Problems
1. Use the properties of logarithms above to either combine the log terms or to decompose the logarithm terms into a sum or product of simpler log terms.
A. πππ27 + πππ29 = ________________
B. πππππ β πππππ = ____________________
C. ππππ(π π‘) = ____________________
D. ππππ (π
π‘) = _____________________
E. ππππ(ππ) = ____________________
F. πππππ(π) = ____________________
G. Change of base formula. Write in terms of the natural or common logarithm.
ππππ(π) =
2. Use properties of logarithms to do the problems below.
A. Fill in the missing values to make the statement a true statement.
i. πππ712 β πππ7(_______) = πππ74
ii. πππ65 + πππ68 = πππ6(______)
iii. β2πππ53 = πππ5(_____)
iv. πππ316 = (_______)πππ32
v. ππ5
ππ6= πππ6(_______)
B. Expand the following. Each logarithm in your answer should involve only one variable. Assume that all variables are positive.
I. log(π₯5π¦7) =____________________
II. log3 (π₯7π¦3
βπ§3 ) =____________________
III. log (π₯5
βπ§7π¦3) =____________________
IV. ln((4 β π₯)(π₯ + 2)) =____________________
V. ln (π₯6β7
5π§2) = ____________________
Functions and Relations Page 62
C. Write the following as one term. i. 6πππ3π₯ + 3πππ3π¦ = ______________
ii. 1
2ππππ₯ β 3ππππ¦ β 2ππππ§ = _______________
D. Compute the values below exactly
i.πππ2 (1
8) = ______________
ii.log (0.000001) = ______________
iii.ln(π6) = _____________
iv.β ln(βπ) = __________
v.πππ3 (1
27) = ____________
E. Use a scientific calculator evaluate the following, and round your answers to 2 digits.
i. 2000
log (1+0.04
5)= ________________
ii. πππ5
πππ3= ____________
iii. ln (0.03)
2= ____________
iv. πππ54 = _____________
Practice Problems Solutions
1. Properties of logarithms: Please fill out the missing values.
A. πππ27 + πππ29 = πππ2(7 Γ 9) = πππ263 B. πππππ β πππππ = ππππ (π
π )
C. ππππ(π π‘) = πππππ + πππππ‘
D. ππππ (π
π‘) = πππππ β πππππ‘
E. ππππ(ππ) = πππππ(π) F. πππππ(π) = ππππ(π
π)
G. Change of base formula: ππππ(π) =ππππ
ππππ=
πππ
πππ
2. Use properties of logarithms to do the problems below.
A. Fill in the missing values to make the statement a true statement.
i. πππ712 β πππ7(3) = πππ74
ii. πππ65 + πππ68 = πππ6(40)
iii. β2πππ53 = πππ5 (1
32) = πππ5 (
1
9)
iv. πππ316 = (4)πππ32
v. ππ5
ππ6= πππ6(5)
B. Expand the following. Each logarithm in your answer should involve only one variable. Assume that all variables are positive.
I. log(π₯5π¦7) = 5ππππ₯ + 7ππππ¦
II. log3 (π₯7π¦3
βπ§3 ) = 7log3π₯ + 3log3π¦ β
1
3log3π§
III. log (π₯5
βπ§7π¦3) = 5log3π₯ β
1
2(7log3π§ + 3log37)
= 5log3π₯ β7
2log3π§ β
3
2log3π¦
IV. ln((4 β π₯)(π₯ + 2)) = ππ(4 β π₯) + ππ(π₯ + 2)
V. ln (π₯6β7
5π§2) = 6lnπ₯ +
1
2ππ7 β (ππ5 + 2πππ§)
Functions and Relations Page 63
= 6lnπ₯ +1
2ππ7 β ππ5 β 2πππ§
C. Write the following as one term.
i. 6πππ3π₯ + 3πππ3π¦ = πππ3(π₯6π¦3)
ii. 1
2ππππ₯ β 3ππππ¦ β 2ππππ§ = πππ (
βπ₯
π¦3π§2)
D. Compute the values below exactly
i. πππ2 (1
8) = β3
ii.log(0.000001) = β6
iii.ln(π6) = 6
iv.β ln(βπ) = β1
2
v.πππ3 (1
27) = β3
E. Use a scientific calculator evaluate the following, and round your answers to 2 digits.
i. 2000
log(1+0.04
5)= 250998. 67
ii. πππ5
πππ3= 1. 46
iii. ln (0.03)
2= β1. 75
iv. πππ54 =πππ4
πππ5= 0.86
Functions and Relations Page 64
Worksheet Section1.3a Date:______________ Name:__________________________
Concept Meaning in words and/or examples as required
1. State the Properties of logarithms
2. List two uses each for exponential functions and logarithmic functions.
3. How are exponential and logarithmic functions related to each other?
4. How do we find domains of logarithmic functions?
5. In your own words describe how to change logarithmic equations into an exponential equation.
6. In your own words describe how to change exponential equations into logarithmic equations.
Difficulties encountered in the section:
Functions and Relations Page 65
Exercises 1.3a
1. Rewrite the exponential equations in logarithmic form and logarithmic equations in exponential form. If possible simplify your answers.
Exponential Equation
Logarithmic Equation
ππ₯ = 5
2π₯+1 = 8
πππ2(π₯) = β1
log(π₯ + 1) = 2
ln(π₯ + 1) = 3
51βπ₯ = 3
πππ12
(π₯) = β3
Exponential Equation
Logarithmic Equation
πππ13
(81) = β4
52 = 25
πππ3 (1
9) = β2
π3 = π₯
ln(π2) = 2
10β2 = 0.01
πππ1000 = 3
2. Find the domain of the functions below.
A. π(π₯) = log (π₯ + 1) Domain:
B. π(π₯) = πππ (3
π₯β4)
Domain:
C. β(π₯) = 3π₯β1 Domain:
D. π(π₯) = ππ(1 β π₯)
Domain:
3. Properties of logarithms: Please fill out the missing values.
A. πππππ₯ + πππππ¦ = ________________
B. πππππ₯ β πππππ¦ = ____________________
C. ππππ(π₯π¦) = ____________________
D. ππππ (π₯
π¦) = _____________________
E. ππππ(π₯π) = ____________________
F. πππππ(π₯) = ____________________
G. Change of base formula: Write in terms of the common or natural logarithm function. ππππ(π₯) =
4. Use properties of logarithms to do the problems below.
Functions and Relations Page 66
F. Fill in the missing values to make the statement a true statement.
vi. πππ58 β πππ5(_______) = πππ54
vii. πππ23 + πππ25 = πππ2(______)
viii. 3πππ72 = πππ7(_____)
ix. πππ549 = (_______)πππ57
x. ππ5
ππ4= πππ4(_______)
G. Expand the following. Each logarithm in your answer should involve only one variable. Assume that all variables are positive.
VI. log(π₯3π¦2) =____________________
VII. log2 (π₯3π¦2
βπ§) =____________________
VIII. log (π₯3
βπ§5π¦) =____________________
IX. ln((4 + π₯)(π₯ β 2)) =____________________
X. ln (π₯5 βπ¦
3
3π§) = ____________________
H. Write the following as one term. i. 4πππ2π₯ + 2πππ2π¦ = ______________
ii. 1
3ππππ₯ β 2ππππ¦ + 3ππππ§ = _______________
I. Compute the values below exactly
i. πππ2(8) = ______________
ii.log (0.000001) = ______________
iii.ln(π5) = _____________
iv.ln(βπ) = __________
v.πππ5 (1
25) = ____________
J. Use a scientific calculator to evaluate the following, and round your answers to 2 digits.
i. 20000
log (1+0.02
5)= ________________
ii. πππ3
πππ2= ____________
iii. ln (0.5)
3= ____________
iv. πππ311 = _____________
5. State a formula for the value of each compound interest account after π‘ years.
a. An account starts with $5000 and earns 6.2%/yr interest compounded annually.
Functions and Relations Page 67
b. An investment is initially worth $20,000 and earns 6% interest/yr compounded
quarterly.
c. An investment of $2000 earns interest at 7.5%/yr compounded continuously.
d. A credit card debt has an interest rate of 18%/yr and is compounded monthly. Find the
function for the debt after π‘ years if it is initially at $10,000 and no payments are made.
e. You purchase a pair of shoes for $200 with your credit card but make no payments. The
interest rate then gets moved to 24% (2%/month). State a formula for the balance due
after π years and state list the balance due at 1
2 , 1, 2, 3 and 10 years.
6. Population growth is modeled with exponential functions in several equivalent formats where
π0 is the population at π‘ = 0.
β’ If the percentage growth over a one-year period is known, then we treat it like annual
compounding and get π(π‘) = π0(1 + π)π‘.
β’ If the continuous growth rate is known, then we use π(π‘) = π0πππ‘ just like continuous
compounding interest problems.
β’ If the time it takes a population to double is known, then we write the exponential function
in terms of the base-two exponential as π(π‘) = π0 β 2π‘
ππ
State formulas for each population growth problem below.
a. Wisconsinβs population was 5,000,000 at π‘ = 0 in 2010 and is growing 0.35% each year.
b. The U.S. population was 310,000,000 in 2010 at π‘ = 0 and has a continuous growth rate
of 0.97% per year.
c. A blood infecting bacteria grows rapidly and has a doubling time of 30 minutes. Find the
exponential function that gives the number of bacteria π‘ minutes after a puncture
wound introduces 50 of the microbes into the blood stream.
7. Exponential decay models are modeled much like growth problems above but with the π being
negative.
β’ If the percentage decline over a one-year period is known, then we treat it like annual
compounding and get π(π‘) = π0(1 β π)π‘. E.g., 10% decline per year in the value of a car
leads to the value function of a car originally worth $25,000 as π(π‘) = 25000(0.9)π‘
β’ If the continuous decay rate is known, then we use π(π‘) = π0πβππ‘ much like continuous
compounding interest problems. E.g., A 20 gram sample of radioactive tritium declines
continuously by 5.6% per year. The Amount after π‘ years is π΄(π‘) = 20πβ0.056π‘.
Functions and Relations Page 68
β’ If the time it takes for a quantity to decline to half its original amount (πβ), then we write
the exponential function in terms of the base-(1
2) exponential as π΄(π‘) = π΄0 β (
1
2)π‘
πβ
State formulas for each exponential decay problem below.
a. Ukraineβs population was 42,690,000 in 2016 and decreasing by 8.4% each year. State
π(π‘) =________________ where π‘ is years after 2016.
b. The light intensity under water in Lake Superior decreases continuously at 3% per foot
of depth. If the surface intensity (π₯ = 0) is 180π
π2. State the intensity function as a
function of the number of feet below the surface. πΌ(π₯) =
Also predict the light intensity at a depth of 100 feet.
c. Radioactive Carbon-14 occurs naturally in all plant-derived materials. Once the plant
dies, the carbon-14 atoms in the plant are slowly converted back to nitrogen-14. The
half-life for this conversion is about 5,730 years. Give the exponential function that
describes the amount of carbon-14 (π‘) years after the plant died if it started with 5
trillion atoms of carbon-14.
Also predict how many atoms would remain after 200,000 years.
8. Consider the population growth function for the world π(π‘) = 7.4π0.011π‘ where π‘ = 0
corresponds to 2015.
a. What is the continuous growth rate?
b. Determine by what percent the population increases over the course of one year. Then give
a formula for π(π‘) in the βannual compoundingβ format.
c. By guess and check or other methods, try to use π(π‘) = 7.4π0.011π‘ to determine how many
years it takes for the population to double and give the formula for π(π‘) in the βdoubling
timeβ format.
9. Evaluate the measures of acidity, sound intensity and earthquake magnitude using the
appropriate logarithmic definitions of these measures.
a. Find the pH of blueberries which have a [π»+] concentration of 0.00076 πππ
πππ‘ππ
b. Most foods have pH less than 7 and are thus acidic. Tofu has a pH of 7.2. Write down the
pH statement for this and convert it to its exponential form to obtain the [π»+]
concentration of tofu.
c. A very loud speaker outputs sound that you can feel a 5 feet from the speaker. The sound
intensity there is at 5 π€ππ‘π‘π
π2 . Determine the decibel level of this sound.
d. A weed whipper produces noise at 95 dB for the operator. Determine the sound intensity
level in πππ‘π‘π
π2 .
Functions and Relations Page 69
e. Oklahoma has seen experienced unprecedented earthquakes since the advent of Fracking
for natural gas. In 2015 there were more than 30 of magnitude 4 or greater on the Richter
scale. Determine the seismograph reading in ππ of a magnitude 4 and the most recent
magnitude 9 quake that occurred 6 years ago in Japan. What is the ratio of the amount of
movement between a level 4 and 9 earthquake?
10. Create an exponential function with π(0) = 100 and π(1) = 200. Do you think your function is
unique? Explain your answer.
11. Create a logarithmic function with π(1) = 100, and π(2) = 200. Do you think your function is
unique? Explain your answer.
Another Class of Functions: Sequences
A sequence is an ordered collection of objects. There is a first object, a second object, third
object, β¦ etc. The list can be finite or infinite. A class roster of the first names would be a finite
sequence that may have repeats in the list. Sequences can be interpreted as functions from the
natural numbers to the set of objects where e.g., the natural number 3 as the input
corresponds to the third item in the ordered list.
If we take a real function and restrict the domain to the natural numbers π = {1, 2, 3,β¦ } the ordered
list of outputs π(1), π(2), π(3), β¦.is one way to generate interesting sequences. In the study of
sequences, we typically donβt use the function notation above, but instead use subscript notation. E.g.,
the sequence of squares of natural numbers might be denoted by {ππ} = {π2} = π1, π2, π3, β¦ =
1, 4, 9, 16,β¦ Here weβd typically define the sequence using ππ = π2 instead of π(π) = π2, for π β π.
A sequence {ππ} = π1, π2, π3, β¦ where ππ is often given by some formula in terms of π
Examples of Sequences generated by functions with domain restricted to the natural numbers π΅.
1. The constant function π(π₯) = 50 has all elements of the sequence being equal to 50. With
ππ = π(π) = 50, we have {ππ} = 50, 50, 50, 50, β¦ for π = 1,2,3,4,β¦ is a sequence of numbers
with all outputs being 50. This is called a constant sequence. So when we write π43, it is
referring to the term in the sequence that is the 43rd number in the sequence. In our case it
would be 50 since all terms are 50.
2. Consider a linear function restricted to the domain of natural numbers say ππ = 3 + 7π.
Starting with π = 1, this sequence is: π1 = 10, π2 = 17, π3 = 24, π4 = 31,β¦. The fiftieth
element in this sequence is π50 = 3 + 7(50) = 353.
Sometimes we start a sequence with π0 instead of π1. If asked to find the 50th element in the
sequence {3 + 7π}π=0 β = π0, π1, π2, β¦ = 3, 10, 17,β¦ since we started at zero the fiftieth term
would actually correspond to π = 49 thus the fiftieth element would be π49 = 3 + 49(7) =
346.
Functions and Relations Page 70
Sequences obtained from linear functions are called arithmetic sequences. ππ = π β π + π0
The common difference in the sequence is the slope and usually denoted as π and π0 is the
π§ππππ‘β element.
3. A sequence obtained from exponential function restricted to the domain of whole numbers say
ππ = 3(2)π is called a geometric sequence. In this example, increasing π by one, means we
have one more 2 bieng multiplied and thus each element in the sequence is 2 times larger than
the one before. π0 = 3, π1 = 6, π2 = 12, π3 = 24, π4 = 48,β¦.
If asked to find the 50th term since we started at zero we would have π49 = 3(2)49 =
56294953421312.
Whenever, each successive term is obtained by multiplying by the same constant ratio (π), the
sequence is really just an exponential function with domain restricted to π or π . You can
recognize this by observing that the ratio of any two consecutive elements is the same.
Sequences obtained from exponential functions are called geometric sequences. ππ = π0ππ
The common ratio π is the base of the exponential function and π0 is the π§ππππ‘β element.
Practice Problems
1. Find the first 4 terms of the sequences given below.
ππ = ππ‘β term of the sequence
First term
Second term
Third term
Fourth term
ππ = 5(1
2)2π+1
, π =
0,1,2,β¦
Arithmetic Geometric Neither
ππ = 4π + 3, π = 4,5,β¦ Arithmetic Geometric Neither
ππ =2πβ1
π+2, π = 1,2,3,β¦ Arithmetic
Geometric Neither
ππ =(β1)π
2π, π = 1,2, β¦ Arithmetic
Geometric Neither
2. For the sequences below determine if they are arithmetic or geometric. Then find a formula for the ππ and evaluate: π5 = , and π10 = .
Sequence Type π(π) = ππ = ππ‘β term
Evaluate these elements.
{13,17,21,25,β¦ . } Arithmetic Geometric Neither
π5 =
π10 =
{7,21,63,189, . . . } Arithmetic Geometric Neither
π5 =
π10 =
{12,10,7,1,β¦ . } Arithmetic π5 =
Functions and Relations Page 71
Geometric Neither
π10 =
{1,β11,1,β11,β¦ . } Arithmetic Geometric Neither
π5 =
π10 =
3. For a given arithmetic sequence, the 82ππ term,
π82 = β373 and the 6π‘β term, π6 = 7. Find the 42nd term π42. (One way is to think of this as a linear function, so find the slope and use the point (π = 6, π¦ = 7) to find the intercept. Thus ππ = π β π + π0 ))
4. For a given geometric sequence, the 7π‘β
term, π7 = 15 and the 9π‘β term, π9 =
135. Find the 10π‘β term π10. (Recall that geometric sequences are exponential functions of (π) and the formula for ππ can be written in the form: ππ = π0 β π
π.)
Practice Problems Solutions
1. Find the first 4 terms of the sequences given below.
ππ = ππ‘β term of the sequence
First term
Second term
Third term
Fourth term
ππ = 5(1
2)2π+1
, π =
0,1,2,β¦
π0 =5
2 π1 =
5
8 π2 =
5
32 π3 =
5
128
X Arithmetic Geometric Neither
ππ = 4π + 3, π = 4,5,β¦ π4 = 19 π5 = 23 π6 = 27 π7 = 31 Arithmetic X Geometric Neither
ππ =2πβ1
π+2, π = 1,2,3,β¦ π1 =
1
3 π2 =
3
4 π3 =
5
5= 1
π4 =7
6
Arithmetic Geometric X Neither
ππ =(β1)π
2π, π = 1,2, β¦ π1
= β1
2
π2 =1
4
π3
= β1
6
π4 =1
8
Arithmetic Geometric X Neither
2. For the sequences below determine if they are arithmetic or geometric. Then find a formula for the ππ and evaluate: π5 = , and π10 = .
Sequence Type π(π) = ππ = ππ‘β term Evaluate these elements.
{13,17,21,25,β¦ . } Arithmetic
If we start at π0 = 13, with π = π = 4, we have:
ππ = 4π + 13
π5 = 33
π10 = 53
{7,21,63,189, . . . } Geometric
If we start at π0 = 7 and the ratio is π = 3, we have: ππ = 7 β 3π
π5 = 7 β 35 = 1701
π10 = 7 β 310 = 413343
{12,10,7,1,β¦ . } Neither Many possibilities for continuing a patter here.
π5 =
π10 =
{1,β11,1,β11,β¦ . } Neither It seems to just alternate, so we could just say ππππ =1 and πππ£ππ = β11
π5 = 1
π10 = β11
Functions and Relations Page 72
3. For a given arithmetic sequence, the 82ππ
element, π82 = β373 and the 6π‘β element, π6 =7. Find the 42nd element π42.
βArithmeticβ means a ππ is a linear function of π. First we compute the slope which is the common
difference as = π = πβππππ ππ ππ’π‘ππ’π‘
πβπππ ππ ππππ’π‘ π=
π82βπ6
82β6=
β373β7
76= β5.
Then using the element (π = 6, π6 = 7 we have: 7 = β5 β 6 + π0 β π0 = 37 Thus the formula is: ππ = β5 β π + 37. And π42 = β5 β 42 + 37 = β173
4. For a given geometric sequence, the 7π‘β
term, π7 = 15 and the 9π‘β term, π9 = 135.
Find the 10π‘β term π10.
First we determine the common ratio (π), i.e. the base of the exponential function. Going from π7 = 15, to π9 = 135, we must multiply by π two times, thus 15 β π2 = 135
from which π2 =135
15= 9 β π = 3.
Now we can use either point to determine π0 in the formula ππ = π0 β 3
π. Take (π = 7, π7 =
15) to get: 15 = π0 β 37 β π0 =
5
36.
Thus ππ =5
36β 3π and π10 =
5
36β 310 = 405.
Fibonacci Sequence and the Golden Ratio
The Fibonacci sequence is an interesting sequence that has applications to many areas. It is usually
defined in a recursively which means the value of the ππ‘β element of the sequence is give in terms of
one or more of the elements previous to the ππ‘β. A recursive formula for a sequence would thus look
like: ππ =(some expression in the variables ππβ1, ππβ2, β¦
Fibonacci sequence recursively defined is given by: ππ = ππβ1 + ππβ2 and π1 = 1, π2 = 1 .
With this recursive formula it is easy to make a list
of the elements of the sequence. Thus:
π1 = 1, π2 = 1,
π3 = 1 + 1 = 2,
π4 = 1 + 2 = 3,
π5 = 2 + 3 = 5,
π6 = 3 + 5 = 8β¦
Arithmetic and geometric sequences can also be defined recursively. For arithmetic, the ππ‘β element is
obtained by adding π to the (π β 1)π‘β element. Thus a recursive formula for an arithmetic sequence is
given by; ππ = ππβ1 + π, and the starting value π0 or π1 must be given. Similarly, the ππ‘β element of
a geometric sequence is obtained by multiplying the previous element by the common ratio (π) and a
recursive formula is given by: ππ = ππβ1 β π where π0 or π1 must be given.
One question one might ask is whether it is possible to obtain the Fibonacci sequence as an explicit
function that just involves the index of the sequence π. I.e., can we get a formula for the ππ‘β term
Fibonacci Sequence in Optimal Branching Pattern
Functions and Relations Page 73
ππ = (some expression involving only π)? Well, it turns out that it can be done and the formula is a bit
complicated, but comes down to the difference of two separate geometric sequences.
Explicit formula for Fibonacci Sequence ππ =1
β5((
1+β5
2 )π
β (1ββ5
2)π
)
It is easy to see that with π = 1 the formula does produce π1 = 1. One can also compute π0 = 0 from
the formula. Use your calculator with π = 6 to show that the above formula produces the correct value
of π6 = 8.
The base 1+β5
2 in the Fibonacci formula actually shows up in another context where it is called the
Golden Ratio. It happens to show up in βpleasingβ dimensions for openings in architecture. It is the
length to width ratio π
π€ of a rectangle with the special property that when the rectangle is truncated by
slicing off a π€ by π€ square, the rectangle that remains has this same ratio of length to width.
Thus π
π€=
π€
πβπ€=
1π
π€β1 . If we call
π
π€= π, we have π =
1
πβ1. We can solve this for π to get: π(π β 1) =
1 β π2 β π β 1 = 0. Solving with the quadratic formula, we have π =1Β±β5
2 and these are the two
bases in the explicit formula for the Fibonacci sequence. The positive of these π =1+β5
2β 1.618 is
called the Golden Ratio.
Golden Ratio is π
π€
Golden Spiral where the radius increases by the golden ratio for each quarter turn.
Functions and Relations Page 74
Section1.3b Worksheet Date:______________ Name:__________________________
Concept Meaning in words and/or examples as required
What is a sequence?
What is an arithmetic sequence?
What is a geometric sequence?
Give an example of a sequence that is neither an arithmetic sequence nor a geometric sequence.
What is a Fibonacci sequence?
What is the golden ratio?
Give an example of an arithmetic sequence.
Give an example of an geometric sequence
Given an example of a sequence that is neither arithmetic nor geometric.
Difficulties encountered in the section:
Functions and Relations Page 75
Exercises 1.3b
1. For the sequences below determine if they are arithmetic or geometric. Then find the formula for the ππ and then fill in the missing terms column with value of that term.
Sequence Type π(π) = ππ = ππ‘β term {22, 26, 30, 34,β¦ . } Arithmetic
Geometric Neither
{6, 12, 24, . . . } Arithmetic Geometric Neither
{9, 13, 17,β¦ . } Arithmetic Geometric Neither
{7, 11, 19,β¦ . } Arithmetic Geometric Neither
2. Find the first 4 terms of the sequences given below.
ππ = ππ‘β term of the sequence
First term
Second term
Third term
Fourth term
ππ = 4(1
3)2π+1
, π =
0,1,2,β¦
Arithmetic Geometric Neither
ππ = 3π + 5, π =3,4,5,β¦
Arithmetic Geometric Neither
ππ =2π
π+3, π = 3,4,5,β¦ Arithmetic
Geometric Neither
ππ =(β1)π
π, π = 1,2, β¦ Arithmetic
Geometric Neither
3. For a given arithmetic sequence, the 82ππ
element, π82 = β370 and the 6π‘β element, π6 = 10. Find the linear function
for ππ and find the 33ππ element π33.
4. For a given geometric sequence, the 7π‘β
element π7 =23
25 and the 10π‘β element π10 =
115. Find a formula for this exponential
sequence and evaluate the14π‘β element π14.
5. Find the 20th and 21st elements of the Fibonacci sequence. Also compute the ratio π21
π20 to show
that this ratio is very near in value to the Golden Ratio.
6. Consider the sequence of continued fractions: π1 = 1, π2 = 1 +1
1 , π3 = 1 +
1
1+1
1
,
π4 = 1 +1
1+1
1+11
, β¦ππ+1 = 1 +1
ππ. Simplify the first 5 elements and express how the results
relate to the Fibonacci sequence. What number does this sequence tend to?
Functions and Relations Page 76
7. Create your own sequence where π5 = 32. Is your sequence unique? Explain your answer.
8. Create a sequence of each type listed below a. Geometric b. Arithmetic c. Constant d. Neither
Even and Odd Functions
A function is called even if and only if π(βπ₯) = π(π₯).
This says that the function output π¦ is the same at any positive number π₯ = π and the negative number
π₯ = βπ . On the graph of an even function the points (π, π) and (βπ, π) always show up directly
across the π¦-axis from each other. Geometrically speaking it means that the graph of the function π¦ =
π(π₯) is symmetric with respect to the π¦-axis.
Examples
1. All polynomials functions of the type π(π₯) = π₯π ,where π is an even number, and π₯ is any real
number are even functions.
The reason for why the above function is even is that (βπ₯)π = π₯π, when π is an even since
(βπ₯)π = (β1)π(π₯)π and (β1) multiplied by itself an even number of times equals one.
2. The function π¦ = |π₯|, where π₯ is any real number is also an even function.
The reason π(π₯) = |π₯| is an even function is the fact that |βπ₯| = |π₯|, for all real numbers π₯.
3. All the functions below are even functions.
a.
b.
c.
A function is called odd if and only if π(βπ₯) = βπ(π₯).
This says that the function output π¦ at any negative number π₯ = βπ and is the negative of the π¦-value
at the positive number π₯ = π . On the graph of an odd function the points (π, π) and (βπ,βπ) always
show up opposite each other across the point (0, 0) since (0, 0) is the midpoint of the line segment
joining these two points. Geometrically speaking it means that the graph of the function π¦ = π(π₯) is
Functions and Relations Page 77
symmetric with respect to the origin. This symmetry also means that the graph of an odd function
remains the same when rotated by 180 degrees.
Examples
1. All polynomials functions of the type π(π₯) = π₯π, where π is an odd number are odd functions.
The reason that π(π₯) = π₯πππ is even is the fact that π¦ = (βπ₯)π = (β1)ππ₯π, and (β1)πππ =
β1 and hence π¦ = β(π₯π).
2. All the functions below are odd functions since the graphs are symmetric with respect to the
origin.
a.
b.
c.
Practice Problems
1. Determine if the functions below are odd, even, or neither.
a. π(π₯) = 2π₯2 + 4
b. π(π₯) = 3π₯2 β 2π₯ + 1
c. π(π₯) = π₯3 β π₯2
d. π(π₯) = π₯3 β π₯
e. π(π₯) = π₯3 + 3π₯ + 1
f.
g.
h.
Solutions to Practice Problems
1. Determine if the functions below are odd, even, or neither.
a. π(π₯) = 2π₯2 + 4
π(βπ₯) = 2(βπ₯)2 + 4 = 2π₯2 + 4 = π(π₯) so the function is even by the definition.
b. π(π₯) = 3π₯2 β 2π₯ + 1
Functions and Relations Page 78
π(βπ₯) = 3(βπ₯)2 β 2(βπ₯) + 1 = 3π₯2 + 2π₯ + 1. This is different from π(π₯) = 3π₯2 β
2π₯ + 1 and also different from βπ(π₯) = β3π₯2 + 2π₯ β 1. Therefore the function is
neither even or odd.
c. π(π₯) = π₯3 β π₯2
π(βπ₯) = (βπ₯)3 β (βπ₯)2 = βπ₯3 β π₯2. This is also different from π(π₯) = π₯3 β π₯2 and
from βπ(π₯) = βπ₯3 + π₯2 , therefore the function is neither odd or even.
d. π(π₯) = π₯3 β π₯
π(βπ₯) = (βπ₯)3 β (βπ₯) = βπ₯3 + π₯ = β(π₯3 β π₯) = βπ(π₯), therefore the function is
an odd function.
e. π(π₯) = π₯3 + 3π₯ + 1
π(βπ₯) = (βπ₯)3 + 3(βπ₯) + 1 = βπ₯3 β 3π₯ + 1 β π(π₯) ππ β π(π₯), therefore the
function is neither odd nor even.
f.
This would be an odd function since for (π₯, π¦) on the graph so is (βπ₯,βπ¦) on the graph. The graph is symmetric with respect to the origin.
g.
Neither. Since the graph is not symmetric with respect to origin or the π¦-axis.
h.
This would be an even function since for (π₯, π¦) on the graph so is (βπ₯, π¦) on the graph. The graph is symmetric with respect to the π¦-axis.
(π₯, π¦)
(βπ₯,βπ¦)
(π₯, π¦) (π₯, π¦)
Functions and Relations Page 79
Section1.3c Worksheet Date:______________ Name:__________________________
Concept Meaning in words and/or examples as required
What is an even function?
What is an odd function?
What kinds of functions are symmetric with respect to π-axis?
What kinds of functions are symmetric with respect to π-axis?
What kinds of functions are symmetric with respect to the origin?
Give graphical examples of an a) Even function b) Odd Function c) Neither
What does it mean algebraically when a functions an even function? Give examples and explain.
What does it mean algebraically when a functions an odd function? Give examples and explain.
Difficulties encountered in the section:
Functions and Relations Page 80
Exercises 1.3c
1. Determine if the functions below are odd, even, neither.
a. π¦ = π₯2 β π₯ Odd Even Neither
b. π¦ = |π₯| Odd Even Neither
c. π¦ = π₯3 β 3π₯ Odd Even Neither
d.
Odd Even Neither
f.
Odd Even Neither
g.
Odd Even Neither
h.
Odd Even Neither
i. π(π₯) = 5π₯2 + 3 Odd Even Neither
j. π(π₯) =3
π₯
Odd Even Neither
2. Determine whether: a. An exponential function could be even or odd.
b. A Linear function could be even or odd.
c. Is there any function that could be both even and odd?
3. Consider the function π(π₯) = π₯2 β 3π₯ + 5 and try to write it as a sum of an even and an odd function, i.e., find formulas for π(π₯) = , and β(π₯) = such that π(π₯) + β(π₯) = (π + β)(π₯) = π₯2 β 3π₯ + 5.
4. Show that for π(π₯) = π₯2 β 3π₯ + 5 then π(π₯) =π(π₯)+π(βπ₯)
2 is an even function.
Functions and Relations Page 81
5. Create a function of each type listed below. a. Odd b. Even c. Neither
Which of these functions are one-to-one? Do you think an even function can be one-to-one? Explain your answer
1.4 Arithmetic and Composition of Functions In the previous sections weβve seen many kinds of functions. As a mathematician any time we get new
objects to play with (which in this case is functions), we want to see if we can do arithmetic with it.
Playing
Below are some functions and we will see what happens when we do arithmetic with them. Attempt the
problems using your instinct before looking at answers on the next page.
Practice Examples
1. Evaluate the following
A. π(π₯) = 8, πππ β 3 β€ π₯ β€ 11,
π(π₯) = 2, πππ β 8 β€ π₯ β€ 3
I. π(π₯) + π(π₯) =
II. π(π₯) β π(π₯) =
III. π(π₯) Γ π(π₯) =
IV. π(π₯) Γ· π(π₯) or π(π₯)
π(π₯)=
B. π(π₯) = π₯ β 1, for all real numbers π₯ π(π₯) = 2, for all real numbers π₯
I. π(π₯) + π(π₯) = II. π(π₯) β π(π₯) = III. π(π₯) Γ π(π₯) =
IV. π(π₯) Γ· π(π₯) or π(π₯)
π(π₯)=
V. π(4) + π(4) =
Please attempt before looking at the solutions ahead.
Functions and Relations Page 82
Practice Examples Solutions
1. Evaluate the following
A. π(π₯) = 8, πππ β 3 β€ π₯ β€ 11, π(π₯) = 2, πππ β 8 β€ π₯ β€ 3
Picture for parts I. and II.
Picture for parts III. and IV.
1. You can see from the pictures above that adding, subtracting, multiplying or dividing two functions
together is not even possible unless we restrict our domains for both functions π(π₯) and π(π₯) to the
interval β3 β€ π₯ β€ 3 (which is the intersection of the two domains). This is the interval where both
functions are defined and we can do arithmetic with the output-values of the two given function. So
domain of all the functions below is [β3,3].
I. π(π₯) + π(π₯) = 8 + 2 = 10, for β3 β€ π₯ β€ 3
II. π(π₯) β π(π₯) = 8 β 2 = 6, for β3 β€ π₯ β€ 3
III. π(π₯) Γ π(π₯) = 8 Γ 2 = 16, for β3 β€ π₯ β€ 3
IV. π(π₯) Γ· π(π₯) or π(π₯)
π(π₯)=
8
2= 4, for β3 β€ π₯ β€ 3
Functions and Relations Page 83
B. π(π₯) = π₯ β 1, for all real numbers π₯
π(π₯) = 2, for all real numbers π₯
I. π(π₯) + π(π₯) = π₯ β 1 + 2 = π₯ + 1 for all real
numbers (note that, the effect of adding a 2
to the function π₯ β 1 is all the points on the
line π¦ = π₯ β 1 shifted up 2 units)
Domain here is all real numbers
II. π(π₯) β π(π₯) = π₯ β 1 β 2 = π₯ β 3 for all real
numbers (note that, the effect of subtracting
2 from all the outputs of the function π₯ β 1 is
all that all points on the line π¦ = π₯ β 1 move
vertically down 2 units)
Domain here is all real numbers
III. π(π₯) Γ π(π₯) = (π₯ β 1)2 = 2π₯ β 2
Domain here is all real numbers
IV. π(π₯) Γ· π(π₯) or π(π₯)
π(π₯)=
2
π₯β1, domain is all real numbers not equal to 1. Since when π₯ = 1 we get zero
in the denominator. So we can see that division of functions is only possible as long as the
denominator remains non-zero.
Domain is (ββ, 1) βͺ (1,β)
V. π(4) + π(4) = 4 β 1 + 2 = 5
As you can see adding, subtracting, multiplying or dividing the outputs of two functions creates new
functions. We use the following notation for them
π(π₯) + π(π₯) = (π + π)(π₯)
π(π₯) β π(π₯) = (π β π)(π₯)
π(π₯) Γ π(π₯) = (π β π)(π₯) π(π₯)
π(π₯)= (
π
π) (π₯)
The notation of π + π, π β π, π β π,π
π are the names of the function that we get by doing the arithmetic
operations respectively. Remember the domain of these new functions is the intersection of the
domains of the individual functions involved in the operations. In the case of division we need to also be
sure to exclude any inputs (π₯) that make the denominator function = 0.
In applications, when adding or subtracting functions, the outputs of the two functions must have the
same units in order that addition is meaningful. For products of two functions, the product of the two
outputs must be meaningful too. Thus we would not multiply the Wisconsin and Minnesota population
Picture for part I. and II.
Functions and Relations Page 84
functions, be we might add them to get the total population function for both states. We could
multiply the Wisconsin population function by the Average annual income of Wisconsin residents to
obtain the total Wisconsin income as a function of the year (π‘). I.e. πππ‘ππ ππΌ πΌπππππ (π‘) =
[ππΌ πππ(π‘)] β [ππΌ ππ£πππππ πΌπππππ(π‘)].
Functions and Relations Page 85
Section1.4a Worksheet Date:______________ Name:________________________
Concept Meaning in words and/or examples as required
How is the sum function π Β± π defined? I.e. how do you compute outputs of this function?
How do you determine the domain of the sum function π Β± π ?
For the sum function π Β± π, what is necessary about the units of the output for π and for π so that π Β± π is meaningful?
How is the product function π Γ π defined? I.e. how do you compute outputs of this function?
How do you determine the domain of the product function π Γ π ?
How is the quotient function π
π defined? I.e.
how do you compute outputs of this function?
How do you determine the domain of the
quotient function π
π ?
Difficulties encountered in the section:
Functions and Relations Page 86
Exercises 1.4a
1 π(π₯) = 3π₯ β 1 and π(π₯) = π₯2 + 2 i. (π + π)(π₯) = ____________________
ii. Domain of (π + π)
iii. (π + π)(3) = ____________________
iv. (ππ)(π₯) = ____________________
v. Domain of (ππ) = ________________
vi. (ππ)(0) = ____________________
2 π(π₯) = 5π₯2 + 1 and π(π₯) = π₯ β 2
i. (π
π) (π₯) = ____________________
ii. Domain of (π
π)
iii. (π
π) (3) = ____________________
iv. (π β π)(π₯) = ____________________
v. Domain of (π β π) = ________________
vi. (π β π)(0) = ____________________
3. If π(π₯) = (2 + π₯)(β4 + π₯) and π(π₯) = (1 β π₯)(1 + π₯). Find all values that are NOT in the domain
of π
π. If there are more than one value separate them with commas.
4. EXTRA CREDIT: If π(π₯) = βπ₯ β 1 and π(π₯) = ln (π₯ β 3). Evaluate the following A. Domain of π + π
B. Domain of π
π
5. Let the number of bushels of corn produced in each county of Wisconsin at year π‘ be denoted by π΅πππ’ππ‘π¦(π‘) and the number of acres planted in corn in the county by π΄πππ’ππ‘π¦(π‘).
a. Describe what the function π΅πππ’ππ‘π¦(π‘)
π΄πππ’ππ‘π¦(π‘) represents.
b. Describe what the sum over all the counties π΅πππππ + π΅ππππ +π΅πππππ‘βππ +β― over all the counties in the state function represents.
Functions and Relations Page 87
Composition of Functions
Composition of functions Part 1 https://www.youtube.com/watch?v=q3KcPmQfACo
Composition of functions Part 2 https://www.youtube.com/watch?v=HnKUOLC_PKs
Another important way to combine two or more functions is through composition. What this means is
that we start with an input to one function, then take the output of that function an use it as the input
of another function. This could go on for several levels of composition.
Playing
If we let the function π(π₯) = π₯ β 1 and π(π₯) = βπ₯ then if we start with π₯ as the input to π and then
take the output βπ₯ and use that as the input to π we call this the composition of the functions π and π
with π acting first. The notation for this would look like:
π(π(π₯)) = π(βπ₯) = βπ₯ β 1
When evaluating a composition of functions as π(π(π₯)), we start the evaluation from the inside
working our way out. We might say here that π is the inner function and π the outer function. This
two-step process is denoted as above or by the notation π β π(π₯) which means the same as π(π(π₯)).
The domain of π β π is all those inputs of π that will produce outputs that are in the domain of π.
Composition of Functions : Given two functions π(π₯) and π(π₯), a third function called the
composite function with notation (π β π)(π₯) is a function defined by evaluating π(π(π₯)) for
appropriate values of π₯, or you can think of it as π₯πβπ(π₯)
πβπ(π(π₯)).
The domain of (π β π)(π₯) : {π₯|π₯ is in the domain of π and also π(π₯) is in the domain of π}
So you can think about (π β π)(π₯) as evaluating function of a function starting with the input of π₯,
getting an output from computing π(π₯), this output then becomes the input in the function π and the
output π(π(π₯)) is the answer to the composite function (π β π)(π₯). The domain of (π β π)(π₯) function
contains only those values of domain of π for which the output values π(π₯) are in domain of π.
Practice Examples
1. Evaluate the composite functions below. Find the domain of the composite functions.
a. Let π(π₯) = 2π₯ β 1 and π(π₯) = βπ₯. Find i. (π β π)(π₯)
π·πππππ ππ (π β π)(π₯)
ii. (π β π)(π₯)
iii. π·πππππ ππ (π β π)(π₯)
b. Let π(π₯) =2
π₯β1 and π(π₯) = π₯ + 5. Find
i. (π β π)(π₯)
ii. (π β π)(5)
iii. π·πππππ ππ (π β π)(π₯)
Functions and Relations Page 88
c. Let π(π₯) = 2π₯ β 1 and π(π₯) =π₯+1
2. Find
i. (π β π)(3)
ii. (π β π)(5)
iii. (π β π)(π₯)
iv. (π β π)(π₯)
v. πβ1(π₯)
vi. πβ1(π₯)
d. Let π(π₯) =1
π₯β1 and π(π₯) =
1
π₯+ 1. Find
i. (π β π)(6)
ii. (π β π)(2)
iii. (π β π)(π₯)
iv. (π β π)(π₯)
v. πβ1(π₯)
vi. πβ1(π₯)
e. After solving problems 1c and 1d: i. What can you say when you have two functions π and π and π(π(π₯)) = π₯ ?
ii. If π is the same function as πβ1 then what is the function πβ1 = (πβ1)β1, i.e. what is the inverse of the inverse of a function?
Functions and Relations Page 89
f. Use the graph to evaluate: i. π(1) =
ii. π(π(0)) =
iii. πβ1(β1) = iv. πβ1(0) = v. Plot the graph of π¦ = πβ1(π₯)
vi. Find a formula for π¦ = π(π₯) vii. Find a formula for π¦ = πβ1(π₯)
Practice Examples Solutions
1. Evaluate the composite functions below. Find the domain of the composite functions.
a. Let π(π₯) = 2π₯ β 1 and π(π₯) = βπ₯. Find
i. (π β π)(π₯) = π(π(π₯)) = π(βπ₯) =
2βπ₯ β 1 (see below for alternate explanation)
π₯πββπ₯
πβ2βπ₯ β 1
Domain:[0,β)
ii. (π β π)(π₯) = π(π(π₯)) = π(2π₯ β 1) =
β2π₯ β 1
π₯πβ2π₯ β 1
πββ2π₯ β 1
Domain:[1
2, β) since otherwise the end
result would not make sense.
b. Let π(π₯) =2
π₯β1 and π(π₯) = π₯ + 5. Find
i. (π β π)(π₯) = π(π(π₯)) = π(π₯ + 5) =2
π₯+5β1=
2
π₯+4 (see below for alternate
explanation) Domain of π β π requires π₯ β β4.
π₯πβπ₯ + 5
πβ
2
π₯ + 5 β 1=
2
π₯ + 4
ii. (π β π)(5) = π(π(5)) = π (2
5β1) = π (
2
4) =
π (1
2) = 5 +
1
2= 5
1
2
π₯πβ
2
5 β 1=2
4=1
2
πβ
1
2+ 5
c. Let π(π₯) = 2π₯ β 1 and π(π₯) =π₯+1
2.
Find
i. (π β π)(3) = π(π(3)) = π (3+1
2) =
π(2) = 2(2) β 1 = 4 β 1 = 3
3πβ3 + 1
2=4
2= 2
πβ2(2) β 1 = 4 β 1 = 3
Notice input was 3 and output is 3 also.
ii. (π β π)(5) = π(π(5)) = π(10 β 1) =
d. Let π(π₯) =1
π₯β1 and π(π₯) =
1
π₯+ 1. Find
i. (π β π)(6) = π(π(6)) = π (1
6+ 1) =
11
6+1β1
=11
6
= 6
You can see that π is undoing what π did to 6.
ii. (π β π)(2) = π(π(2)) = π (1
2β1 ) = π(1) =
1
1+ 1 = 2
You can see that π is undoing what π did to
Functions and Relations Page 90
π(9) =9+1
2= 5
5πβ2(5) β 1 = 10 β 1 = 9
πβ9 + 1
2=10
2= 5
Again notice input was 5 and output was 5.
iii. (π β π)(π₯) = π (π₯+1
2) = 2 (
π₯+1
2) β 1 =
π₯
iv. (π β π)(π₯) = π(2π₯ β 1) =2π₯β1+1
2= π₯
v. πβ1(π₯) π¦ = 2π₯ β 1 π₯ = 2π¦ β 1 π₯ + 1
2= π¦
πβ1(π₯) =π₯ + 1
2
vi. πβ1(π₯)
π¦ =π₯ + 1
2
π₯ =π¦ + 1
2
2π₯ = π¦ + 1 ππ 2π₯ β 1 = π¦ πβ1(π₯) = 2π₯ β 1
2.
iii. (π β π)(π₯) = π (1
π₯+ 1 ) =
11
π₯+1β1
=1
1π₯
= π₯
iv. (π β π)(π₯) = π (1
π₯β1) =
11
π₯β1
+ 1
= π₯ β 1 + 1 = π₯
v. πβ1(π₯)
π¦ =1
π₯β1 or π₯ =
1
π¦β1
π¦ β 1 =1
π₯ or π¦ =
1
π₯+ 1
πβ1(π₯) =1
π₯+ 1
vi. πβ1(π₯)
π¦ =1
π₯+ 1 or π₯ =
1
π¦+ 1
π₯ β 1 =1
π¦ or π¦ =
1
π₯β1
πβ1(π₯) =1
π₯ β 1
e. After solving problems 1c and 1d: i. What can you say when you have two functions π and π and π(π(π₯)) = π₯ ?
It seems when this happens that π and g are inverse functions of each other!
ii. If π is the same function as πβ1 then what is the function πβ1 = (πβ1)β1, i.e. what is the inverse of the inverse of a function?
The inverse of the inverse of a function is the original function!
Functions and Relations Page 91
f. Use the graph to evaluate:
i. π(1) = β1
2 , i.e., π₯ = 1 β π¦ = β1/2
ii. π(π(0)) = π(β1) = β3/2
iii. πβ1(β1) = 0 , since π¦ = β1 β π₯ = 0. iv. πβ1(0) = 2, since π¦ = 0 β π₯ = 2 v. Plot the graph of π¦ = πβ1(π₯)
vi. Find a formula for π¦ = π(π₯)
π¦ = π(π₯) = {1
2π₯ β 1 πππ β 2 β€ π₯ β€ 2
3π₯ β 6 πππ 2 < π₯ β€ 4}
vii. Find a formula for π¦ = πβ1(π₯)
π¦ = πβ1(π₯) = {
2π₯ + 2 πππ β 2 β€ π₯ β€ 01
3π₯ + 2 πππ 0 < π₯ β€ 6
}
For any one-to-one function π¦ = π(π₯), πβ1 β π (π₯) = π₯ and also π β πβ1(π₯) = π₯.
This makes sense in that if we start with π₯ in the domain of π and compute π¦ = π(π₯), then acting on this
π¦ with πβ1 will take us back to π₯ in the domain of π. Also for π β πβ1(π₯) = π₯, we start with some value
(π₯ = π) in the range of π and πβ1 takes us back to the value π = πβ1(π) in the domain of π that
produces π . Finally we act on this that π with π to get back to π₯ = π in the range of π. These
compositions just go back and forth between a domain and range value of a one-to-one function π.
π = πβπ(π)
Functions and Relations Page 92
Section1.4b Worksheet Date:______________ Name:________________________
Concept Meaning in words and/or examples as required
What is the composite of two functions?
How do you find domain and range of a composite function? Explain.
What is the result of the composition of a one-to-one function and its inverse?
Give example of a composite function that is its own inverse.
Difficulties encountered in the section:
Functions and Relations Page 93
Exercises 1.4b
1. Two functions π and π are defined in the figure below. Find the domain and range of the compositions πoπ(π₯) = π(π(π₯)), and πoπ(π₯) = π(π(π₯)). Then evaluate the function values below.
Domain of π
Domain of π
Domain of (π β π)(π₯) = π(π(π₯))
Domain of (π β π)(π₯) = π(π(π₯))
Range of π
Range of π
Range of (π β π)(π₯) = π(π(π₯))
Range of (π β π)(π₯) = π(π(π₯))
a. π β π(3)
b. π β π(9)
c. π β π(7)
d. π β π(0)
Functions and Relations Page 94
2. Consider the functions π¦ = π(π₯) and π¦ = π(π₯) which are given by their graphs below. a. Evaluate π(4) = b. Evaluate π(4) = c. Evaluate (π + π)(4) = d. Evaluate (π + π)(2) = e. Evaluate (π + π)(1) = f. Plot the graph of π¦ = (π + π)(π₯) g. Evaluate (ππ)(3) =
h. Evaluate π
π(4) =
i. Evaluate π β π(4) = j. Evaluate π β π(2) = k. Evaluate πβ1(4) = (Approximately)
Functions and Relations Page 95
4. For the real valued functionsπ(π₯) = βπ₯ + 8, and π(π₯) = π₯2 + 7 find the compositions listed below and specify domains of these functions using interval notation. Then evaluate the values of the function listed below.
(πoπ)(π₯) = π(π(π₯)) =
(πoπ)(π₯) = π(π(π₯)) =
Domain of π Domain of π
Range of π Range of π
Domain of (πoπ)(π₯) = π(π(π₯)) Domain of (πoπ)(π₯) = π(π(π₯))
Range of (πoπ)(π₯) = π(π(π₯)) Range of (πoπ)(π₯) = π(π(π₯))
e. πoπ(π₯)
f. πoπ(1)
g. πoπ(π₯)
h. πoπ(1)
3. For the real valued functions π(π₯) =π₯+6
π₯β5, and π(π₯) = 2π₯ β 7 find the compositions listed below
and specify domains of these functions using interval notation. Then evaluate the values of the function listed below.
(π β π)(π₯) = π(π(π₯)) =
(π β π)(π₯) = π(π(π₯)) =
Domain of π Domain of π
Range of π Range of π
Domain of (πoπ)(π₯) = π(π(π₯)) Domain of (πoπ)(π₯) = π(π(π₯))
Range of (πoπ)(π₯) = π(π(π₯)) Range of (πoπ)(π₯) = π(π(π₯))
a. (πoπ)(π₯)
b. (πoπ)(6)
c. (πoπ)(6)
d. (πoπ)(0)
Functions and Relations Page 96
5. For each of the pairs of functions below find π(π(π₯)) andπ(π(π₯)). Then determine whether π and π are inverses of each other. Simplify your answers as much as possible. (Assume that your expressions are defined for all π₯ in the domain of the composition. It is a good idea to write the domain of each of the functions to make sure you know what π₯ values make sense in the functions.
a. π(π₯) = 6π₯ + 3 and π(π₯) = 6π₯ β 3 π(π(π₯)) π(π(π₯))
o π and π are inverses of each other o π and π are not inverses of each other
b. π(π₯) =2
π₯ and π(π₯) =
2
π₯
π(π(π₯)) π(π(π₯))
o π and π are inverses of each other o π and π are not inverses of each other
c. π(π₯) = βπ₯ and π(π₯) = π₯2, for π₯ β₯ 0 π(π(π₯)) π(π(π₯))
o π and π are inverses of each other o π and π are not inverses of each other
d. π(π₯) = βπ₯ and π(π₯) = π₯2 π(π(π₯)) π(π(π₯))
o π and π are inverses of each other o π and π are not inverses of each other
Functions and Relations Page 97
6. Given the formula for the composite function
π β π(π₯) = π(π(π₯)) = β2π₯2 β 4 , state possible
formulas for what each of the functions π and π. π(π₯) =
π(π₯) =
7. Come up with formulas for two functions π and π so that when you
compose them, then π(π(π₯)) = π₯.
π(π₯) = π(π₯) =
8. Evaluate the following for each of the one-to-one functions given by their graphs below.
a. π(1)
b. πβ1(1)
c. π(2)
d. πβ1(β4)
e. π(πβ1(β3))
f. πβ1(π(β4))
a. π(2)
b. πβ1(0)
c. π(β5)
d. πβ1(β2)
e. π(πβ1(β2))
f. πβ1(π(β5))
Functions and Relations Page 98
9. Find the inverses of the following one-to-one functions. Then find the domains and ranges of the functions and their inverses.
a) π(π₯) =7π₯+1
2π₯β1
b) π(π₯) = β2π₯ β 1 for π₯ β₯1
2
Domain of π
Range of πβ1
Domain of π
Range of πβ1
Domain of πβ1
Range of π
Domain of πβ1
Range of π
(π β πβ1)(π₯) =
(πβ1 β π)(π₯) =
(π β πβ1)(π₯) = (πβ1 β π)(π₯) =
Functions and Relations Page 99
c) β(π₯) = 2π₯
d) π(π₯) = πππ₯
Domain of β
Range of ββ1
Domain of π
Range of πβ1
Domain of ββ1
Range of β
Domain of πβ1
Range of π
(β β ββ1)(π₯) = (ββ1 β β)(π₯) = (π β πβ1)(π₯) = (πβ1 β π)(π₯) =
Functions and Relations Page 100
10. For the function π¦ = π(π₯) graphed below, Use the graph to evaluate: a. π(1) =
b. π(π(0)) =
c. πβ1(β1) =
d. πβ1(0) =
e. Plot the graph of π¦ =
πβ1(π₯)
f. Find a formula for π¦ =π(π₯) (Piecewise defined!)
g. Find a formula for π¦ =πβ1(π₯)
11. Show that π΄ = π(π‘) = 200π0.055π‘ and π(π΄) =1
0.055ln
π΄
200 are inverses of each other by
showing that both π β π(π΄) = π΄ and also that π β π(π‘) = π‘.
Graphing Functions and Relations Page 101
Chapter 2: Graphing Functions and Relations
2.1 Library of Functions Lecture
Graphing Part 1 (14 min) https://www.youtube.com/watch?v=oxILQDhM3yM
Graphing Part 2 (15min) https://www.youtube.com/watch?v=zdO_bttea_E
Graphing Part 3 (15 min) https://www.youtube.com/watch?v=4xwIHBko9RA
In this chapter we will focus on the graphs of relations and functions. We already have some basic
understanding of specific functions from last chapter. You graphed versions of the basic polynomial,
rational, square root, absolute value, exponential and logarithmic functions. One way to graph functions
and relations is to plot enough points to give you a good sense of the main features of the graph. On the
other hand as a mathematician we can gain deeper understanding of functions and relations by
exploring their generic forms and observe how certain modifications of these forms or coefficients give
rise to predictable changes in the graphs.
Before we begin, let us review a basic library of functions that we have already encountered along with
their graphs. This is necessary so we can recognize how modifications of the basic formulas account for
changes in graphs and the other way too. This will allow us to recognize from a graph the basic type of
function formula for that graph and guide us in what the modifications of the basic equation form are.
Library of graphs of functions
1. π(π₯) = π₯
π₯ π¦ = π₯
β2 β2
β1 β1
0 0
1 1
2 2
This is an odd function with the graph being symmetric with respect to the origin. And (0,0) is the π₯-intercept and π¦-intercept.
Graphing Functions and Relations Page 102
2. π(π₯) = |π₯|
π₯ π¦ = |π₯|
β2 2
β1 1
0 0
1 1
2 2
Note that the graph of π¦ =|π₯| is symmetric with respect to the π¦-axis, i.e., π(π) = π(βπ) and π(π₯) =|π₯| is an Even Function. Also, (0,0) is the π₯-intercept and π¦-intercept.
Notice how this graph differs from the graph of π¦ = π₯ on the left. You can see that all the absolute value changed in the graph of π¦ = π₯ is to invert bottom left part and make it positive.
3. π(π₯) = π₯2
π₯ π¦ = π₯2 β2 4
β1 1
0 0
1 1
2 4
This is an even function with the graph being symmetric with respect to the π¦-axis. Also, (0,0) is the π₯-intercept and π¦-intercept.
Graphing Functions and Relations Page 103
4. π(π₯) = π₯3
π₯ π¦ = π₯3 β2 β8 β1 β1 0 0 1 1 2 8
This is an odd function with its graph symmetric with respect to the origin. And (0,0) is the π₯-intercept and the π¦-intercept.
5. π(π₯) = βπ₯
π₯ π¦ = βπ₯
0 0
1 1
4 2
9 3
This function is neither odd nor even. And (0,0) is the π₯-intercept and the π¦-intercept.
Graphing Functions and Relations Page 104
6. π(π₯) =1
π₯
π₯ π¦ =
1
π₯
π₯ π¦ =
1
π₯
β0.1 1
β0.1= β10
β10 1
β10= β0.1
β0.01 1
β0.01= β100
β100 1
β100= β0.01
β0.001 1
β0.001= β1000
β1000 1
β1000= β0. .001
0.1 1
0.1= 10
10 1
10= 0.1
0.01 1
0.01= 100
100 1
100= 0.01
0.001 1
0.001= 1000
1000 1
1000= 0. .001
This is an odd function as the graph is symmetric with respect to the origin. There are no π₯ or π¦ intercepts. The lines π¦ = 0 and π₯ = 0 are horizontal and vertical asymptotes respectively.
Graphing Functions and Relations Page 105
7. π(π₯) = ππ₯
π₯ π(π₯) = ππ₯
β1 1
π
0 1
1 π
Domain (ββ, β) Range (0, β)
8. π(π₯) = πππππ₯
π₯ π(π₯) = πππππ₯
1
π
β1
1 0
π 1
Domain (0, β) Range (ββ, β)
Playing
Now equipped with the library of graphs of functions above we try to figure out how these graphs
change when we change a few parameters or coefficients in the formulas for π. In other words, we are
trying to figure out how the graph of the function π¦ = ππ(π₯ β β) + π relates to that of π¦ = π(π₯). So
letβs play with functions below to figure out how the numbers π, β and π actually lead to predictable
modifications of the graph of π¦ = π(π₯).
Graphing Functions and Relations Page 106
Horizontal and Vertical Shift of Function Graphs
Examples
Complete the table of values for each transformation of the functions 1-4 below and plot the respective
graph of each transformation, and please make sure you label each graph as shown for 1a below. Please
use colored pencils with a different color for each parts a.-e.
1. π(π₯) = |π₯|
π₯ π¦ = |π₯|
β2 2
β1 1
0 0
1 1
2 2
a. π¦ = π(π₯) + 1 = |π₯| + 1
π₯ π¦ = |π₯| + 1
β2
β1
0
1
2
b. π¦ = π(π₯) β 1 = |π₯| β 1
π₯ π¦ = |π₯| β 1
β2
β1
0
1
2
c. π¦ = π(π₯ β 2) = |π₯ β 2|
π₯ π¦ = |π₯ β 2|
0
1
2
3
4
d. π¦ = π(π₯ + 2) = |π₯ + 2|
π₯ π¦ = |π₯ + 2|
β4
β3
β2
β1
0
e. π¦ = π(π₯ β 2) + 1 = |π₯ β 2| + 1
π₯ π¦ = |π₯ β 2| + 1
0
1
2
3
4
Observations The graph of π¦ = π(π₯ β β) + π = |π₯ β β| + π is the same shape as the graph of
π¦ = |π₯| but is shifted ______ units up/down, and shifted ________ units left/right.
Graphing Functions and Relations Page 107
2. π(π₯) = π₯2
π₯ π¦ = π₯2 β2 2
β1 1
0 0
1 1
2 2
a. π¦ = π(π₯) + 1 = π₯2 + 1
π₯ π¦ = π₯2 + 1 β2
β1
0
1
2
b. π¦ = π(π₯) β 1 = π₯2 β 1
π₯ π¦ = π₯2 β 1 β2
β1
0
1
2
c. π¦ = π(π₯ β 2) = (π₯ β 2)2
π₯ d. π¦ = (π₯ β 2)2
0
1
2
3
4
e. π¦ = π(π₯ + 2) = (π₯ + 2)2
π₯ π¦ = (π₯ + 2)2 β4
β3
β2
β1
0
f. π¦ = π(π₯ + 2) + 1 = (π₯ + 2)2 + 1
π₯ π¦ = (π₯ + 2)2 + 1 β4
β3
β2
β1
0
Observations
The graph of π(π₯) = (π₯ β β)2 + π is the same shape as the graph of π¦ = π₯2 but is shifted ______ units
up/down, and shifted ________ units left/right.
Graphing Functions and Relations Page 108
3. π(π₯) = 3π₯ π₯ π¦ = 3π₯
β2
β1
0
1
2
Horizontal Asymptote _______
a. π¦ = π(π₯) + 1 = 3π₯ + 1 π₯ π¦ = 3π₯ + 1
β2
β1
0
1
2
Horizontal Asymptote _______
b. π¦ = π(π₯) β 1 = 3π₯ β 1 π₯ π¦ = 3π₯ β 1
β2
β1
0
1
2
Horizontal Asymptote _______
a. π¦ = βπ(π₯) = β3π₯ π₯ π¦ = β3π₯
β2
β1
0
1
2
Horizontal Asymptote _______
b. π¦ = π(π₯ β 2) = 3π₯β2 π₯ π¦ = 3π₯β2 0
1
2
3
4
Horizontal Asymptote _______
c. π¦ = π(π₯ + 2) = 3π₯+2 π₯ π¦ = 3π₯+2
β4
β3
β2
β1
0
Horizontal Asymptote _______
Observations: The graph of π¦ = π(π₯ β β) + π = ππ₯ββ + π is the same shape as the graph of
π¦ = ππ₯ but is shifted ______ units up/down, and shifted ________ units left/right. The
horizontal asymptote of the graph of π(π₯) = ππ₯ββ + π is given by π¦ =_____________.
Graphing Functions and Relations Page 109
4. π(π₯) = πππ3π₯ π₯ π¦ = πππ3π₯ 1
9
1
3
1
3
9
Vertical Asymptote π₯ =___
a. π¦ = π(π₯) + 2 = πππ3π₯ + 2 π₯ π¦ = πππ3π₯ + 2 1
9
1
3
1
3
9
Vertical Asymptote π₯ =_____
b. π¦ = π(π₯) β 2 = πππ3π₯ β 2 π₯ π¦ = πππ3π₯ β 2 1
9
1
3
1
3
9
Vertical Asymptote π₯ =____
c. π¦ = βπ(π₯) = βπππ3π₯ π₯ π¦ = βπππ3π₯ 1
9
1
3
1
3
9
Vertical Asymptote π₯ =___
d. π¦ = π(π₯ β 2) = πππ3(π₯ β 2) π₯ π¦ = πππ3(π₯ β 2)
1
9+ 2
1
3+ 2
1 + 2 = 3
3 + 2 = 5
9 + 2 = 11
Vertical Asymptote π₯ =___
e. π¦ = π(π₯ + 2) = πππ3(π₯ + 2) π₯ π¦ = πππ3(π₯ + 2)
1
9β 2
1
3β 2
1 β 2 = β1
3 β 2 = 1
9 β 2 = 7
Vertical Asymptote π₯ =___
Observations:
β’ The graph of π¦ = π(π₯ β β) + π = ππππ(π₯ β β) + π is the same shape as the graph of π¦ = πππππ₯ but is shifted ______ units up/down, and shifted ________units left/right.
β’ The vertical asymptote of the graph of π¦ =ππππ(π₯ β β) + π is given by π¦ =_____.
β’ In general our observations are that the graph of π¦ = π(π₯ β β) + π is the same shape as the graph of π¦ = π(π₯) but is shifted ______ units up/down, and shifted ________units left/right.
β’ If the graph of π¦ = π(π₯) has vertical or horizontal asymptotes of the type π₯ = π, or π¦ = π, then in the new graph of π¦ = π(π₯ β β) + π the vertical asymptote of the graph is at π₯ =____________ and the horizontal asymptote of the graph is at π¦ =_____________.
Graphing Functions and Relations Page 110
Horizontal and Vertical Stretch/Compression and Reflections
Examples
Complete the table of values for each transformation of the functions 5-6 below and plot the respective
graphs of the transformations as in the previous problems, and please make sure you label each graph.
Please use colored pencils with a different color for each of the transformations b. through g.
Plot the graphs of the transformations of functions indicated below based on the original graphs
of π(π₯) = β4 β π₯2 , and π¦ = π(π₯) shown.
Function Transformation
π(π₯) = β4 β π₯2
π₯ π¦ = β4 β π₯2
β2 0
0 2
2 0
A. π¦ = 2π(π₯) = 2β4 β π₯2
π₯ π¦ = 2β4 β π₯2
β2
0
2
B. π¦ = 4π(π₯) = 4β4 β π₯2
π₯ π¦ = 4β4 β π₯2
β2
0
2
C. π¦ =1
2π(π₯) =
1
2β4 β π₯2
π₯ π¦ =
1
2β4 β π₯2
β2
0
2
D. π¦ =1
4π(π₯) =
1
4β4 β π₯2
π₯ π¦ =
1
4β4 β π₯2
β2
0
2
E. π¦ = π(2π₯) = β4 β 4π₯2
π₯ π¦ = β4 β 4π₯2
β1
0
1
F. π¦ = π (1
2π₯) = β4 β
π₯2
4
π₯
π¦ = β4 βπ₯2
4
β4
0
4
G. = π (1
4π₯) = β4 β
π₯2
16
π₯
π¦ = β4 βπ₯2
16
β8
0
8
Graphing Functions and Relations Page 111
Observations: The graph of π¦ = ππ(π₯) = πβ4 β π₯2 has a similar shape as the graph of π¦ =
π(π₯) = β4 β π₯2 but is stretched by a factor of ______ in the _____ direction when π > 0.
The graph of π¦ = ππ(π₯) = πβ4 β π₯2 is the same shape as the graph of π¦ = π(π₯) = β4 β π₯2
but is stretched by a factor of ______ in the _____ direction and reflected across ___-axis,
when π < 0.
5. π¦ = π(π₯)
π₯ π¦ = π(π₯)
0 0
1 1
2 0
3 β1
4 0
a. π¦ = 2π(π₯)
π₯ π¦ = 2π(π₯)
0
1
2
3
4
b. π¦ = 3π(π₯)
π₯ π¦ = 3π(π₯)
0
1
2
3
4
c. π¦ = βπ(π₯)
π₯ π¦ = βπ(π₯)
0
1
2
3
4
d. π¦ = β2π(π₯)
π₯ π¦ = β2π(π₯)
0
1
2
3
4
e. π¦ =1
2π(π₯)
π₯ π¦ =
1
2π(π₯)
0
1
2
3
4
f. π¦ = π(1
2π₯)
π₯ π¦
= π(1
2π₯)
0
1
2
3
4
Graphing Functions and Relations Page 112
Observations: The graph of π¦ = ππ(π₯) has a similar shape as the graph of π¦ = π(π₯) but is
stretched by a factor of ______ in the _____ direction when π > 0.
The graph of π¦ = ππ(π₯) is the same shape as the graph of π¦ = π(π₯) but is stretched by a factor
of ______ in the _____ direction and reflected across the ___-axis, when π < 0.
6. π(π₯) = βπ₯
π₯ π¦ = βπ₯
0 0
1 1
4 4
a. π¦ =1
2π(π₯) =
1
2βπ₯
π₯ π¦ =
1
2βπ₯
0
1
4
b. π¦ =1
4π(π₯) =
1
4βπ₯
π₯ π¦ =
1
4βπ₯
0
1
4
c. π(π₯) = ββπ₯
π₯ π¦ = βπ₯
0
1
4
d. π¦ = β2π(π₯) = β2βπ₯
π₯ π¦ = β2βπ₯ 0
1
4
e. π¦ = β1
2π(π₯) = β
1
2βπ₯
π₯ π¦ = β
1
2βπ₯
0
1
4
Graphing Functions and Relations Page 113
Observations: The graph of π¦ = ππ(π₯) = πβπ₯ is the same shape as the graph of π¦ = βπ₯ but is
streched______ units _______________ when 0 < π < 1.
The graph of π¦ = βπ(π₯) = ββπ₯ is the same shape as the graph of π¦ = βπ₯ but is reflected
across the ___-axis.
The graph of π¦ = ππ(π₯) = πβπ₯ is the same shape as the graph of π¦ = βπ₯ but is reflected
across the ___-axis and streched______ units _______________ when π < β1.
The graph of π¦ = ππ(π₯) = πβπ₯ is the same shape as the graph of π¦ = βπ₯ but is reflected
across the ___-axis and streched______ units _______________ when β1 < π < 0.
7. π¦ = π(π₯) π₯ π¦ = π(π₯)
0 0
1 1
2 0
3 β1
4 0
a. π¦ =1
2π(π₯)
π₯ π¦ = 2π(π₯)
0
1
2
3
4
b. π¦ = π(βπ₯) π₯ π¦ = βπ(π₯)
0
β1
β2
β3
β4
c. π¦ = π(2π₯)
π₯ π¦ = π(2π₯)
0
1
2
1
3
2
2
d. π¦ = π (1
2π₯)
π₯ π¦ = π (
1
2π₯)
0
2
4
6
8
e. π¦ = π (β1
2π₯)
π₯ π¦ = π (β
1
2π₯)
0
β2
β4
β6
β8
Graphing Functions and Relations Page 114
Observations: The graph of π¦ = ππ(π₯) is the same shape as the graph of π¦ = π(π₯) but is
stretched______ units ______________ when 0 < π < 1.
The graph of π¦ = π(ππ₯) is the same shape as the graph of π¦ = π(π₯) but is stretched______
units _______________ when π > 1.
The graph of π¦ = π(ππ₯) is the same shape as the graph of π¦ = π(π₯) but is stretched____ units
____________ and reflected across ___-axis, when 0 < π < 1.
The graph of π¦ = π(βπ₯) is the same shape as the graph of π¦ = π(π₯) but is reflected across
___-axis.
You probably noticed by doing examples above that with the vertical and horizontal stretch
transformations, when the factor π is negative, the stretch also induces a reflection in the direction of
the stretch. Thus π¦ = β1
2βπ₯ stretches all the π¦-coordinates from the graph of π¦ = βπ₯ by a factor of
1
2
and then also makes them negative. Thus π¦ = ππ(π₯) involves a stretch and a reflection across the
horizontal axis. Likewise the graph of π¦ = π(ππ₯) is obtained by horizontally stretching by the factor 1
π
and when π is negative, it also reflects the graph across the π¦-axis.
For the graph of π¦ = π(π₯) that we plotted in the examples 6 and 8 above, the graphs of π = π(ππ) and
of π = π (βπ
ππ) are given below in case you struggled with it. Note that the graph of π (β
1
2π₯)
stretches the graph of π¦ = π(π₯) by a factor of 1
1/2= 2, and reflects across the π¦-axis.
Graphing Functions and Relations Page 115
Once you do enough examples and find the patterns. We must investigate why is the effect of the
constants in the different parts of π¦ = π(π₯ + β) + π what they are.
Another way to write π¦ = π(π₯ + β) + π is π¦ β π = π(π₯ + β). If a coordinate (π, π) is on the original
graph π¦ = π(π₯), that means π(π) = π in the new graph then the π¦ β π = π(π₯ + β), we must have
π β β as our input and the output must be π + π. That would give us π + π β π = π(π β β + β) giving
us π = π(π). So on the new graph the coordinate must be (π β β, π + π)explaining why the graph
translates up/down π units (up if π > 0, and down if π < 0), and left/right β units (left if β > 0, and
right if β < 0).
A similar argument can tell us what happens to the functions π¦ = ππ(π₯) and π¦ = π(ππ₯) (for all π β 0).
If (π, π) is a point on the original π¦ = π(π₯) graph, again π(π) = π. Then the coordinate (π, ππ) would be
on the new graph. Making the transformed graph stretch vertically since the π¦-coordinate is getting
multiplied by a factor of π (if π < 0 the reflection across the π₯-axis is predicted along with the vertical
stretch). Now in π¦ = π(ππ₯) if the π₯ =π
π we get π (π β
π
π) = π(π) = π so (
π
π, π) is a coordinate on the
graph of π¦ = π(ππ₯) this explains the fact that it is a horizontal stretch (if π < 0 the graph will also have
reflection across the π¦-axis).
Practice Problems
1. Identify the standard (original) functions that was transformed to the equations below. Explain what
kind of transformation results in each of the functions below compared to their standard functions.
Sketch the graphs of the original and the transformed function on the graph.
a. π¦ = 3|π₯|
b. π¦ = (π₯
3)
2= (
1
3π₯)
2 or viewed another way π¦ =
1
9π₯2
c. π¦ = β3βπ₯
d. π¦ = ββ1
2π₯
Solutions
Graphing Functions and Relations Page 116
a. π¦ = 3|π₯| π = π|π|, This is a vertical stretch by a factor of 3 of the π¦ = |π₯| graph. Its graph is the V-shaped graph to the right.
b. π¦ = (π₯
3)
2= (
1
3π₯)
2 or viewed another way
π¦ =1
9π₯2
As π = (π
π π)
π, this can be viewed as a
horizontal stretch by a factor of 1
1/3= 3 of the
standard graph π¦ = π₯2. Alternatively, viewed
as π¦ =1
9 π₯2 this can be viewed as a vertical
stretch by a factor of 1/9 of the graph of π¦ =π₯2. Its graph is the parabola graph to the right.
c. π¦ = β3βπ₯
π = βπβπ, This is a vertical stretch by a factor of three and also reflection across the
π₯-axis of the graph of π¦ = βπ₯.
Graphing Functions and Relations Page 117
d. π¦ = ββ1
2π₯
π = ββπ
ππ, This is a
horizontal stretch by 1
1/2= 2
and reflection across the π¦-
axis of the graph of π¦ = βπ₯.
Combination of Translations, Stretches and Reflections
By using a combination of these transformation methods, we are able to move, stretch and reflect our
basic function shapes anywhere in the coordinate plane. Four parameters will determine the location
and stretch/reflection of the new graph relative to the standard graph of any function type as indicated
below.
Function Transformation Summary:
The transformation of a function π¦ = π(π₯) to π¦ = ππ(π(π₯ β β)) + π
corresponds to transforming the graph of π¦ = π(π₯) by:
1. A vertical stretch by a factor of π (if π is negative, in addition we will have a vertical reflection
across the π₯-axis too)
2. A horizontal stretch by a factor of 1
π (if π is negative, in addition we will have a reflection
across the π¦-axis too)
3. A vertical shift of π units
4. A horizontal shift of β units
Note that the order of operations requires we do parenthesis first, then followed with multiplication
before addition or subtraction. That means we start with start by doing the stretch/reflections when
transforming a standard function graph. We need to know our standard function graphsβ main features
in order that we can apply these transformations correctly. The standard functions we should be
familiar with were discussed in at beginning of this section when we listed the library of functions, which
include: π¦ = π₯, π¦ = π₯2, π¦ = π₯3, π¦ =1
π₯, π¦ = βπ₯, π¦ = ππ₯ , π¦ = loga π₯ , π¦ = |π₯|, π¦ =
1
π₯. If you are
unsure of the general shape get some points to help solidify your mental imagery of the basic shapes.
Examples:
1. Describe how each function is a transformation of a basic function and sketch its graph.
a. π¦ =1
2(π₯ + 4)2 β 5
b. π¦ = β3βπ₯ β 2 β 1
c. π¦ = β2 β 2π₯β3 + 7
Graphing Functions and Relations Page 118
d. π¦ = log4(π₯ β 1) + 3
e. π¦ = β21
π₯+4+ 5
2. Obtain a formula for each of the functions a-d shown in the graphs below as a transformation of one
of the standard functions listed below. Explain your reasoning.
π¦ = π₯, π¦ = π₯2, π¦ = π₯3, π¦ =1
π₯, π¦ = βπ₯, π¦ = ππ₯ , π¦ = loga π₯ , π¦ = |π₯|, π¦ =
1
π₯.
a.
b.
c.
d.
Solutions
1. Describe how each function is a transformation of a basic function and sketch its graph.
a. π¦ =1
2(π₯ + 4)2 β 5
The standard function is π¦ =π(π₯) = π₯2 and this function
is: π¦ =1
2π(π₯ + 4) β 5. It is a
β’ Vertical stretch by Β½ β’ Horizontal shift of 4 to the
left, β’ Vertical shift of 5 down
Graphing Functions and Relations Page 119
b. π¦ = β3βπ₯ β 2 β 1 The standard function is π¦ =
π(π₯) = βπ₯ and this function is: π¦ = β3π(π₯ β 2) β 1. It is a
β’ Vertical stretch by 3 β’ A reflection across the π₯-
axis β’ Horizontal shift of 2 to the
right β’ Vertical shift of 1 down
c. π¦ = β2 β 2π₯β3 + 7 The standard function is π¦ =π(π₯) = 2π₯ and this function is: π¦ = β2π(π₯ β 3) + 7.
β’ Vertical stretch by 2 β’ Reflection across the π₯-axis β’ Horizontal shift of 3 to the
right β’ Vertical shift of 7 up β’ Horizontal Asymptote is at
π¦ = 7
d. π¦ = log4(π₯ β 1) + 3 The standard function is
π¦ = π(π₯) = log4 π₯ and this function is: π¦ = π(π₯ β 1) + 3. β’ Horizontal shift of 1 to the
right β’ Vertical shift 3 up. β’ Vertical asymptote is at
π₯ = 1
Graphing Functions and Relations Page 120
e. π¦ = β21
π₯+4+ 5
The standard function is π¦ =
π(π₯) =1
π₯ and this function is:
π¦ = β2π(π₯ + 4) + 5. β’ Vertical stretch by 2 β’ Reflection across the
π₯-axis. β’ Horizontal shift of 4
to the left β’ Vertical shift of 5 up β’ Asymptotes are: π₯ =
β4 & π¦ = 5.
2.
a. Original graph is π¦ = |π₯|. The graph is shifted to the left 3, up 4 and stretched
vertically by a 1
2 . So the transformed graph
would be π¦ =1
2|π₯ + 3| + 4
b. Original graph is π¦ = π₯3. The graph is shifted to the right 3, up 4 and a vertical
stretch by a 1
2. So the transformed graph
would be π¦ =1
2(π₯ β 3)3 + 4
c. Original graph is π¦ = βπ₯. The graph is shifted right 2, down 3, reflected across the π₯-axis and a vertical stretch of 2. So the transformed graph would be π¦ = β2βπ₯ β 2 β 3
d. Original graph is π¦ =1
π₯. The graph is
shifted left 2, down 3. The vertical asymptote is at π₯ = β2 and horizontal asymptote is π¦ = β3. So the
transformed graph would be π¦ =1
π₯+2β
3
Graphing Functions and Relations Page 121
Section2.1 Worksheet Date:______________ Name:__________________________
Concept Meaning in words and/or examples as required
1. Vertical Shift
2. Horizontal Shift
3. Vertical Stretch
4. Horizontal Stretch
5. Reflection across π-axis
6. Reflection across y-axis
7. Transformation of a function
Difficulties encountered in the section:
Graphing Functions and Relations Page 122
Exercises 2.1
1. Please fill in the blanks below.
a. The graph of the function π¦ = π(π₯) + π is shifted ______________________________ from the
graph of the original function π¦ = π(π₯).
b. The graph of the function π¦ = π(π₯) β π is shifted ______________________________ from the
graph of the original function π¦ = π(π₯).
a. Statements 1 and 2 can also be written as statements 3 and 4 below.
c. The graph of the function π¦ β π = π(π₯) is shifted ______________________________ from the
graph of the original function π¦ = π(π₯).
d. The graph of the function π¦ + π = π(π₯) is shifted ______________________________ from the
graph of the original function π¦ = π(π₯).
e. The graph of the function π¦ = π(π₯ β β) is shifted ______________________________ from the
graph of the original function π¦ = π(π₯).
f. The graph of the function π¦ = π(π₯ + β) is shifted ______________________________ from the
graph of the original function π¦ = π(π₯).
g. The graph of the function π¦ = ππ(π₯), π > 1 is _________________________ from the graph of
the original function π¦ = π(π₯).
h. The graph of the function π¦ = ππ(π₯), 0 < π < 1 is _________________________ from the
graph of the original function π¦ = π(π₯).
i. The graph of the function π¦ = βπ(π₯) is _____________________________________________
from the graph of the original function.
j. The graph of the function π¦ = π(βπ₯) is _____________________________________________
from the graph of the original function.
k. For quadratic functions, completing the squares allows you to find the vertex of the parabola. So
if we had the function π¦ β π = π(π₯ β β)2, the point (β, π) is referred to as the ______________
of the parabola.
l. For quadratic functions, completing the squares allows you to find the vertex of the parabola. So
if we had the function π¦ + π = π(π₯ + β)2, the point (ββ, βπ) is referred to as the
______________ of the parabola.
m. An even function is symmetric with respect to the ______________.
n. An odd function is symmetric with respect to the ______________________.
Graphing Functions and Relations Page 123
2. Please find the following for all functions below.
I. Sketch the graph of the functions and relations below. Explain clearly how you decided the
graph was the shape you drew. Show all relevant parts of the graph.
II. Find the domain and range of the functions and relations that you graphed based on your graph.
a. π¦ = |π₯| + 5
b. π¦ = |π₯| β 5
c. π¦ = |π₯ + 3|
d. π¦ = |π₯ β 3|
e. π¦ = |π₯ β 3| + 5
f. π¦ = |π₯ + 3| + 5
g. π¦ = |π₯ β 3| β 5
h. π¦ = |π₯ β 3| β 5
i. π¦ = 2|π₯|
j. π¦ = β2|π₯|
k. π¦ = 2|π₯| + 5
l. π¦ = β2|π₯| + 5
m. π¦ = 2|π₯ β 3|
n. π¦ = β2|π₯ + 3| β 5
o. π¦ = βπ₯ β 4 β 5
p. π¦ = ββπ₯
q. π¦ = 2βπ₯ β 4 β 5
r. π¦ = 2ββπ₯ β 4 β 5
s. π¦ = β2ββπ₯ + 4 β 5
t. π¦ = π₯2 + 3
u. π¦ = β2π₯3
v. π¦ = βπ₯4+1
w. π¦ = 3(π₯ β 1)2 β 4
x. π¦ = |2π₯| β 1
y. π¦ = 2βπ₯ + 5 β 4
z. π¦ β 5 = β3(π₯ β 1)2
aa. π¦ = β1
2(π₯ β 4)2 + 5
bb. π¦ = ββ1
2(π₯ + 2) β 1
Graphing Functions and Relations Page 124
3. Please find the following for all functions below.
I. Sketch the graph of the functions and relations below. Explain clearly how you decided the
graph was the shape you drew. Show all relevant parts of the graph.
II. Find the domain and range of the functions and relations that you graphed based on your
graph.
III. List all the vertical and horizontal asymptotes if any.
a. π¦ = 2π₯ + 3
b. π¦ = 2π₯ β 3
c. π¦ = 2π₯+3
d. π¦ = 2π₯β3
e. π¦ = ππ₯ + 3
f. π¦ = ππ₯ β 3
g. π¦ = ππ₯+3
h. π¦ = ππ₯β3
i. π¦ = 2π₯β1 + 3
j. π¦ = ππ₯β2 β 3
k. π¦ = (1
2)
π₯+1β 3
l. π¦ = 2ππ₯β1 + 3
m. π¦ = 2πβπ₯ + 1
n. π¦ = 300 (1
2)
π‘
4
o. π¦ =1
2πβ2π₯ + 1
p. π¦ = 300 (1
2)
π‘
4
q. π¦ = log(π₯) + 3
r. π¦ = πππ2(π₯)
s. π¦ = β3 log(π₯)
t. π¦ = ln(π₯ + 1)
u. π¦ = β log2(π₯ + 3) + 4 v. π¦ = 2 ln(βπ₯) + 5 w. π¦ = β2 log(π₯ β 1) + 4
x. π¦ =1
π₯β2
y. π¦ =1
π₯β2 + 5
z. π¦ = β 3
π₯β2β 5
Graphing Functions and Relations Page 125
4. Write a formula for each of the functions a-d shown in the graphs below so that they are a
transformation of one of the standard functions listed below. Explain your reasoning.
π¦ = π₯, π¦ = π₯2, π¦ = π₯3, π¦ =1
π₯, π¦ = βπ₯, π¦ = ππ₯ , π¦ = loga π₯ , π¦ = |π₯|, π¦ =
1
π₯.
a.
b.
c.
d.
e.
f.
Graphing Functions and Relations Page 126
5. For each original graph π¦ = π(π₯) please graph the transformations, and label all relevant parts of the new graph appropriately.
a. π¦ = π(βπ₯) b. π¦ = βπ(π₯) c. π¦ = π(π₯ + 3) + 2 d. π¦ = π(βπ₯ + 2) β 3
e. π¦ = β1
2π(π₯)
a. π¦ = π(π₯ + 1) + 3 b. π¦ = βπ(βπ₯)
c. π¦ =1
2π(π₯)
d. π¦ β 2 = π(βπ₯ β 1)
a. π¦ = π(βπ₯ β 3) + 1 b. π¦ = βπ(π₯) c. π¦ = βπ(π₯) d. π¦ = 2(π(π₯) β 1)
Graphing Functions and Relations Page 127
a. π¦ = π (βπ₯) β 3 b. π¦ = βπ (π₯) + 2 c. π¦ = π (π₯ β 1) d. π¦ = π (π₯) + 2
6. Sketch the graph of the function below. Please show all your work and clearly show relevant
points.
π¦ = π(π₯) has the graph below, use that to find the graph of π¦ =1
3π(βπ₯).
7. Given the two graphs below, find a algebraic relationship between the two so that one function
is the result of a transformation of the other.
In other words, write a formula connecting π(π₯) and π(π₯).
Graphing Functions and Relations Page 128
8. The graph of a function π is shown. Use it to sketch the graph of π¦ = βπ(π₯ + 1) β 2 on the
same axes. Show all your intermediary transformations using colored pens and label all the
transformations.
9. Investigate the difference between the graphs of the following functions. What is the conclusion you draw from it? You can use a graphing utility to explore these relations.
a. π¦ = π₯2 + 2
b. π¦ = π₯2 + 2.5
c. π¦ = π₯2 + 2.7
d. π¦ = π₯2 + 3
e. π¦ = π₯2 + 5
f. π¦ = π₯2 β 2
g. π¦ = π₯2 β 2.5
h. π¦ = π₯2 β 2.7
i. π¦ = π₯2 β 3
j. π¦ = π₯2 β 5
k. π¦ = π₯2 + π₯
Graphing Functions and Relations Page 129
2.2 Graphing Conic Sections
Conic Sections Part 1 (17 min) https://www.youtube.com/watch?v=UZlB9Bs8hQg
Conic Section Part 2 (12 min) https://www.youtube.com/watch?v=ofZn7v1jJ0U
Part 3 Equations of Circles (10 mins) http://www.youtube.com/watch?v=fzNXmoCHRCk
Conic Sections: Conic sections are the curves formed by the intersection of a plane and a cone. A
mathematical cone can be imagined as two ice-cream cones attached at their tips as shown below. The
angle that the intersecting plane makes with the axis of the cone determines which of three basic types
of conic curve you get. These are among the earliest curves to be studied back to ~350 B.C. Their initial
definition in terms of a cone and a plane precedes the development of algebra by nearly 2000 years.
Since then, several equivalent ways to define conic curves have been discovered. The βLocusβ definition
below is one, while we will mostly work with graphs of algebraic equations in π₯ and π¦ to define conic
curves.
A conic is the "Locus" (which is another word for "pathβ) of a point π in the plane so that its
distance from a fixed point πΉ (called the βfocusβ) in the plane has a constant ratio βπβ (called
the βeccentricityβ) to its distance from a fixed straight line π (called the βdirectrixβ) in the plane.
β’ π = 1 the conic is called a parabola
β’ π < 1 the conic is called an ellipse
β’ π > 1 the conic is called a hyperbola.
Mathematical Cone
Intersections of a plane and a cone.
Graphing Functions and Relations Page 130
2.2a Parabolas
Parabola: A parabola is collection of all points equidistant from a fixed point (focus) and a fixed
line (directrix).
Consider a fixed point πΉ and line πΏ as shown below. The locus or collection of all points equidistant from
the fixed point focus and the fixed line Directrix are shown below. You may notice this shape as you
have seen it before in form of quadratic function in one variable.
The distance ππΏ and ππΉ are the same in a parabola. You can see from the picture that the collection of
all points (shown in red here) form the parabola. In the picture above, the vertex is at some point (β, π),
the focus is above this π units at (β, π + π), and the directrix is π units below the vertex at π¦ = π β π.
The point π at (π₯, π¦) represents a general point on the parabola and thus the distance from π to the
focus at (β, π + π) must be equal to the distance straight down to the directrix at π¦ = π β π.
The distance ππΏΜ Μ Μ Μ is the difference in the π¦-coordinates from π at (π₯, π¦) to the line π¦ = π β π, or
(π¦ β (π β π)).
The distance ππΉΜ Μ Μ Μ from π at (π₯, π¦) to πΉ at (β, π + π) is obtained by the distance formula or Pythagorean
Theorem by (ππΉΜ Μ Μ Μ )2 = (π₯ β β)2 + (π¦ β (π + π))2
Squaring the distance (ππΏΜ Μ Μ Μ )2 = (π¦ β (π β π))2
and setting this equal to
(ππΉΜ Μ Μ Μ )2 = (π₯ β β)2 + (π¦ β (π + π))2
we get: (π¦ β (π β π))2
=(π₯ β β)2 + (π¦ β (π + π))2
.
Rearranging to (π¦ β π + π)2 β (π¦ β π β π)2 = (π₯ β β)2 and factoring the left side as a difference of
squares, we obtain:
[ (π¦ β π + π) β (π¦ β π β π) ] [ (π¦ β π + π) + (π¦ β π β π) ] = 2π β 2(π¦ β π) = (π₯ β β)2
4π(π¦ β π) = (π₯ β β)2 ππ π¦ =1
4π(π₯ β β)2 + π
You may notice that this is the graph of π¦ = π₯2 stretched vertically by the factor 1
4π and translated to the
vertex of the parabola (β, π). Thus the π¦ = π₯2 graph is a parabola!
Graphing Functions and Relations Page 131
Putting the focus below the directrix is accomplished by letting π < 0 which causes the vertical stretch
factor to be negative and thus the parabola opens down.
Playing
As a mathematician we would logically ask the questions about what changes might occur if we make
the directrix vertical? As we might predict, this simply reverses the roles of π₯ and π¦ in the form of the
equation and we get π₯ =1
4π(π¦ β π)2 + β. This implies that the parabolas facing up/down and left/right
with vertex at (β, π) and distance from vertex to focus equal to π can be expressed in Vertex Form as
one of the equations: π¦ =1
4π(π₯ β β)2 + π ππ π₯ =
1
4π(π¦ β π)2 + β respectively.
In general then,
π¦ =1
4π(π₯ β β)2 + π , for π > 0
π¦ =1
4π(π₯ β β)2 + π , for π < 0
π₯ =
1
4π(π¦ β π)2 + β, for π > 0
π₯ =1
4π(π¦ β π)2 + β, for π < 0
Graphing Functions and Relations Page 132
Example
1. Consider the parabola with equation π¦ =1
8(π₯ β 2)2 β 3. Find the focus, directrix, and the vertex of
this parabola and sketch the graph.
Solution: We see the vertex is shifted 2 Right and 3 down and 1
4π=
1
8 or π = 2 > 0 which means the
parabola faces up. The directrix is π units below the vertex at π¦ = β5. The focus is π units above the
focus at (2, β1).
Reflective Property of Parabola: One of the properties of a parabola is that if an energy source
located far away transmits energy and the parabola is made of a reflective surface the energy
bounces of the parabola and merges onto the focus. Similarly if an energy source is located at
the focus it will bounce off the parabola along the lines perpendicular to the point where it hits
the parabola.
This reflective property use can be seen in satellite dishes, telescopic mirrors, car headlights, and search
beams, making parabolas useful in real life in many different domains.
Graphing Functions and Relations Page 133
All quadratic equations with only one of π₯ or π¦ being squared and the other to the first power can be
transformed into one of the standard Vertex-Forms of a parabola. Thus we recognize these curves from
their equations by the presence of either π₯ or π¦ being squared. When there are both quadratic and
linear terms present, we complete the square to put the equation in vertex form.
Equation of an up/down opening parabola in vertex form is given by: π¦ = π(π₯ β β)2 + π
Where (β, π) = Vertex of the parabola, π determines the direction of the parabola (up, or down) and contributes the width of the parabola. Also from above,
π =1
4π where π is the distance from the vertex to the
focus. β’ If π > 0, the parabola opens up. β’ If π < 0, the parabola opens down. β’ If |π| > 1, the parabola is vertically stretched or is
narrower. β’ If |π| < 1, the parabola is vertically compressed, or
is wider. β’ The graph moves up/down π units as π₯ moves one
unit right or left from the vertex.
Equation of a left/right opening parabola in vertex form is given by: π₯ = π(π¦ β π)2 +h
Where (β, π) = Vertex of the parabola and π determines the direction of the parabola (left or right) and impacts the width of the parabola. The distance
from the vertex to focus is π and π =1
4π
β’ If π > 0, the parabola opens to the right. β’ If π < 0, the parabola opens to the left. β’ If |π| > 1, the parabola is narrower. β’ If |π| < 1, the parabola is wider. β’ The graph moves left/right π units as π¦ moves one
unit up or down from the vertex.
Graphing Functions and Relations Page 134
Example
1. Sketch the graph of π¦ = 5π₯2 + 3π₯ + 2
Solution: We will try to get the equation in the standard form which will help in the graphing and
finding the vertex. Use completing the square process.
π¦ = 5 (π₯2 +3
5π₯) + 2
To complete the square, add half of the middle term squared and subtract the same so the net
result is no change on the right side.
π¦ = 5 (π₯2 +3
5π₯ + (
3
2(5))
2
) + 2 β 5 (3
2(5))
2
π¦ = 5 (π₯ +3
10)
2
+ 111
20
Vertex is at (β3
10, 1
11
20)
π π
β6
10
2
Vertex β3
10 31
20
0 2
Graphing Functions and Relations Page 135
In general, for any quadratic function in standard form y = aπ₯2 + ππ₯ + π we can complete square
and get the equation into vertex-form. If we apply this to a quadratic in π₯ in standard form, we
obtain general results as to where the vertex and focus are located in terms of the coefficients π, π,
and π.
π¦ = aπ₯2 + ππ₯ + π
π¦ = a (π₯2 +π
ππ₯) + π add the square of half of the π₯-coefficient and also subtract this outside the ().
π¦ = a (π₯2 +π
ππ₯ + (
π
2π)
2
) + π β π (π
2π)
2
π¦ = a (π₯ +π
2π)
2
+ π βπ2
4π
Vertex is at (βπ
2π, π β
π2
4π). Recall that β
π
2π is the first part of the quadratic formula. This is very useful
in that βπ
2π very quickly locates the π₯-coordinate of the vertex. It is perhaps easiest to obtain the π¦-
coefficient by plugging π₯ = βπ
2π into the original formula. Two additional points can be obtained by
moving one unit in either direction from the vertex and π¦ will move π units. If one needs to locate the
focus, we have that π =1
4π which solved for π gives the focus-vertex distance as π =
1
4π .
The graph of π¦ = π(π₯) = ππ₯2 + ππ₯ + π, is a parabola with vertex at π₯ = βπ
2π and focus a
distance of π =1
4π from the vertex. The π¦-coordinate of the vertex is at π (β
π
2π).
When π > 0, the function output at the vertex, π¦ = π (βπ
2π), is the minimum value of the function.
If π < 0, the parabola opens downward and π¦ = π (βπ
2π) is the maximum value of the function.
Quadratic functions are used to model many situations where either a maximum or minimum output of
the function is desired. Some examples include finding optimum revenue by controlling the number of
items that are to be sold and determining the maximum height of an object that is thrown in the air.
For parabolas opening left/right we can solve for π₯ = ππ¦2 + ππ¦ + π with π β 0. The vertex is
located at = π = βπ
2π . Plug this π¦ into the equation to get: = π (β
π
2π)
2
+ π (βπ
2π) + c.
The focus is π =1
4π units horizontally from the vertex. (Opens right if π > 0, left if π < 0.) Also,
the stretch factor π is how far left or right we move from the vertex π₯-value, as π¦ moves up or
down one unit from the vertex value.
Graphing Functions and Relations Page 136
Practice Problems
1. For the problems below find the vertex, and the focus. Then sketch the graph of the function.
a. π¦ =1
3π₯2 β 8π₯ + 2
b. π¦ = β0.05π₯2 + π₯ + 2
c. π₯ = β2π¦2 β 12π¦ β 14
d. π₯ =1
12π¦2 β π¦ + 5
2. Find the equation of the parabola whose graph is given below.
3. Find an equation of the parabola that goes through the points (3, 3), (-1, 3), and (1, 5).
Solutions
1.
a. π¦ =1
3π₯2 β 8π₯ + 2
Vertex is at π₯ =8
2/3= 12, π¦ = β46
Focus is at π =1
4/3=
3
4 up from vertex.
b. π¦ = β0.05π₯2 + π₯ + 2
Vertex is at π₯ = β1
2(β0.05)= 10, π¦ = 7
Focus is at π =1
4(β0.05)= β5; 5 down from
vertex. Note that π¦ moves π units as π₯ goes one in either direction from the vertex. Here as π₯ moves from 10 to 11 or to 9, y moves from 7 down 0.05 units to 6.95.
Graphing Functions and Relations Page 137
c. π₯ = β2π¦2 β 12π¦ β 14
Vertex is at π¦ =12
β4= β3, π₯ = 4
Focus is at =1
β8 ;
1
8 left from vertex.
d. π₯ =1
12π¦2 β π¦ + 5
Vertex is at π¦ =1
1/6 = 6, π₯ = 2
Focus is at π =1
1/3 = 3; 3 right of
vertex.
2. Find the equation of the parabola whose graph is given below.
The vertex is at (8, β2) and π¦ goes up ΒΌ unit as π₯
moves one unit left/right from π₯ = 2, so π = 1/4
and π¦ =1
4(π₯ β 8)2 β 2
3. Find an equation of the parabola that goes through the points (3, 3), (-1, 3), and (1, 5).
First plot the points.
From the points and symmetry, the point ( 1,5) must be the vertex. Also we move
down 1
2 units when π₯ moves 1 left/right, so
π = β1
2 and π¦ = β
1
2(π₯ β 1)2 + 5.
Graphing Functions and Relations Page 138
Section2.2a Worksheet Date:______________ Name:__________________________
Concept Meaning in words and/or examples as required
1. What is a parabola?
2. What are all the different forms in which we can write equation of a parabola?
3. What are the different kinds of parabolas?
4. What equation forms are possible for a parabola with vertex at (π, βπ). There is one parameter π that is not given.
π₯ = Or π¦ =
5. What is the maximum value of a function that is quadratic?
6. What is the minimum value of a quadratic function?
Difficulties encountered in the section:
Graphing Functions and Relations Page 139
Exercises 2.2a
1. Sketch the graphs of the following relations. Find the vertex of the parabola and other relevant
information if asked.
a. π¦ = 2(π₯ β 3)2 β 5 b. π¦ = β2(π₯ β 3)2 β 5 c. π₯ = 2(π¦ β 3)2 β 5 d. π₯ = β2(π¦ β 3)2 β 5 e. π¦ = 3(π₯ + 1)2 + 2 f. π₯ = β3(π¦ + 1)2 + 2 g. π₯2 β 4π₯ + 1 = π¦
(Hint: Use completing the squares)
h. βπ₯2 β 4π₯ + 1 = π¦ (Hint: Use completing the squares)
i. 2
3π₯2 β 4π₯ +
1
3= π¦
j. β2
3π¦2 β 4π¦ β
1
3= π₯
k. π¦ = 4π₯2 l. π¦ = 4(π₯ β 1)2 β 4 m. π¦ = 4π₯2 β 8π₯ n. π₯2 + 8π₯ + 3π¦ + 22 = 0 o. 2π¦2 β 16π¦ β π₯ + 29 = 0 p. 3π₯2 β 5π₯ + 4 β 2π¦ = 10
2. Graph and find an equation of the parabola with vertex (4,3) and focus (4,6).
3. Graph and find an equation of the parabola passing through (β1,3), (3,3) and (1,2). What is
the vertex of this parabola?
4. Graph and find an equation of the parabola with vertex (β4,3) and focus (4,3).
5. An old satellite dish is has a cross-section that is in the shape of a parabola. The dish is 48 inches across and 8 inches deep at its center. The vertex is at this lowest point in the parabola. Put the vertex at (0, 0) and find an equation for the parabola cross section. Also determine the value of π and state how high above the vertex the focus is.
6. Create a parabola for a parabolic trough solar cooker that can be contained inside of a copier paper box. These boxes are 17 inches long and 10 inches deep. Make it so your parabola has vertex within one inch of the bottom of the box and so that the focus is within 3 inches of the bottom center of the box. Find an equation for this parabola when the origin is at the center line of the bottom of the box.
Graphing Functions and Relations Page 140
2.2b Circles and Ellipses
A locus definition for a circle would be all points that lie a fixed distance called the radius from a
fixed point called the center of the circle. We saw earlier that all circles lead to equations of the
form (π₯ β β)2 + (π¦ β π)2 = π 2 when the center and radius are known to be (β, π) & π .
An ellipse can be obtained by stretching the graph of a circle to make it oblong in shape. The locus
definition for an ellipse requires two fixed points each called a focus point and a length (2π) called the
major diameter of the ellipse.
A locus definition for an ellipse is all points π such that the sum of the two distances from π to
each focus is equal to the major diameter of the ellipse.
This provides an easy way to draw an ellipse when the two foci and the major diameter are given. You can place pins at each focus and connect them with a string that is the major diameter long. Then take your pencil and use it to keep the string taught as you move it along the ellipse.
The locus definition leads to the standard form of the equation of an ellipse as:
(π₯ β β)2
π2+
(π¦ β π)2
π2= 1
Here (β, π) is the center and π and π are each half of the major and minor diameters which are
horizontal and vertical depending on whether π > π, or π < π. The larger of 2π or 2π is
referred to as major diameter. The two focus points are located on the major diameter at a
distance π = β|π2 β π2| from the center. The endpoints of the major diameter are called the
vertices of an ellipse.
β’ If π < π, the ellipse major diameter is vertical, and the Foci are given by (β, π Β± π) and
vertices by (β, π Β± π).
β’ If π > π the major diameter is horizontal and the Foci are given by (β Β± π, π) and vertices
by (β Β± π, π).
β’ If π = π this ellipse equation reduces to that of a circle of radius π and center(β, π).
Graphing Functions and Relations Page 141
Reflective Property of Ellipse: A light or sound wave leaving in any direction from one focus of
an ellipse will reflect off of the ellipse and arrive at the other focus of the ellipse.
This property is used in whispering galleries where the walls of a room are in the shape of an ellipse and if a person speaks softly at one focus, a listener at the other focus point will receive all of the sound from the whisperer from all parts of the wall. Elliptical reflectors are also used in medicine to treat kidney stones without surgery. Shockwaves are generated at one focus outside of the body and an elliptical shell is adjusted so that the kidney is positioned at the other focus of the elliptical shell where it receives the concentrated energy of the waves and gets pulverized.
Examples:
1. Sketch the graph of (π₯+2)2
64+
(π¦β3)2
25= 1. Find the center, major and minor axis, and the foci.
2. Sketch the graph of the ellipse and find the center of the ellipse.
4π₯2 + 16π₯ + 9π¦2 + 18π¦ = 119
3. Sketch the graph of the ellipse with equation 25π₯2 β 150π₯ + 9π¦2 + 36π¦ = β36 and locate its
center, vertices and foci.
4. Find equations and the location of the foci for the ellipses which are graphed below.
a.
b.
Graphing Functions and Relations Page 142
Solutions
1. Sketch the graph of (π₯+2)2
64+
(π¦β3)2
25= 1. Find the center, major and minor axis, and the foci.
Solution: Center is (β2,3). π = 8,
π = 5, π = β|64 β 25| = β39 and major axis is of length 16 and minor axis of length 10. Foci are
located at (β2 + β39, 3), (β2 β
β39, 3). The vertices are located
at (β2 + 8, 3) = (6, 3) and at (β2 β 8, 3) = (β10, 3)
2. Sketch the graph of the ellipse and find the center of the ellipse.
4π₯2 + 16π₯ + 9π¦2 + 18π¦ = 119
Solution: 4(π₯2 + 4π₯ + π) + 9(π¦2 + 2π¦ + π) =119+ππ + π 4(π₯ + 2)2 + 9(π¦ + 1)2 = 144 (π₯ + 2)2
36+
(π¦ + 1)2
16= 1
Center is (β2, β1), π = 6, π = 4. Vertices are (β2 + 6,1) = (4,1) and (β2 β 6,1) = (β8,1)
Graphing Functions and Relations Page 143
3. Sketch the graph of the ellipse with equation 25π₯2 β 150π₯ + 9π¦2 + 36π¦ = β36 and locate its
center, vertices and foci.
Solution: Complete the squares: 25(π₯2 β 6π₯ + π) + 9(π¦2 + 4π¦ + π) = β36 + πππ + ππ 25(π₯ β 3)2 + 9(π¦ + 2)2 = 225 (divide both sides by 225)
(π β π)π
π+
(π + π)π
ππ= π
Center is at (3, β2) major diameter is vertical and 2π =10 units long, while the minor diameter is 2π = 6 units
long. The foci are vertically π = β|9 β 25| = β16 = 4 units up and down from the center since π = 3 < π = 5 and the vertices are 5 units up and down from the center.
4. Find equations and the location of the foci for the ellipses which are graphed below.
a.
Center is at (β2, 3), π = 5, π = 2,
π = β|52 β 22| = β21
Foci at (β2 Β± β21, 3),
Vertices at (β2 Β± 5, 3 )
Equation is: (π₯+2)2
25+
(π¦β3)2
4= 1 .
b.
Center is at (3, β4), π = 4, π = 3,
π = β42 β 32 = β7
Foci at (3, β4 Β± β7),
Vertices at (3, Β±4)
Equation is: (π₯β3)2
9+
(π¦+4)2
16= 1
Graphing Functions and Relations Page 144
Circles
Equations of Circles
http://www.youtube.com/watch?v=fzNXmoCHRCk (10 mins)
We derive the equation of a circle with a given the center and radius using the distance formula.
Hint: The distance between two points; the center at (π₯1, π¦1) = (π, π), and a point on the circle (π₯2, π¦2) = (π₯, π¦) is given by:
π = β(π₯2 β π₯1)2 + (π¦2 β π¦1)2 π2 = (π β π)2 + (π¦ β π)2
With this distance set to π = π , the equation of the circle centered at (β, π) = (0, 0) with radius of π = 4 becomes: π₯2 + π¦2 = 16.
Equation of a Circle in Standard Form
The equation (π β π)π + (π β π)π = πΉπ represents the circle with center (β, π) and radius π .
Examples
1. Sketch the graph of (π₯ β 2)2 + (π¦ + 1)2 = 9
2. Find equation of a circle with
a. Center at (3,5) and radius of 2.
b. Center (β3,7) and radius of 5.
3. Find the center and radius of the circle given by (π₯ β 6)2 + (π¦ + 3)2 = 49
4. Complete the squares to establish that the equation π₯2 β 12π₯ + π¦2 + 8π¦ = 12 represents a
circle and determine its center and radius.
5. Find the equation of the circle whose graph is given below.
Graphing Functions and Relations Page 145
Solutions
1. Sketch the graph of (π₯ β 2)2 + (π¦ + 1)2 = 9 So the graph of the equation (π₯ β 2)2 + (π¦ + 1)2 = 32 would be a circle with radius of 3 and center of (2, β1) since (π₯ β 2)2 + (π¦ + 1)2 = 32 can be written as (π₯ β 2)2 + (π¦ β (β1))2 = 32
2. Find equation of a circle with
a. Center at (3,5) and radius of 2.
Solution: An equation for the circle centered at (3, 5) with radius 2 is
(π₯ β 3)2 + (π¦ β 5)2 = 22
b. Center (β3,7) and radius of 5.
Solution: The circle of radius 5 centered at (β3, 7) is represented by
(π₯ β (β3))2
+ (π¦ β 7)2 = 52 or (π₯ + 3)2 + (π¦ β 7)2 = 25.
3. Find the center and radius of the circle given by (π₯ β 6)2 + (π¦ + 3)2 = 49
Solution: The equation (π₯ β 6)2 + (π¦ + 3)2 = 49 represents a circle centered at (6, β3) of
radius 7.
4. Complete the squares to establish that the equation π₯2 β 12π₯ + π¦2 + 8π¦ = 12 represents a
circle and determine its center and radius.
π₯2 β 12π₯ + ππ + π¦2 + 8π¦ + ππ = 12 + ππ + ππ
(π₯ β 6)2 + (π¦ + 4)2 = 64.
Thus the graph of π₯2 β 12π₯ + π¦2 + 8π¦ = 12 is the circle of radius 8 centered at (6, -4).
5. Find the equation of the circle whose graph is given below.
The equation for the circle graphed to the right is obtained by reading off the center as (3, β2) and radius as π = 3 from the graph. Thus the equation is (π₯ β 3)2 + (π¦ + 2)2 = 9.
Graphing Functions and Relations Page 146
Section2.2b Worksheet Date:______________ Name:__________________________
Concept Meaning in words and/or examples as required
1. What is a circle?
2. What is an ellipse? How is it different than a circle?
3. What are the major axis and minor axis of an ellipse?
4. How do we find the center and the foci of an ellipse if we are given its equation?
5. How do we find the center and radius of a circle from its equation?
Difficulties encountered in the section:
Graphing Functions and Relations Page 147
Exercises 2.2b
1. Sketch the graph of each circle and state its center and radius.
a. π₯2 + π¦2 = 36 b. (π₯ β 3)2 + (π¦ β 4)2 = 25 c. (π₯ + 2)2 + (π¦ β 1)2 = 9 d. 4π₯2 + 8π₯ + 4π¦2 β 24π¦ = 24 e. Plot the graph of(π₯ + 5)2 + (π¦ β 3)2 = 9.
f. Plot the graph of(π₯ β 2)2 + (π¦ + 1)2 =25
4.
g. π₯2 + 4π₯ + π¦2 β 6π¦ = 12
h. 2π₯2 + 8π₯ + 2π¦2 β 14π¦ = β29
2
2. Sketch graphs and write equations that represent each circle described below.
a. The center is at (2, 7) and the radius is π = 9.
b. Two endpoints of a diameter are the points (1, 2) and (5,2).
c. Find equation of the circle with center (β3, 5) and radius 5.
d. Find the equation of a circle with center (2,5), and passing through (β2,2).
e. Find equation of the circle whose diameter has endpoints (β1,4), and (4, β2).
f. Find an equation of the circle with center (-3, 5) that goes through the point (5, -1).
3. Write the equation in standard form for the graph of the circle shown below.
4. Write equation of the circle in standard form for the graph of the circle shown below.
5. Plot the ellipses below and locate their center, vertices, foci, and the values of π and π.
a. (π₯+3)
16
2+
(π¦β1)
25
2= 1
b. 16π₯2 + π¦2 β 16 = 0
c. 4π₯2 β 16π₯ + 9π¦2 β 18π¦ = 11
d. (π₯β3)
9
2+ (π¦ + 2)2 = 1
Graphing Functions and Relations Page 148
6. Find an equation of each ellipse described or plotted below and find the location of their foci.
a. The ellipse is centered at (3, 6) with a horizontal major diameter of 10 and minor diameter of 6.
b. The ellipse has vertices at (β3, 6) and (β3, β2) and (β1, 2) is the endpoint of a minor diameter.
c. The ellipse plotted below:
d. The ellipse plotted below:
2.2c Hyperbolas One final type of conic section call a hyperbola can be constructed when given two fixed points called
foci and either a single point on the hyperbola or a distance 2π which will again be the distance between
two vertices of a hyperbola.
A locus definition for a hyperbola is all points π such that the difference between the distance
from π to one focus is 2π more than or less than the distance from π to the other focus.
General form of the equation of the hyperbola is given by:
(π β π)π
ππ β
(π β π)π
ππ = π, ππ β
(π β π)π
ππ +
(π β π)π
ππ = π
Here (β, π) is the center, the two vertices are at distance π units from the center in the
direction of the positive variable, asymptotes are diagonal lines through a box with sides π and
π units away from the center in the direction of the positive and negative variables respectively.
The foci are located a distance of π = βπ2 + π2 from the center and lie beyond the vertices.
Vertices for (π₯ββ)2
π2β
(π¦βπ)2
π2= 1 are located at (β Β± π, π).
Vertices for β(π₯ββ)2
π2 +(π¦βπ)2
π2 = 1 are located at or (β, π Β± π) respectively.
For extra credit, find out about the reflective property of hyperbolas and an application of the reflective
property.
Graphing Functions and Relations Page 149
Examples
1. Sketch the graph of (π₯β2)2
9β
(π¦β3)2
4= 1. Find the center, the foci and the asymptotes.
2. Plot the graph of hyperbola and find the center, asymptotes, and foci.
β4π₯2 β 16π₯ + 9π¦2 + 18π¦ = 43
Solutions
1. Sketch the graph of (π₯β2)2
9β
(π¦β3)2
4= 1. Find the center, the foci and the asymptotes.
Solution: Center is (2, 3), π = 3, π = 2, π = β9 + 4 = β13, foci are located at (2 Β± β13, 3).
Asymptotes are given by π¦ = 3 Β±2
3(π₯ β 2). Vertices are at (2 Β± 3, 3) = (5, 3)ππ (β1, 3).
2. Plot the graph of hyperbola and find the center, asymptotes, and foci.
β4π₯2 β 16π₯ + 9π¦2 + 18π¦ = 43
β4(π₯2 + 4π₯) + 9(π¦2 + 2π¦) =43 (Complete the squares β (π₯ + 2)2 & (π¦ + 1)2)
β4(π₯ + 2)2 + 9(π¦ + 1)2 = 43 β 16 + 9
β4(π₯ + 2)2 + 9(π¦ + 1)2 = 36
β(π₯ + 2)2
9+
(π¦ + 1)2
4= 1
Center is (β2, β1), π = 2, π = 3, asymptote box is 3 R/L and 2 U/D from center and asymptote
equations are π¦ = β1 Β±2
3(π₯ + 2). Vertices are located at (β2, β1 Β± 2) = (β2,1)ππ (β2, β3).
The foci are at π = Β±βπ2 + π2 Up/Down from the center at (β2, β1 Β± β5).
Graphing Functions and Relations Page 150
In general then, an equation of the type ππ₯2 + ππ¦2 + ππ₯π¦ + ππ₯ + ππ¦ + π = 0 will represent a
conic section.
β’ If π = 0 and either π or π is zero but not both, we will get a parabola.
β’ If π = 0 and both π and π are both nonzero real numbers with the same sign, we will get an
ellipse with its diameters parallel to the π₯ and π¦ axes.
β’ If π = 0 and both π and π are both nonzero real numbers with the opposite sign, we will get
a hyperbola aligned with the coordinate axes.
In the case β 0 , we still end up with one of these three curves, but they will be rotated through some
angle from the cases we studied here with π = 0. Some examples are shown below. Finding the
center, foci, and even what type of conic an equation represents requires a little more advanced
knowledge of trigonometry and is a motivation for you take more math if you like what you see below.
1. π₯2 β π₯π¦ + π¦ β 2 + 2π¦2 = 0
2. π₯2 β π₯π¦ + π¦ β 2 β 2π¦2 = 0
3. 2π₯2 β π₯π¦ + π¦ β 2 + π¦2 = 0
4. β2π₯2 β π₯π¦ + π¦ β 2 + π¦2 = 0
Graphing Functions and Relations Page 151
Summary and tips on Identifying Conic Sections
Equations of the type ππ₯2 + ππ¦2 + ππ₯ + ππ¦ + π = 0 have graphs that are one of the three conic section
types. (There are a few pathological cases that can happen where the equation has no solutions, e.g.,
π₯2 + π¦2 = β5.) The type of conic can be determined by the presence and sign of the coefficients π and
π. The table below summarizes what weβve done in this chapter to identify type and plot these conic
section graphs.
Conic Section Tips to Identify
Parabola When only one variable has a squared term, it is a parabola.
If the equation contains an π₯2 term this means it is an up/down parabola. Put the π¦ term on one side and move the rest to the other side.
If equation contains π¦2 term this means it is a left/right parabola. Put the π₯ term on one side and move the rest to the other side.
If the coefficient of the π₯2 term is > 0, then the parabola faces up. Example 3π₯2 β 4π¦ + π₯ β 1 = 0
or 3
4π₯2 +
1
4π₯ β
1
4= π¦
If the coefficient of the π₯2 term is < 0, then the parabola faces down. Example 3π₯2 + 4π¦ + π₯ β 1 = 0
or π¦ = β3
4π₯2 β
1
4π₯ +
1
4
If the coefficient of the π¦2 term is > 0, then the parabola faces right. Example
π₯ =3
4π¦2 β
1
4π¦ +
1
4
If the coefficient of the π₯2 term is < 0, then the parabola faces down. Example
π₯ = β3
4π¦2 β
1
4π¦ +
1
4
Focus is π units from vertex where π =1
4π
Ellipse Get all terms on the same side. Both π₯2, and π¦2 terms are of the same sign.
Circle Ellipse that is not a circle
Both π₯2, and π¦2 terms are of the same sign and the same coefficient. Example 4π₯2 + 4π¦2 β 8π₯ + 4π¦ = 10
Both π₯2, and π¦2 terms are of the same sign and different coefficients. Example 8π₯2 + 3π¦2 β 8π₯ + 4π¦ = 10
Graphing Functions and Relations Page 152
Hyperbola Get all terms on the same side, complete the squares if there are π₯ or π¦-terms and finally make the other side equal to one. To be a hyperbola, the (π₯ β β)2, and (π¦ β π)2 terms have coefficents of opposite sign.
When both π₯2, and π¦2 terms are on the same side the (π₯ β β)2 is positive, the hyperbola opens left/right. Example π₯2 β 4π¦2 + 6π₯ + 8π¦ = 11 β
(π + π)π
ππβ
(π β π)π
π= π
When both π₯2, and π¦2 terms are on the same side the (π¦ β π)2 is positive, the hyperbola opens up/down. Example β π₯2 + 4π¦2 β 6π₯ β 8π¦ = 21 β
β(π₯ + 3)2
16+
(π¦ β 1)2
4= 1
Examples
Determine the type of conic, locate vertex(s) and focus(s) and asymptotes (if any) and graph.
1. π₯2 + 5π¦2 β 4π₯ + 30π¦ = β29 Solution: Complete Squares
π₯2 β 4π₯ + 5(π¦2 + 6π¦ ) = β29 (π₯β2)2
20+
(π¦+3)2
4= 1
This is the equation of an ellipse with a horizontal
major axis, center at (2, β3), π = β20, π = 2
π = β20 β 4 = β16 = 4
Vertices at (2 Β± β20, β3) Foci at (2 Β± 4, β3).
2. π¦2 β 6π¦ β 8π₯ β 15 = 0 Solution: Complete Squares π¦2 β 6π¦ = 8π₯ + 15 (π¦ β 3)2 = 8π₯ + 24
π₯ =1
8(π¦ β 3)2 β 3
The equation is a parabola with vertex at (β3, 3).
π =1
4β 1
8
=11
2
= 2 > 0 means it opens to the right.
Focus is at (-1, 3).
Graphing Functions and Relations Page 153
3. 9π₯2 β 16π¦2 + 36π₯ + 96π¦ = β36 Solution: Complete Squares
9(π₯2 + 4π₯ + 4) β 16(π¦2 β 6π¦ + 9) = β36 + 36 β 144 9(π₯ + 2)2 β 16(π¦ β 3)2 = β144 Divide both sides by β144
β(π₯ + 2)2
16+
(π¦ β 3)2
9= 1
Hyperbola! Center at (β2, 3), opens up/down since the negative sign is with the π₯ term, π = 3, π = 4. Vertices at (β2, 3 Β± 3).
π = β9 + 16 = 5 Foci are at (β2, 3 Β± 5) = (β2, β2) and (β2,8).
Asymptotes are given by π¦ = Β±3
4(π₯ + 2) + 3 are shown in
the dotted red lines. The purple dotted box is used a guide to plot the asymptotes.
Graphing Functions and Relations Page 154
Section2.2c Worksheet Date:______________ Name:__________________________
Concept Meaning in words and/or examples as required
1. What is a hyperbola?
2. What are the different kinds of hyperbolas?
3. How do you find center and foci of a hyperbola?
4. How do we find the asymptotes of a hyperbola?
Difficulties encountered in the section:
Graphing Functions and Relations Page 155
Exercises 2.2c
1. In problems a through f, locate the center, vertices, foci, and the asymptotes. Then sketch the graph
and show all the information you found in your graphs.
a. (π¦+3)
9
2β
(π₯β1)
16
2= 1
b. (π₯+3)
16
2β
(π¦β1)
9
2= 1
c. 16π₯2 β π¦2 β 16 = 0
d. 4π₯2 β 16π₯ β 9π¦2 + 18π¦ = 29
e. 4π¦2 β 16π¦ β 25π₯2 β 150π₯ = 309
f. β4π₯2 β 16π₯ + 9π¦2 + 18π¦ = 43
2. For problems a through h, complete the square as needed to identify what conic section the
equation represents, and locate the vertex(center), vertices, and focus(foci) and asymptotes as
appropriate.
a. π¦2 + 6π¦ β 2π₯ + 5 = 0 Ellipse Hyperbola
Circle Parabola
b. π₯2 + 6π₯ β 8π¦ β 31 = 0
Ellipse Hyperbola
Circle Parabola
c. 16π₯2 β 64π₯ + 9π¦2 + 108π¦ = β244 Ellipse Hyperbola
Circle Parabola
d. 16π¦2 β 96π¦ β 9π₯2 β 18π₯ = 9 Ellipse Hyperbola
Circle Parabola
Graphing Functions and Relations Page 156
e. π¦2 = 2π₯ β π₯2 Ellipse Hyperbola
Circle Parabola
f. π₯2 + π¦2 β 2π₯ + 4π¦ = 4 Ellipse Hyperbola
Circle Parabola
g. 3π¦2 β 4π₯ + 6π¦ = 4
Ellipse Hyperbola
Circle Parabola
h. 3π¦2 + 4π₯ + 6π¦ = 4
Ellipse Hyperbola
Circle Parabola
3. In problems a through d find an equation of the conic section for the given graphs
a. b.
Graphing Functions and Relations Page 157
c. d.
4. Two friends live 10 miles from away from each other along an east-west highway. One day as
they are chatting on their cell-phones, the eastern friend hears a loud clap of thunder over the
cell phone and 40 seconds later hears the sound of the same thunder reaching her house
directly. Since the sound of thunder travels a mile in 5 seconds, this means that the thunder
traveled an additional 40 seconds to reach the eastern friend. Thus the distance from the
lightning strike to the eastern friend was 8 miles longer than the distance to the western friend.
Use the locus definition of a hyperbola to locate the possible locations of the lightning strike.
(Hint: Draw a graph with the friends located on the π₯-axis at Β± 5 and then draw an appropriate
hyperbola. )
Also, if the western friend notices that the flash of lightning that produced this thunder was
directly north of her, determine exactly where the strike was.
Graphing Functions and Relations Page 158
5. Match each relation to its appropriate graph. If there is no match, please state so.
Match all the quantities in Column B that are equivalent to quantities in Column A. Some of the
column B quantities may not have any corresponding items in column A, but all items in column
A have at least one or more corresponding items in column B. Explain your reasoning for the
choices you made. (3 pt each)
Column A Answer Column B
i. (π₯ + 2)Β²
16+
(π¦ β 2)Β²
4= 1
Reasoning:
a.
ii.
(π₯ β 2)Β²
4β
(π¦ β 3)Β²
16= 1
Reasoning:
b.
iii.
β(π₯ β 2)2
4+
(π¦ β 3)Β²
16= 1
Reasoning:
c.
iv. π₯2 + π¦2 = 4 d.
Graphing Functions and Relations Page 159
2.3 Graphing Polynomial Functions We concluded the last chapter with graphs of second degree polynomials in π₯ and π¦ which graphically
are represented by parabolas, ellipses and hyperbolas. While these curves have been studied for over
2000 years, about 400 years ago with the development of algebra came infinitely many new curves to
study by looking at graphs of equations. In this section we look at general polynomial functions in one
variable. Back in chapter 1 we saw examples of some simple polynomial functions which we review
below.
1. Polynomial function: A function defined as π(π₯) = π0 + π1π₯ + π2π₯2 + π3π₯3 + β― + πππ₯π, where π0, π1, π2, π3, β¦ ππ are real numbers and n β₯ 0 is a whole number. The domain of these functions is all real numbers.
Examples:
i) Constant Function: A function defined as π(π₯) = π, where π is any real number. This is a polynomial function of degree zero.
Generic Graph
Domain: All real numbers Range: {π} One-to-One: NO
Example If π(π₯) = 5, then find
a. π(3) b. π(β2) c. π(β3456) d. π(π + β) e. Sketch the graph
Solutions a. π(3) = 5
b. π(β2) = 5 c. π(β3456) = 5 d. π(π + β) = 5
e. As you can see from the parts a,b,and c, no matter what π₯-coordinate you plot the π¦-coordinate is always 5 so its graph is a horizontal line as shown above.
Graphing Functions and Relations Page 160
ii) Linear Function: A function defined as (π₯) = ππ₯ + π . With π β 0, this is a first degree polynomial function. Recall that π =slope of the line, π = π¦-intercept. This is a polynomial function of degree one.
Example 1. If π(π₯) = 2π₯ β 3, then
a. Find π(β1) b. Find π(0) c. Find π(4) d. Find the inverse function e. Sketch the graph of π¦ = π(π₯)
Solution a. π(β1) = 2(β1) β 3 = β2 β
3 = β5 b. π(0) = β3 c. π(4) = 2(4) β 3 = 8 β 3 =
5 d. Inverse function
π¦ = 2π₯ β 3 π₯ = 2π¦ β 3
π₯ + 3 = 2π¦ π₯ + 3
2= π¦
Inverse function is
πβ1(π₯) =π₯+3
2
e. Graph of the function is to the right with slope of 2 and π¦-intercept of β3.
2. An Olympic size swimming pool holds 2,500,000 liters of water. If the pool currently holds 100,000 liters of water, and water is being pumped at 400,000 liters/hour into the tank.
a. Write a function that represents the amount of water in the tank after π‘ hours. b. Find the domain and range of this function. c. Sketch the graph of this function.
Solution: a. π΄(π‘) = 100000 + 400000π‘
Liters. The domain restriction 0 β€ π‘ β€ 6 for this application reflects the fact that the water inflow starts at π‘ = 0 and at π‘ = 6 the poo is filled with 2,500,000 liters of water.
b. Domain of π΄ = [0,6] and Range of π΄ = [100000,2500000]
c. Graph is to the right. The scale is each tick mark represents 1 million liters on the π¦-axis and 1 hour on the π‘-axis.
Graphing Functions and Relations Page 161
iii) Square Function: A function defined as (π₯) = ππ₯2 + ππ₯ + π . This is a general degree-two polynomial function (π β 0) and is also called a quadratic function.
Example If π(π₯) = π₯2, then
a. Find π (2
3)
b. Findπ(β2) c. Find π(2) d. Find π(π + β) e. Sketch the graph the function π¦ = π₯2 f. Is the function one-to-one?
Solution:
a. π (2
3) = (
2
3)
2=
4
9
b. π(β2) = (β2)2 = 4 c. π(2) = (2)2 = 4 d. π(π + β) = (π + β)2 = π2 + 2πβ + β2 e. Plot a few points to sketch the graph of
the function. See to the left f. No. The function is not one-to-one as it
does not pass the horizontal line test.
π₯ π¦ = π₯2 β2 (β2)2 = 4 β1 (β1)2 = 1 0 (0)2 = 0 1 (1)2 = 1 2 (2)2 = 4
iv) Cubic Function: A function defined as π(π₯) = ππ₯3 + ππ₯2 + ππ₯ + π
A special case of a polynomial function of degree three (π β 0).
Example If π(π₯) = π₯3, then
a. Find π (2
3)
b. Findπ(β2) c. Find π(2) d. Sketch the graph the function π¦ = π₯3 e. Is the function one-to-one?
Solution:
a. π (2
3) = (
2
3)
3=
8
27
b. π(β2) = (β2)3 = β8 c. π(2) = (2)3 = 8 d. Plot a few points to sketch the graph of
the function. See to the left e. Yes. The function is one-to-one as it does
pass the horizontal line test. f. Finding inverse function write π¦ = π₯3,
then we get π₯ = π¦3 solving for π¦ we get
that inverse function is πβ1(π₯) = βπ₯3
π₯ π¦ = π₯3 β2 (β2)3 = β8 β1 (β1)3 = β1 0 (0)3 = 0 1 (1)3 = 1 2 (2)3 = 8
Graphing Functions and Relations Page 162
So we now have a fairly complete understanding of the graphs of constant functions, linear functions,
quadratic functions, and the basic cubic function π(π₯) = π₯3 . I.e., constant functions and linear
polynomial graphs are lines. In conic sections, we saw that quadratic functions of the type π(π₯) =
ππ₯2 + ππ₯ + π have parabola graphs with vertex (βπ
2π, π (
βπ
2π)) and open up when π > 0 and down when
π < 0.
We will use this information and the information we learned about transformation of functions to
explore the behavior of more complicated polynomial function graphs.
Let us start with exploring functions of the type π(π₯) = π₯π, where π is a whole number.
1 When π = 0 we have a constant function π(π₯) = 1 which is a horizontal straight line.
2 When π = 1 we have π(π₯) = π₯ which is also a straight line passing through (0,0)
3 When π = 2 we have a quadratic π(π₯) = π₯2 which is a parabola with vertex (0,0) facing up.
Graphing Functions and Relations Page 163
4 In general then when π > 2 and an even number we know it will behave similar to the graph of π(π₯) = π₯2 in its generic shape. Plotting a few points we notice that a) When β1 < π₯ < 1, π₯π < π₯2. For
example, 0.14 = 0.0001 < 0.12 =0.01.
b) When β1 > π₯, ππ π₯ > 1, π₯π > π₯2. For example, 104 = 10000 > 102 =100 or (β10)4 = 10000 >(β10)2 = 100.
See graphs of some of the functions of the type π(π₯) = π₯π in comparison to each other on the right.
5 When π = 3 we have a cubic function π(π₯) = π₯3 which is passes through (0,0) and is symmetric with respect to origin.
6 In general then when π > 3 and an odd number we know it will behave similar to the graph of π(π₯) = π₯3 in its generic shape. Plotting a few points we notice that a) When 0 < π₯ < 1, π₯π < π₯3 . For
example, 0.15 = 0.00001 < 0.13 =0.001
b) When β1 < π₯ < 0, π₯π > π₯3. For
example, (β0.1)5 = β0.00001 >(β0.1)3 = β0.001
c) When < β1, π₯π < π₯3. For example,
(β10)5 = β100000 < 103 = 1000. d) When π₯ > 1, π₯π > π₯3. For example,
105 = 100000 > 103 = 1000
See graphs of some of the functions of the type π(π₯) = π₯π in comparison to each other below.
Graphing Functions and Relations Page 164
All polynomial functions are continuous (~no breaks or holes in the graph) on their
domain (ββ, β).
Intermediate Value Theorem: For a polynomial function π(π₯), if π(π) < π(π), with
π < π, then for every number πΆ for which π(π) β€ πΆ β€ π(π), there exists a value of π₯
with π β€ π₯ β€ π so that π(π₯) = πΆ. That means π attains all π¦-values between π(π), and
π(π) at least at one value π₯ in the interval [π, π].
How can we use this fact to our benefit? The Intermediate Value Theorem graphically means that every
horizontal line between the π¦-values at π₯ = π and π₯ = π intersects the polynomial graph at least one
time.
In scenario at least once the function attains the
value of πΆ.
In scenario the function attains the value of πΆ
several times.
As you can see, even from the graphs of the basic polynomial functions if you travel on the graph from
left to right, sometimes the graph of the function does not rise or fall as in the case of the constant
function, and other times it rises or falls like riding a roller coaster.
The definitions below are for any function and not just polynomial functions.
A function π(π₯) is constant on an interval (π, π) iff for all π₯1 < π₯2 in the interval (π, π),
π(π₯1) = π(π₯2).
A function π(π₯) is strictly increasing on an interval (π, π) iff for all π₯1 < π₯2 in the interval
(π, π), π(π₯1) < π(π₯2). Graphically, π¦-coordinates get higher as π₯ moves to the right.
A function π(π₯) is strictly decreasing on an interval (π, π) iff for all π₯1 < π₯2 in the interval
(π, π), π(π₯1) > π(π₯2). Graphically, π¦-coordinates go down as π₯ moves to the right.
Graphing Functions and Relations Page 165
Example
1. Determine all the intervals where the function below is increasing, decreasing and or constant.
The function is increasing on the interval (ββ, β1) βͺ (1,1.7)
The function is decreasing on the interval (β1,1)
The function is constant on the interval (1.7,7)
Practice problems:
Try to come up with rough sketches of the graphs the following functions. Try using the knowledge you
have accumulated so far (i.e., transformation of functions, arithmetic of functions). Do not peek after
this page to see what the answers are. Cultivating your intuition will aid in your deeper understanding of
the material. You can always resort to plotting points that might be of interest to you.
1. π(π₯) = (π₯ β 1)2(π₯ + 1)3
= π₯5 + π₯4 β 2π₯3 β 2π₯2 + π₯ + 1
2. π(π₯) = (π₯ β 1)2(π₯ + 1)3 + 2
= π₯5 + π₯4 β 2π₯3 β 2π₯2 + π₯ + 3
3. π(π₯) =1
10(π₯ β 4)(π₯ + 4)(π₯ β 2)3(π₯ + 2)2 4. π(π₯) = β
1
10(π₯ β 4)(π₯ + 4)(π₯ β 2)3(π₯ + 2)2
Graphing Functions and Relations Page 166
Solutions to practice problems:
1. π(π₯) = (π₯ β 1)2(π₯ + 1)3 = π₯5 + π₯4 β 2π₯3 β 2π₯2 + π₯ + 1 We know that the function π¦ = (π₯ β 1)2 is a parabola shifted one to the right, and π¦ = (π₯ + 1)3 is a cubic function shifted one to the left. These individual graphs are shown below. Plotting them on the same coordinate axis will give us insight into the graph of their product. Note π¦-intercept is π(0) =(0 β 1)2(0 + 1)3 = 1
To plot the graph of π(π₯) we need to see how the product of the two functions we know behaves.
π₯ < β1 β1 < π₯ < 1 π₯ > 1 Test Point β3 0 1.5
π(π₯) = (π₯ β 1)2(π₯ + 1)3
(β3 β 1)2(β3 + 1)3 = β128 < 0
(0 β 1)2(0 + 1)3 = 1 > 0
(1.5 β 1)2(1.5 + 1)3 3.90625 > 0
Graph of π(π₯) is below the π₯-axis
Graph of π(π₯) is above the π₯-axis
Graph of π(π₯) is above the π₯-axis
π(1) = (1 β 1)2(1 + 1)3 = 0 and π(β1) = (β1 β 1)2(β1 + 1)3 = 0 that means π₯ = 1, β1 are π₯-intercepts. So just knowing the sign of the function we can sketch a rough sketch. The domain of the function π(π₯) is all real numbers. We can see the two functions = (π₯ β 1)2 , and π¦ = (π₯ + 1)3 have no breaks or holes in them and so that is what we would expect in the product function. You can plot several points to see what happens to convince yourself but each of them will preserve their shapes near to where each crosses the π₯-axis. Our rough sketch would look as shown below. We donβt know exactly know how high it will rise before turning back down when β1 < π₯ < 1 but we know it goes through the π¦-axis at (0, 1).
π = (π β π)π
π = (π + π)π
Graphing Functions and Relations Page 167
From the table below we can see that when π₯ approaches β or ββ, the value of just the highest degree
term is a good approximation to the value of the full polynomial output. The lower degree terms are
much smaller at these π₯-values because they have fewer factors of the large number π₯ in them. This
means that terms other than the highest degree term will not contribute much make a dent in the
magnitude of the overall function value. In other words the highest degree terms will take over the
behavior of the function when π₯ β Β±β. This is called the end behavior of the function. See table below.
π₯ π¦ = π₯5 π¦ = π(π₯) = π₯5 + π₯4 β 2π₯3 β 2π₯2 + π₯ + 1
1000 1,000,000 ,000,000,000 1,000,997 ,998,001,001
10000 100 ,000,000,000 ,000,000,000 100, 009,997,999 ,800,010,001
100000 10000000 000000000 000000000 10000099 997999980 000100001
β1000 β1,000,000 ,000,000,000 β998,998 ,002,000,999
β10000 β100 ,000,000,000 ,000,000,000 β99 ,989,998,000, 200,009,999
β100000 β10,000,000,000,000,000,000,000,000 β9,999,899 ,998,000,020 ,000,099,999
The end behavior of π(π₯) is that when π₯ β Β±β, the graph of π¦ = π(π₯) is very much like π¦ = π₯5.
This example illustrates the importance
of using an appropriate viewing window
when using technology to create the
graph of a function. The details in the
top-left plot are totally lost in the lower
graph. We need to know where to look
to see the interesting details. In this
course, weβll develop tools to see how
polynomials behave near their π₯-
intercepts. Other details, such as where
the local maximum is near π₯ = 0.2
require tools from calculus.
Graphing Functions and Relations Page 168
When π(π₯) is a polynomial function with (π₯ β π)π as a factor, we can lump the other factors together as
π(π₯) so π(π₯) = π(π₯)(π₯ β π)π . We can do it so that π(π) β 0. Note that when π₯ = π we get
π(π) = π(π)(π β π)π = 0, and this is a reason why then π₯ = π is said to be a zero of π of multiplicity π.
The multiplicity refers to how many times π₯ = π occurs as a zero in π(π₯) (since(π₯ β π)π = (π₯ β π)(π₯ β
π) β¦ (π₯ β π) (a product of π₯ β π, π times)).
Lets look at the behavior of the function π(π₯) = π(π₯)(π₯ β π)π for π₯ = π, where π very close to π.
Then at this value π(π) will be near to π(π) and we can think of π(π) β π(π)(π β π)π. In other words
near each zero π₯ = π the graph will behave like a stretched version of π¦ = (π₯ β π)π where the stretch
factor is equal to π(π) , i.e., the number when π₯ = π is plugged into all the π₯ locations except (π₯ β π)π.
Thus multiplicity π = 1 means the graph crosses at π₯ = π like a line with slope π(π). With multiplicity
π = 2, the graph behaves like a parabola with vertex at π₯ = π and stretch factor of π(π). For
multiplicity π = 3 the graph crossses at π₯ = π, but in a manner that is flat to the π₯-axis the same as how
π¦ = π₯3 looks likes near π₯ = 0.
2. π(π₯) = (π₯ β 1)2(π₯ + 1)3 + 2 = π₯5 + π₯4 β 2π₯3 β 2π₯2 + π₯ + 3 We know from before that the graph here will be similar to the one before just moved up 2 units.
Graphing Functions and Relations Page 169
Summary
End behavior of a polynomial function is governed by the its degree and the leading
coefficient.
When the leading coefficient is positive, then as π₯ β +β the π¦ = π(π₯) β +β.
If the degree is even, then π¦ = π(π₯) β +β when π₯ β ββ .
If the degree is odd, then π¦ = π(π₯) β ββ when π₯ β ββ.
When the leading coefficient is negative, then as π₯ β +β the π¦ β ββ.
If the degree is even, then π¦ = π(π₯) β ββ when π₯ β ββ.
If the degree is odd, then π¦ = π(π₯) β +β when π₯ β ββ.
A pictorial summary of some of the points made above is below.
Degree of the Polynomial
Leading Coefficient
Positive
Negative
Even
Odd
Graphing Functions and Relations Page 170
3. π(π₯) =1
10(π₯ β 4)(π₯ + 4)(π₯ β 2)3(π₯ + 2)2
Step 1: π₯-intercepts Setting π(π₯) = 0 we get we get π₯ = 4, β4,2, β2 will be the π₯-intercepts.
Near π₯ = β2, π¦ β1
10(β6)(2)(β4)3(π₯ + 2)2 = 76.8(π₯ + 2)2 So the graph is like a steep parabola
opening up near π₯ = β2.
Likewise near π₯ = β4, π¦ β1
10(β8)(π₯ + 4)(β6)3(β2)2 = 691.2(π₯ + 4); a line with slope ~700.
And near π₯ = 2, π¦ β 96(π₯ β 2)3, and near π₯ = 4, π¦ β 230.4(π₯ β 4)
Step 2: π¦-intercept Setting π₯ = 0 in the function we get π (0) =1
10(0 β 4)(0 + 4)(0 β 2)3(0 +
2)2 =256
5= 51.2 will be the π¦-intercept.
Step 3: End behavior
Note that if we were to expand the polynomial the degree of the polynomial would be 7 since we
have two linear terms, a cubic term and a square term. So our polynomial will behave like 1
10π₯7 on
the ends (left hand side falls, right hand side rises).
Step 4: Graph the function using steps 1-3.
We use our knowledge of the end behavior and near π₯-intercepts to finish the graph making use of our
observation of where the function is linear, quadratic, or cubic. See if you can come with the graph
before looking on the next page using the information summarized from steps 1-3 below.
Putting all pieces together we have our rough sketch below.
Graphing Functions and Relations Page 171
Another way to sketch is to investigate signs of the function at different π₯ values.
We know that an odd number of negative numbers multiply to give us a negative number, and an
even number of negative numbers multiply to give us a positive number. An odd number of positive
numbers and an even number of positive numbers multiply to give a positive number.
We also know that a linear polynomial ππ₯ + π, we have ππ₯ + π > 0 for all real numbers π₯ > βπ
π,
and ππ₯ + π < 0 for all real numbers π₯ < βπ
π. Using these two facts we can determine whether the
function π(π₯) > 0 or π(π₯) < 0 for certain values of π₯ by doing a sign chart as shown on the next
page.
Graphing Functions and Relations Page 172
See details below.
π < βπ βπ < π < βπ βπ < π < π π < π < π π > π
Test Pts β5 β3 0 3 5
π + π β + + + +
(π + π)π + + + + +
(π β π)π β β β + +
π β π β β β β +
π(π) β + + β +
π(π₯) =1
10(π₯ β 4)(π₯ + 4)(π₯ β 2)3(π₯ + 2)2
Looking at the sign of the function as shown to the left, and our knowledge of the 7th degree polynomial end behavior we can now draw a rough sketch of the graph. We can see that at π₯ = β2 we will still retain the parabola shape, and at π₯ = 2 we would retain the cubic polynomial shape. See below for the rough sketch of the graph. We still do not know how far the graph rises or falls at the local maximum and minimum points.
Graphing Functions and Relations Page 173
To really make sense of the chart and graph together, see below. The graph of the function was drawn
using a graphing utility since without it we cannot know how high or low the function rises or falls. As
you can see, our rough sketch is pretty accurate for the main features of the graph.
Graphing Functions and Relations Page 174
4. π(π₯) = β1
10(π₯ β 4)(π₯ + 4)(π₯ β 2)3(π₯ + 2)2
We know that putting a negative in front of the graph will just reflect the graph in previous example
about the π₯-axis. So the graph will look like
The peaks and valleys we see have names. They are called local extrema or local maximum and
minimum values of the function output.
Local Maximum: In an interval (π, π) a point (π, π(π)) where π < π < π is called a local
maximum point if π(π) β₯ π(π₯) for all π₯ in the interval (π, π). Weβd say π(π) is a local max.
Local Minimum: In an interval (π, π) a point (π, π(π)) where π < π < π is called a local
minimum if π(π) β€ π(π₯) for all π₯ in the interval (π, π). Weβd say π(π) is a local min.
Now attempt to graph the following functions.
5. π(π₯) = (2π₯ β 9)(π₯ + 3)2(π₯ β 1)2
π₯-intercepts are ______________
π¦-intercept is ____________
End Behavior like _______________
Number of local extremum _______
Graph of the polynomial
6. β(π₯) = π₯(π₯ + 5)(π₯ + 2)(π₯ β 1)(π₯ β 4)
π₯-intercepts are ______________
π¦-intercept is ____________
End Behavior like _______________
Number of local extremum _______
Graph of the polynomial
Graphing Functions and Relations Page 175
Solution:
5. π(π₯) = (2π₯ β 9)(π₯ + 3)2(π₯ β 1)2
π₯-intercepts are π₯ =9
2= 4.5, β3, and 1. The behavior is like a parabola at π₯ = 1 πππ π₯ = β3 and the
graph crosses the π₯-axis at π₯ = 4.5 like a line. The π¦-intercept is π¦ = π(0) = (β9)(3)2(β1)2 = β81
End Behavior like π¦ = 2π₯(π₯2)(π₯2) = 2π₯5 (left hand side falls, right hand side rises).
The number of local maximums/minimums total 4 (as seen from the graph).
The computer generated graph shows how low the function dips between π₯ = 1, π₯ = 4. To locate the
exact location of these local max/min points requires tools from calculus. The knowledge of the
behavior at the intercepts, however, is enough to provide the basic shape and alert us to there being
local max and minimum points. We also recommend that you plot a few points that that lie in-between
the intercepts as a check that your intercept and end-behavior analysis is correct.
Graphing Functions and Relations Page 176
6. β(π₯) = π₯(π₯ + 5)(π₯ + 2)(π₯ β 1)(π₯ β 4)
π₯-intercepts are π₯ = 0, β5, β2, 1 and 4 . π¦-intercept is β(0) = 0(5)(β2)(1)(4) = 0.
End Behavior is π¦ β π₯(π₯)(π₯)(π₯)(π₯) = π₯5 (left hand falls, right hand rises)
Number of local extrema is 4 (as seen from the graph).
Graph of the polynomial:
Graphing Functions and Relations Page 177
Observation:
You can see from the examples we have done so far that a factor of the type (ππ₯ + π)π
where π β 0, π is a counting number has the graph either cross or touch (without crossing) the
π₯-axis at π₯ = βπ
π .
When π is an even number the graph of the polynomial function will touch the π₯-axis at
π₯ = βπ
π.
When π is a odd number the graph of the polynomial function will cross the π₯-axis at
π₯ = βπ
π.
Maximum number of local extrema possible is one less than the degree of the polynomial
function.
As you can see the factors of a polynomial function provide a lot of information about its graph. This
makes the factored form very useful. We can also use this in reverse when a graph is given to
determine the form of an equation of a polynomial function..
7. For each graph below, find the factored form of a polynomial that would produce the graph of that
shape. You can label the axis hash marks as you please but the basic shape should match. You can
then use a graphing utility like a graphing calculator, or websites like www.wolframalpha.com or
https://www.desmos.com/calculator to graph your polynomial to check your answers.
a.
b.
Zoomed
In closer
Graphing Functions and Relations Page 178
Solutions:
a.
Since the two ends are facing up, the polynomial has even degree and a positive leading coefficient. The x-intercepts can be thought of as π₯ = β2, 0, 2, 4 and the powers of the factors that give you those intercepts would be odd, even, odd, even respectively. Note that the graph looks similar to π₯3 shape at 2 and -2, thus π should be at least 3 for those factors. Thus, the simplest form for π would be:
π(π₯) = π₯2(π₯ + 2)3(π₯ β 2)3(π₯ β 4)2 Graph is below and this seems a very good match.
If we had used a 4th order factor at π₯ = 0, we get the graph below which has a similar shape but poorer fit.
π(π₯) =1
100π₯4(π₯ + 2)3(π₯ β 2)3(π₯ β 4)2
b.
We can set the right-most π₯-intercept at π₯ = 3 and this puts the next one at
π₯ =1
2 (factor (2π₯ β 1)) with even
multiplicity. Also π₯ = 0 is another intercept with odd multiplicity. The left-most intercept seems to be at
π₯ β β5
3 (factor (π₯ +
5
3) ππ π’π π (3π₯ +
5) with multiplicity of π = 1. Also, the leading coefficient is negative. The polynomial below is the simplest to do this.
π(π₯) = βπ₯3(2π₯ β 1)2(3π₯ + 5)(π₯ β 3)2 The graph of π below confirms it works.
Zoomed
In closer
Graphing Functions and Relations Page 179
8. Match the functions on the right with their graphs on the left. Explain your answers.
I. π(π₯) = (π₯ β 1)2(π₯ + 1)3 II. π(π₯) = (π₯ β 1)2(π₯ + 1)2
III. β(π₯) = β10π₯2(π₯ β 1)2(π₯ + 1)2 IV. π (π₯) = 10π₯3(π₯ β 1)2(π₯ + 1)2
A.
B.
C.
D.
Graphing Functions and Relations Page 180
Solution:
I. π(π₯) = (π₯ β 1)2(π₯ + 1)3 This graph matches B. Since the end
behavior is like π₯5, and at π₯ = 1 we have a parabola shape, and a cubic shape at π₯ = β1.
II. π(π₯) = (π₯ β 1)2(π₯ + 1)2
End behavior is like π₯4, and parabola shapes at π₯ = 1, β1, so must match graph C.
III. β(π₯) = β10π₯2(π₯ β 1)2(π₯ + 1)2
Both ends down since leading coefficient is β10 and degree is 6. Parabola shapes at π₯ = 0, β1, and 1 so must match graph A.
IV. π (π₯) = 10π₯3(π₯ β 1)2(π₯ + 1)2
There is only one graph left D. It touches at π₯ = 1 and π₯ = β1 and behaves like π₯3 at π₯ = 0.
A.
B.
C.
D.
Graphing Functions and Relations Page 181
Playing
In all the examples we have done so far the polynomial function given to you was written in the factored
form which makes it easier for analyzing its graph. So one might wonder can all polynomial functions be
written in the factored form.
To explore this question please consider the polynomials below and answer the following questions.
a) Sketch the graph of the polynomials functions below.
b) Can any of them be written in factored form? If yes, please factor them, and then show what
information is more visible when the polynomial is written in the factored form.
c) Can any of them be written in a way that is a transformation of a polynomial that can be written in
factored form? If yes, please explain.
1. π₯2 + 2π₯ β 2 2. π₯2 β 4π₯ β 12 3. π₯4 + 16
4. π₯4 β 16 5. π₯3 β 2π₯2 β π₯ + 2 6. 3π₯3 β 2π₯2 + 4
We will answer some of the questions posed here about whether or not all polynomials can be factored
or not in Chapter 3.
Graphing Functions and Relations Page 182
Section2.3 Worksheet Date:______________ Name:__________________________
Concept Meaning in words and/or examples as required
1. Degree of a polynomial
2. Leading coefficient of a polynomial
3. Local extrema
4. π-intercept
5. π-intercept
6. Difference between solving an equation in one variable, and sketching the graph of a function in one variable
7. Difference between solving an inequality in one variable, and sketching the graph of a function in one variable
Difficulties encountered in the section:
Graphing Functions and Relations Page 183
Exercises 2.3
1. Answer the questions below
a. How does the degree of the polynomial affect the end behavior?
b. What does the leading coefficient of a polynomial control in its graph ?
c. What is the maximum number of local extrema you can expect in a polynomial of degree π?
d. How does the exponent π in factors of the type (ππ₯ + π)π influence the shape of the graph near
the π₯-intercept associated with this factor?
e. How do you find all the π₯-intercepts of a polynomial function?
f. How do you find all the π¦-intercepts of a polynomial function?
g. How would you solve a an equation of the type
π(π₯ β π1)π1(π₯ β π2)π2(π₯ β π3)π3 β¦ . (π₯ β ππ)ππ = 0?
h. What is the significance of the solutions to the equation in question 7 to the graph of the
polynomial function π(π₯) = π(π₯ β π1)π1(π₯ β π2)π2(π₯ β π3)π3 β¦ . (π₯ β ππ)ππ?
i. How would you solve an inequality of the type
π(π₯ β π1)π1(π₯ β π2)π2(π₯ β π3)π3 β¦ . (π₯ β ππ)ππ > 0
Or
π(π₯ β π1)π1(π₯ β π2)π2(π₯ β π3)π3 β¦ . (π₯ β ππ)ππ < 0
j. How are the solutions to the inequalities in question 9 related to the graph of the polynomial
function π(π₯) = π(π₯ β π1)π1(π₯ β π2)π2(π₯ β π3)π3 β¦ . (π₯ β ππ)ππ?
k. What is a local extrema of a polynomial function?
l. Does every polynomial function have local extreme points? Explain your answer.
2. Sketch the graph of the functions below label all π₯-intercepts, π¦-intercepts, label all local extrema
points, show end behavior.
a. π(π₯) = π₯2(π₯ β 1)(π₯ + 2)3
π₯-intercepts are ______________
π¦-intercept is ____________
End Behavior like _______________
Number of local extremum _______
Graph of the polynomial
b. β(π₯) = π₯(π₯ + 1)(π₯ + 4)(π₯ β 2)(π₯ β 5)
π₯-intercepts are ______________
π¦-intercept is ____________
End Behavior like _______________
Number of local extremum _______
Graph of the polynomial
c. π(π₯) = (π₯ + 1)3(π₯ β 1)(π₯ β 2) d. β(π₯) = (π₯ β 1)(π₯ + 1)2(π₯ + 2)2(π₯ β 3)
Graphing Functions and Relations Page 184
π₯-intercepts are ______________
π¦-intercept is ____________
End Behavior like _______________
Number of local extremum _______
Graph of the polynomial
π₯-intercepts are ______________
π¦-intercept is ____________
End Behavior like _______________
Number of local extremum _______
Graph of the polynomial
3. Determine the interval(s) on which the function is (strictly) increasing, or decreasing, or constant.
Write your answer in interval notation.
A.
Increasing: ___________ Decreasing: ___________ Constant: ___________
B.
Increasing: ___________ Decreasing: ___________ Constant: ___________
Graphing Functions and Relations Page 185
C.
Increasing: ___________ Decreasing: ___________ Constant: ___________
D.
Increasing: ___________ Decreasing: ___________ Constant: ___________
4. Use the graph of the function π below to find (If there is more than one answer, separate them
with commas)
I. All values of π₯ at which π has local minimum
II. All values of π₯ at which π has local maximums
III. All local minimum values of π
IV. All local maximum values of π.
a.
b.
Graphing Functions and Relations Page 186
c.
d.
5. Find all the π₯-intercepts and π¦-intercepts of the functions below. If there is more than one
answer, separate them with commas.
A. π₯-intercepts _________ π¦-intercepts _________
B. π₯-intercepts _________ π¦-intercepts _________
C. π(π₯) = (π₯ + 3)(π₯ β 1)2(π₯ + 5) π₯-intercepts _________ π¦-intercepts _________
Graphing Functions and Relations Page 187
D. (factor by grouping) π(π₯) = 2π₯3 + 2π₯2 β 18π₯ β 18 π₯-intercepts _________ π¦-intercepts _________
E. π(π₯) = π₯4 β 4π₯2 β π₯3 + 4π₯ π₯-intercepts _________ π¦-intercepts _________
6. Mark the end behavior of the graph of each polynomial function below.
A. π(π₯) = (π₯ + 3)(π₯ β 1)2(π₯ + 5)
Left Right
Falls
Rises
Falls Rises
B. π(π₯) = β2π₯3 + 2π₯2 β 18π₯ β 18
Left Right
Falls
Rises
Falls Rises
E. π(π₯) = 5π₯11 β 4π₯2 β π₯3 β 2π₯
Left Right
Falls
Rises
Falls Rises
F. π(π₯) = β4π₯32 β 5π₯20 + 4π₯ β 1
Left Right
Falls
Rises
Falls Rises
Graphing Functions and Relations Page 188
7. Match the graphs below with the functions listed here.
I. π(π₯) = β3(π₯ + 1)2(π₯ + 3)2
II. β(π₯) = π₯2(π₯ β 2)3(π₯ + 1)
III. π(π₯) =1
2(π₯3 β π₯2 β 6π₯)
IV. π(π₯) = (π₯ β 1)(π₯ + 1)(π₯ β 2)
A.
B.
C.
D.
Graphing Functions and Relations Page 189
8. For the graphs below determine the following
I. What is the sign of the leading coefficient of the polynomial function (positive, negative
or not enough information?
II. Which of the following is a possibility of the degree of the function? Choose all that
apply 4, 5, 6, 7, 8, 9.
A.
B.
C.
9. Find a possible equation that represents the graphs below.
a.
b.
c.
d.
Graphing Functions and Relations Page 190
2.4 Graphing Quotients of Functions
Rational Functions We introduced the rational functions briefly in chapter 1.
Quotient of Function: A quotient of function is defined as the ratio of two functions denoted as
π (π₯) =π(π₯)
π(π₯), forl functions π(π₯), and π(π₯) with domain all real numbers π₯ for which π(π₯) β 0.
Rational Function: A rational function is defined as the ratio of two polynomial functions denoted as
π (π₯) =π(π₯)
π(π₯), for polynomial functions π(π₯), and π(π₯) with domain all real numbers π₯ for which
π(π₯) β 0.
Linear Asymptotes of Rational Functions
Rational function graphs can have asymptotic behavior as π₯ β Β±β and also as π₯ β π for where the
denominator function is zero at π₯ = π. The curves that the graph of π (π₯) gets close to as π₯ or π¦
approach Β±β are called asymptotes of the function. Weβll look primarily at asymptotes that are lines
which can be vertical, horizontal, or oblique (slanted).
Vertical Asymptote: A vertical line of the form π₯ = π is called a vertical asymptote to a function
π¦ = π(π₯) if and only if the function π(π₯) approaches either β or ββ as π₯ approaches π either
from the left or right. These are always caused by the denominator polynomial π(π₯) being zero
at π₯ = π.
Horizontal or Oblique Asymptotes: A line π¦ = ππ₯ + π is called a horizontal (when π = 0) or
oblique (π β 0) asymptote to a function π¦ = π(π₯) if and only if the distance between the
function π(π₯) and π¦ = ππ₯ + π approaches zero as π₯ approaches either β or ββ.
Example: Let us review the graphing of the simplest example of a rational function π (π₯) =1
π₯ .
a. Sketch the graph of π (π₯).
b. What is domain of this function?
c. What linear asymptotes does this function have?
Play with it to see if you can recall what we did before looking on the next page.
Graphing Functions and Relations Page 191
We start by plotting a bunch of points and this will help us understand where and why the function has
asymptotes. As we can see the denominator cannot be zero, so the graph wonβt have a point on the π¦-
axis. Thus the graph will be in two pieces, one on either side of the π¦-axis.
π₯ π¦ =
1
π₯
π₯ π¦ =
1
π₯
β0.1 1
β0.1= β10
β10 1
β10= β0.1
β0.01 1
β0.01= β100
β100 1
β100= β0.01
β0.001 1
β0.001= β1000
β1000 1
β1000= β0. .001
0.1 1
0.1= 10
10 1
10= 0.1
0.01 1
0.01= 100
100 1
100= 0.01
0.001 1
0.001= 1000
1000 1
1000= 0. .001
Domain: All nonzero real numbers
The left table above shows how, as π₯ gets close to where the denominator is zero, the π¦-values blow up. That means π¦ approaches either +β when π₯ is small but positive, or π¦ β ββ when π₯ is small but
negative. The π¦-axis is a vertical asymptote of the function π(π₯) =1
π₯. (There will be a vertical
asymptote for any rational function at every π₯-value where the denominator is zero, but the numerator polynomial β 0.) The right table above shows how as π₯ goes to positive infinity, then the π¦-values are small but positive and as to π₯ β ββ then π¦ is small but negative. The π₯-axis a horizontal asymptote of the function
π(π₯) =1
π₯. Not every rational function has a horizontal asymptote. Also, some rational functions have a
horizontal asymptote not at the π₯-axis, but at some other non-zero π¦-value.
Graphing Functions and Relations Page 192
Playing
We can generalize our observations from plotting the function π (π₯) =1
π₯. Note that for (π₯) =
1
π₯π ,
π¦ =1
π₯π β 0 when π₯ β Β±β for any positive integer π. If you continue your study of functions in a
calculus course, limit notation makes these statements much more succinct as indicated below.
limπ₯ββ
1
π₯π= 0 πππ lim
π₯β ββ
1
π₯π= 0
Also note that
a) When π is even, π(π₯) =1
π₯π β β when π₯ β 0 from both sides of 0.
b) When π is odd, π(π₯) =1
π₯π β β when π₯ β 0 from the right and π(π₯) =1
π₯π β ββ when π₯ β 0 from
the left side .
Observation: π(π₯) =1
π₯π, where π is a counting number has a vertical asymptote is at π₯ = 0 and
horizontal asymptote at π¦ = 0.
Using our knowledge from transformation of functions we can see that for a counting number π, the
following statements can be made-
1. For (π₯) =1
(π₯βπ)π , the vertical asymptote is at π₯ = π, and the horizontal asymptote is at π¦ = 0.
2. For (π₯) =1
π₯π + π , the vertical asymptote is at π₯ = 0, and the horizontal asymptote is at π¦ = π.
3. For (π₯) =1
(π₯βπ)π + π , the vertical asymptote is at π₯ = π, and the horizontal asymptote is at π¦ = π.
4. For π(π₯) =1
(π₯βπ)π + ππ₯ + π, the vertical asymptote is at π₯ = π, and the oblique asymptote is at
π¦ = ππ₯ + π. The reason the oblique asymptote is π¦ = ππ₯ + π is the fact that 1
(π₯βπ)π β 0 when
π₯ β Β±β making the vertical distance between π(π₯) =1
(π₯βπ)π + ππ₯ + π and π¦ = ππ₯ + π go to zero
when π₯ β Β±β.
5. For π(π₯) =1
(π₯βπ)π + π(π₯), where π(π₯) is a polynomial function, the vertical asymptote is at π₯ = π,
and π¦ = π(π₯) is a polynomial function asymptote when π₯ β β or π₯ β ββ. This means that the
graph of π¦ = π(π₯) gets very close to the polynomial graph of π¦ = π(π₯) as π₯ β Β±β.
Graphing Functions and Relations Page 193
Letβs take a look at all the scenarios of what a rational function graph may look like when you only have
linear asymptotes. We want you to explore all possibilities of what the graph of a rational function may
do if you knew the asymptotes.
Sketch the possibilities below for how a rational function could fit the three different asymptote
scenarios shown below.
a) π¦ = π is a horizontal asymptote, what are our choices?
b) π₯ = π is a vertical asymptote, what are our choices?
c) π¦ = ππ₯ + π is an oblique asymptote, what are our choices?
Do you think you exhausted all possibilities?
Graphing Functions and Relations Page 194
a) For the horizontal asymptote π¦ = π the graph of the function eventually must approach the line either from the above or the below. Those are the only choices as shown below.
b) For the vertical asymptote π₯ = π the graph of the function must approach this line from the left or right and shoots up or down. Those are the only choices as shown below.
c) For the oblique asymptote π¦ = ππ₯ + π the graph must get closer and closer from above or below as Those are the only choices as shown below as π₯ β Β±β.
So now lets us look at how we can put these observations together to imagine the different possible
graphs we can have with multiple given linear asymptotes.
Graphing Functions and Relations Page 195
Below are some possible graphs we can imagine if we had a vertical asymptote of π₯ = π, and a horizontal asymptote of π¦ = π.
a)
b)
c)
d)
e)
f)
As you can see there are countless possibilities. The last two may seem not possible but there can be
finite number of intersections of the graph of the rational function and the horizontal asymptotes. In
other words, π¦ = π is a horizontal asymptote as long as the graph eventually stays closer and closer to
the line π¦ = π. You can also see that the left side of the graph in the last two examples gets to the
vertical asymptote a little slower than the right side. So it is actually a lot of fun to graph these. In
graphing polynomial functions we find the asymptotes, the intercepts, the points where the graph
intersects the horizontal asymptotes, and then make a sign chart to see what some of the values of the
function are in the domain, we will get a pretty good rough sketch.
Graphing Functions and Relations Page 196
Below are some possible graphs where π₯ = π is the vertical asymptote, and π¦ = ππ₯ + π is an oblique asymptote.
a)
b)
c)
d)
e)
f)
Again, the number of possible shapes is large. The complexity of the numerator and denominator functions lead to more asymptotes and potentially more local max/min points. The asymptotes are very important and provide a framework for the graph and suggest what π₯-values we would want to include in a small π‘-table of points to plot.
Graphing Functions and Relations Page 197
Reciprocal of a Function
Playing
How should one even begin to understand the graphs of rational functions? Mathematicianβs start with
known facts and work their way up. We know that a rational function is a ratio of two polynomials. We
have already seen (in the previous section) how to get a rough sketch of a polynomial function that is in
the factored form.
We now know how a graph of the function π¦ =1
π₯π , π β₯ 1 looks like. So letβs investigate what happens to
the graph of a polynomial function when we invert it.
Graphing Functions and Relations Page 198
Practice Examples
1. Compare the graphs π(π₯) = (π₯ β 1)(π₯ + 2) and π¦ =1
π(π₯)=
1
(π₯β1)(π₯+2)
The graph of π¦ = π(π₯) = (π₯ β 1)(π₯ + 2) is a parabola with π₯-intercepts at π₯ = 1, β2. To graph π¦ =1
π(π₯)
we can see right away the domain will change since the denominator cannot be zero. The domain of the
function π¦ =1
(π₯β1)(π₯+2) is in three pieces: (ββ, β2) βͺ (β2,1) βͺ (1, β). The sign of π¦ = π(π₯) and
π¦ =1
π(π₯) is the same since the numerator is always positive. That means
1
π(π₯)> 0 for all π₯ < β2 and
π₯ > 1. At values of π₯ closer to 1 but above 1 the graph will approach +β, and the same behavior will
be noticed when values of π₯ are closer to β2 but π₯ < β2. Also 1
π(π₯)< 0 for all β2 < π₯ < 1. So the
graph will approach ββ when values of π₯ are closer to π₯ = 1 and β2 but β2 < π₯ < 1.
We also know that the end behavior of the parabola π(π₯) = (π₯ β 1)(π₯ + 2) is similar to π¦ = π₯2 which
means the end behavior of the function π¦ =1
π(π₯)=
1
(π₯β1)(π₯+2) is going to be like π¦ =
1
π₯2 . The graph of
π¦ =1
(π₯β1)(π+2) is similar to π¦ =
1
π₯2 as π₯ β Β±β making the line π¦ = 0 a horizontal asymptote.
Graphing Functions and Relations Page 199
So you can see that for the graph of π¦ =1
π(π₯) , the places where the graph of π¦ = π(π₯) shoots to infinity,
the graph of π¦ =1
π(π₯) goes to zero, and where the graph of π¦ = π(π₯) is zero then π¦ =
1
π(π₯) will have
vertical asymptotes and the signs are preserved. Thus if π(π₯) β +β, then π¦ =1
π(π₯) will be a small (+)
number.
In general let us investigate what happens to the graph of a function and its reciprocal.
2. Sketch the graphs of π¦ =1
π(π₯) where you are given the graphs of π¦ = π(π₯). The principles we saw
above are valid here even when the function π(π₯) is not a polynomial function.
a.
b.
c.
d.
This function is a periodic function with domain (ββ, β) and is not a polynomial function.
Try to work these out before peeking ahead.
Graphing Functions and Relations Page 200
Solutions to problem 2
2. Sketch the graphs of π¦ =1
π(π₯) where you are given the graphs of π¦ = π(π₯). The principles we
observed below are preserved even of the function π(π₯) is not a polynomial function.
a.
b.
Graphing Functions and Relations Page 201
c.
Since the original function π¦ = π(π₯) has no π₯-intercepts,
the graph of π¦ =1
π(π₯) has no vertical asymptotes.
d.
Zeros of the Numerator and Denominator
We now need to see how the numerator polynomial influences the graph of a rational function. We
would predict that the basic features of the numerator polynomial function for π₯-values that are close to
the π₯-intercepts of the numerator will be preserved. A mathematician should always venture an
intuitive guess based on previous experience and check out a few examples, and then hypothesize a
theory that can be proven using mathematical principles. Let us use graphing utilities to see a few
examples.
Below is the process a mathematician might employ to get a grip on how the graph of
π(π₯) = (π₯β1)2
(π₯+3)3
π₯ might compare to the graph of the function π(π₯) = (π₯ β 1)2(π₯ + 3)3.
In the previous section we saw how to graph π(π₯) = (π₯ β 1)2(π₯ + 3)3 , we also know from the
previous section that if we wanted to graph π(π₯) = 2 (π₯ β 1)2(π₯ + 3)3 or β(π₯) =1
2 (π₯ β 1)2(π₯ + 3)3,
Graphing Functions and Relations Page 202
the basic graph would not change except for a scaling factor of 2 or 1
2 respectively as shown below. If we
wanted negative scaling factors , the original graph would reflect across the π¦-axis along with stretching.
So a natural question to ask is what is the effect of a variable scaling factor of say 1
π₯ . If we can answer
that, we would know how the graph of π(π₯) =1
π₯( (π₯ β 1)2(π₯ + 3)3 ) looks like compared to
graph of the function π(π₯) = (π₯ β 1)2(π₯ + 3)3 . Our intuitive guess would be that the basic
numerator features should be the same except for a variable stretching near the π₯-intercepts, and we
would also have to account for a negative scaling factor when variable scaling factor 1
π₯< 0 which would
then reflect that part of the graph across the π¦-axis. So in the vicinity of the π₯ = 1, and π₯ = β3 we
expect the graph of the rational function π to be much like the numerator graph π, but with stretch
factors at 1
1= 1 near π₯ = 1 and a stretch factor of β
1
3 near π₯ = β3. Also, given our work with
functions π¦ =1
π₯π we know that near π₯ = 0 the graph of π must blow with π¦ β β or π¦ β ββ.
So letβs now look at our graph using a graphing utility to see if our analysis and intuitive guess is
consistent with actual graphs. After this example you might want to play on your own and explore more
functions to strengthen your intuition.
Graphing Functions and Relations Page 203
π(π) = (π β π)π(π + π)π
π(π) =(π β π)π(π + π)π
π
The two functions π(π) =(πβπ)π(π+π)π
π and π(π) = (π β π)π(π + π)π together so you can
compare.
As you can see the basic features of the cubic and squares parts of the numerator got preserved. So the
denominator of π₯ in the rational function can be thought of as providing a local stretch factor to just
stretch or compress or reflect the shape of the graph of the numerator polynomial
Graphing Functions and Relations Page 204
We will see that for all rational functions, near the zeros of the numerator function the graph behaves
much like the graph of the numerator polynomial. Also at each zero of the denominator function the
graph has a vertical asymptote and behaves like 1
(π₯βπ)π with a possible vertical stretch or reflection.
Let us now consider the same numerator polynomial but with the denominator being π₯2 i.e., apply the
variable stretch factor of 1
π₯2 . Note that at π₯ = 0 the graph should have the shape of 1
π₯2 and at
π₯ = 1 πππ β 3 the graph should behave like the numerator π¦ = (π₯ β 1)3(π₯ + 3)2, but with stretch
factors 1
12 = 1 and 1
(β3)2 =1
9.
π(π) = (π β π)π(π + π)π
π(π) =(π β π)π(π + π)π
ππ
The two functions π(π) =(πβπ)π(π+π)π
ππ and π(π) = (π β π)π(π + π)π together so you can
compare.
Graphing Functions and Relations Page 205
We can of course play with more examples to really get a good feel and so we will let you play with it.
Try to be creative when using the graphing utility to explore.
One more example:
π(π) = (π β π)π(π + π)π
π(π) =(π β π)π(π + π)π
(π β π)(π + π)
We plot functions π(π) and π(π) together so to see the similarities near the zeros of π(π).
π = βπ, and π = π are shown as dotted lines as a reference only. This way you can see the
blow up of the graph near those values clearly.
Zoomed in
Graphing Functions and Relations Page 206
End Behavior of a Graph of a Rational Function
The top and bottom polynomials also determine how the graph behaves as π₯ goes off to Β±β. Long
division allows us to rewrite π(π₯) in the form (π₯) =π(π₯)
π(π₯)= π(π₯) +
π (π₯)
π(π₯) , where π(π₯) =Quotient, and
π (π₯) = remainder with degree less than that of π(π₯). The term π (π₯)
π(π₯) will always tend to zero as
π₯ β Β±β because the degree of π (π₯) < degree of π(π₯) and each of these polynomials behave like their
highest degree terms as π₯ β Β±β.
See below for the examples we just finished looking at.
π(π₯) =(π₯ β 1)2(π₯ + 3)3
π₯
= π₯4 + 7π₯3 + 10π₯2 β 18π₯ β 27 +27
π₯
A crude end-behavior is to use just the highest degree
terms of the top and bottom so that π¦ βπ₯5
π₯= π₯4
as π₯ β Β±β.
π(π₯) =(π₯ β 1)2(π₯ + 3)3
π₯2
= π₯3 + 7π₯2 + 10π₯ β 18
+β27π₯ + 27
π₯
Crude end-behavior is π¦ βπ₯5
π₯2= π₯3 as
π₯ β Β±β.
π(π₯) =(π₯ β 1)2(π₯ + 3)3
(π₯ + 1)(π₯ β 2)
= π₯3 + 8π₯2 + 20π₯ + 18
+31π₯ + 63
(π₯ + 1)(π₯ β 2)
Crude end-behavior is π¦ βπ₯5
π₯2= π₯3 as
π₯ β Β±β.
Graphing Functions and Relations Page 207
Practice examples
Use the graphs of the numerators and denominators provided to sketch a rough graph of the function
π¦ =π(π₯)
π(π₯)
See example below
Let π(π₯) = (π₯ + 1)2(π₯ β 3), and π(π₯) = (π₯ β 1)(π₯ + 3). To sketch the graph of π¦ =(π₯+1)2(π₯β3)
(π₯β1)(π₯+3)
we can use what we have learnt above and put it together.
Note that expanding and doing long division gives: π¦ =(π₯+1)2(π₯β3)
(π₯β1)(π₯+3)= π₯ β 3 +
4π₯β12
(π₯β1)(π₯+3).
Thus the end-behavior is π¦ = π₯ β 3 is the oblique asymptote. Note the crude estimate of the end-
behavior is π¦ =π₯3
π₯2 = π₯
In these examples we have accumulated some basic tools for connecting information on the zeros and
degrees of the numerator and denominator polynomials of a rational function to features of its graph.
You might ask why do we need to learn how to graph a rational function without a graphing utility? This
is a good question and one answer is that the effective use of a graphing utility requires selecting an
appropriate scale or viewing window to capture the main features. Understanding what equation
Graphing Functions and Relations Page 208
features cause certain graphical behaviors also allows one to come up with formulas for π(π₯) when
given its graph.
We can summarize our observations as following
The Graph of a rational function is influenced by three factors:
The graph behaves like that of the Numerator polynomial near the zeros of the numerator.
The end behavior as π₯ β Β±β is determined by the quotient function from long division of
the rational function. A crude version of this is the power function that is the ratio of the
highest degree term of the numerator over the highest degree term of the denominator.
The zeros of the denominator polynomial can cause vertical asymptotes at these π₯-values
with the behavior like π¦ = πΆ β 1
(π₯βπ)π near each zero π₯ = π of multiplicity π.
Steps to Graph a Rational Function
Let π(π₯) =π(π₯)
π(π₯) be a rational function in lowest terms. We basically look at the three items above to
plot asymptotes, behavior near π₯-intercepts and near vertical asymptotes and the end-behavior. We
then sketch the graph to accommodate these local features..
Step 1 Use long division to write (π₯) =π(π₯)
π(π₯)= π(π₯) +
π (π₯)
π(π₯) , where π(π₯) =Quotient, and
π (π₯) =remainder.
This tells us that as π₯ β Β±β the rational function π(π₯) will behave very much like π(π₯). The equation
π¦ = π(π₯) will is the horizontal or oblique asymptote, or possibly an asymptotic polynomial function.
Note that whenever the degree of numerator π(π₯) is less than the degree of denominator π(π₯),
then the quotient π(π₯) = 0 and there is a horizontal asymptote at the π₯-axis.
When the degree of the numerator and denominator is the same
π(π₯) =πππππππ πππππππππππ‘ ππ π‘βπ π(π₯)
πππππππ πππππππππππ‘ ππ π‘βπ π(π₯) so there is a horizontal asymptote at this value of π¦.
Step 2 Set denominator S(π₯) = 0 to find all the vertical asymptotes (places where the rational function
blows up).
Step 3 Setting the numerator π(π₯) = 0 and factoring it will give you all of the π₯-intercepts. This will
help you get the basic shape near the π₯-axis intercepts.
Step 4 Setting π₯ = 0 in the function π(π₯) will give you the π¦-intercept.
Step 5 Setting the remainder π (π₯) = 0 will give you points where the function intersects the horizontal
or oblique asymptotes if any. This provides fine detail that may not be needed for a rough sketch.
Step 6 Vertical asymptotes break the graph into separate pieces. It is useful to plot a table of points
with a couple of points from each piece of the graph.
Step 7 Use the previous steps to make a rough sketch and label all important parts of the graph.
Graphing Functions and Relations Page 209
Practice Problems
The first one is done for you, please attempt the others on your own.
1. For all the rational functions below please find the requested information below and then graph the
function. In your graph show all the relevant information and label all parts of the graph. Draw the
asymptotes if any as dotted lines.
a) End behavior by long division.
b) Vertical asymptotes where bottom is zero.
c) All the π₯-intercepts where top is zero.
d) π¦-intercept by plugging in π₯ = 0
e) Solve for π (π₯) = 0 from a) to locate points (if any) where the graph intersects the horizontal or
oblique asymptotes.
f) Plot two points on either side of the vertical asymptotes and any additional points as needed to
finally sketch your graph.
I. π(π₯) =1
π₯+2
a) End behavior by long division. Nothing to do here as the numerator is already of degree less than that of the denominator.
Function rewriten as πππππππππ
π·ππππππππ‘ππ+ ππ’ππ‘ππππ‘
π(π₯) =1
π₯+2+ 0 that means the π₯-axis is
horizontal asymptote.
b) Denominator is zero at π₯ = β2. Vertical Asymptotes: π₯ = β2
c) π₯-intercepts: None since 1 β 0 for all π₯.
d) π¦-intercept is at π(0). π¦-intercept π¦ = π(0) =1
0+2=
1
2
e) Points of intersection with the horizontal or oblique asymptotes: None since the remainder is π (π₯) = 1 β 0 ever.
f)
π₯
π¦ = π(π₯) =1
π₯ + 2
β4 1
β4 + 2= β
1
2
β3 1
β3 + 2= β1
0 1
0 + 2=
1
2
1 1
1 + 2=
1
3
Graphing Functions and Relations Page 210
II. π(π₯) =3π₯2+3π₯β18
π₯+1
Before we can answer anything let us do long division so it will allows to find the oblique and
horizontal asymptotes easily
3π₯
π₯ + 1 3π₯2 +3π₯ β18 β (3π₯2 +3π₯)
β18
a) Horizontal or Oblique Asymptote Oblique Asymptote is π¦ = 3π₯ Domain of the function: All real numbers except for π₯ = β1.
Function rewrite as πππππππππ
π·ππππππππ‘ππ+ ππ’ππ‘ππππ‘
π(π₯) =3π₯2 + 3π₯ β 18
π₯ + 1= 3π₯ β
18
π₯ + 1
b) Bottom is zero at π₯ = β1 Vertical Asymptotes: π₯ = β1
c) d) c) 3π₯2 + 3π₯ β 18 = 3(π₯ β 2)(π₯ + 3) = 0 giving us π₯ = β3 or π₯ = 2
π₯-intercepts: (β3, 0), and (2, 0)
d) π¦-intercept is (0, β18), π¦ = π(0) =3(0)2+3(0)β18
0+1=
β18
1= β18
e) Points of intersection with the horizontal or oblique asymptotes: None since π (π₯) = β18.
π(π₯) =3π₯2 + 3π₯ β 18
π₯ + 1= 3π₯ β
18
π₯ + 1
Long division gives us
Graphing Functions and Relations Page 211
Plot two points on either side of the vertical
asymptote.
π₯ π¦ =
3π₯2 + 3π₯ β 18
π₯ + 1
β2 3(β2)2 + 3(β3) β 18
β2 + 1=
β12
β1= 12
β3 3(β3)2 + 3(β3) β 18
β3 + 1= 0
0 3(0)2 + 3(0) β 18
0 + 1=
β18
1= β18
1 3(1)2 + 3(1) β 18
1 + 1=
β12
2= β6
2 3(2)2 + 3(2) β 18
2 + 1=
0
3= 0
f) Plot two points to graph the
horizontal asymptote π¦ = 3π₯.
π₯ π¦ = 3π₯ 0 0 2 6
Graphing Functions and Relations Page 212
iii. π(π₯) =2π₯2+1
(π₯β2)(π₯+3)=
2π₯2+1
π₯2+π₯β6
a) End Behavior: Horizontal asymptote at π¦ = π(π₯) = 2.
We can also get this from knowing that the
π(π₯) β2π₯2
π₯2 = 2 for when π₯ approaches Β±β.
Function rewriten as πππππππππ
π·ππππππππ‘ππ+ ππ’ππ‘ππππ‘
π(π₯) = 2 +β2π₯ + 13
(π₯ β 2)(π₯ + 3)
b) Bottom = 0 at π₯ = 2 πππ π₯ = β3 Vertical Asymptotes: π₯ = 2, π₯ = β3
c) π₯-intercepts: None since 2π₯2 + 1 β 0 for any real π₯.
d) π¦-intercept π¦ = π(0) =1
(β2)(3)=
1
β6
e) Points of intersection with the horizontal asymptotes: β2π₯ + 13 = 0 2π₯ = 13
π₯ =13
2
π (13
2) = 2
f) Points on the graph
π₯ π¦ β5 2(β5)2 + 1
(β5)2 β 5 β 6=
51
14
β4 2(β4)2 + 1
(β4)2 β 4 β 6=
33
6
β1 2(β1)2 + 1
(β1)2 β 1 β 6= β
3
6
= β1
2
0 2(0)2 + 1
(0)2 + 0 β 6= β
1
6
1 2(1)2 + 1
(1)2 + 1 β 6= β
3
4
3 2(3)2 + 1
(3)2 + 3 β 6=
19
6
4 2(4)2 + 1
(4)2 + 4 β 6=
33
14
Graphing Functions and Relations Page 213
g) Graph with relevant information
Graphing Functions and Relations Page 214
Section2.4 Worksheet Date:______________ Name:__________________________
Concept Meaning in words and/or examples as required
1. Quotient Function
2. Rational Function
3. Domain of a rational function
4. Range of rational function
5. Vertical Asymptote
6. Oblique or Horizontal Asymptote
7. Points of intersection of a function with its horizontal, or oblique asymptote
Difficulties encountered in the section:
Graphing Functions and Relations Page 215
Exercises 2.4
1. In your own words describe what role the numerator function and the denominator function
play in a quotient function. Give examples to demonstrate your ideas.
2. How do you locate vertical asymptotes? Why does your method work?
3. Can the graph of a rational function intersect a vertical asymptote? Why or why not?
4. How do you determine the horizontal or oblique asymptotes?
5. What would the graph of the function π(π₯) =1
π₯+ π₯2 look like?
6. If the numerator and denominator have common factors in a rational function, how does that
affect the graph of the rational function?
7. How do you find π₯-intercepts and π¦-intercepts of a rational function?
Graphing Functions and Relations Page 216
8. If you knew all the asymptotes of the rational function, would there be a unique function that
corresponds to these asymptotes? What other information would you need?
9. For the functions π¦ = π(π₯) below, draw rough sketches their reciprocal function graphs π¦ =1
π(π₯)
and clearly label any horizontal and vertical asymptotes.
a.
b.
This function with a repeating pattern is called a βPeriodic functionβ and is defined on (ββ, β).
c.
d.
Graphing Functions and Relations Page 217
10. Draw a rough graph the function =π(π₯)
π(π₯) , given the graphs of π¦ = π(π₯), and π¦ = π(π₯). Locate all
vertical asymptotes and π₯-intercepts.
π¦ = π(π₯)
π¦ = π(π₯)
a.
b.
This is not a polynomial but a periodic function defined on (ββ, β)
This is not a polynomial but a periodic function defined on (ββ, β)
c.
Graphing Functions and Relations Page 218
11. For all the rational functions below find the requested information below and then graph the
function. In your graph show all the relevant information and label all parts of the graph. Draw
the asymptotes if any as dotted lines.
I. Perform long division and write the rational function ππ’πππππ‘ππ
π·ππππππππ‘ππ=
πππππππππ
π·ππππππππ‘ππ+ ππ’ππ‘ππππ‘ to
get the end-behavior.
II. Locate the vertical asymptotes.
III. Locate the π₯-intercepts and the π¦-intercept if possible
IV. Make a T-table to include a couple of points from each piece of the graph in between vertical
asymptotes.
V. Find all points (if any) where the graph intersects the horizontal or oblique asymptotes.
a) π(π₯) =1
π₯β3
b) π(π₯) =1
(π₯β3)2
c) π(π₯) =1
(π₯β3)3
d) π(π₯) =2
π₯β3
e) π(π₯) =2
(π₯β3)2
f) π(π₯) =β2
(π₯β3)3
g) π(π₯) =π₯β1
π₯β3
h) π(π₯) =π₯β1
(π₯β3)2
i) π(π₯) =1
(π₯β3)(π₯+3)
j) β(π₯) =π₯+1
(π₯β3)(π₯+3)
k) π (π₯) =π₯+1
(π₯+2)2(π₯β1)
l) π(π₯) =1
π₯2β4
m) π(π₯) =(π₯+2)2(π₯β4)
(π₯β1)
n) π¦ =3π₯2+4π₯+1
π₯+2= 3π₯ β 2 +
5
π₯+2
o) π(π₯) =2π₯2+1
π₯β1= 2π₯ + 2 +
3
π₯β1
p) π(π₯) =3π₯2+4π₯+1
(π₯+2)2 = 3 +β8π₯β11
(π₯+2)2
q) π(π₯) =3π₯3+π₯2+3π₯+1
π₯2β4= 3π₯ + 1 +
15π₯+5
π₯2β4
r) π(π₯) =3π₯3β3π₯β1
(π₯+1)2 = 3π₯ β 6 +6π₯+5
(π₯+1)2
s) π(π₯) =π₯+1
π₯2β4
t) π(π₯) =3π₯β2
π₯2β4
u) π(π₯) =π₯2+2π₯+1
π₯2β1
v) π(π₯) =π₯2β1
π₯2+2π₯+1
w) π»(π₯) =π₯2β1
π₯+1
x) π(π₯) =π₯3βπ₯2+4π₯
π₯β1
y) π (π₯) = π₯2 β1
π₯
z) π(π₯) =ππ₯
π₯
Graphing Functions and Relations Page 219
12. Choose the graphs below that match the graphs of the rational functions below.
I. π(π₯) =3π₯+3
π₯2+2π₯β3
II. β(π₯) =3
π₯2+2π₯β3
Graph A
Graphing Functions and Relations Page 220
13. For all the graphs below find a rational function representing it. Use the lowest degree numerator
and denominator to achieve the given vertical asymptotes and π₯-axis intercepts. Is there a unique
formula for all of them, or can some of the graphs below could have more than one formula?
Explain your reasoning clearly.
a.
b.
Graphing Functions and Relations Page 221
c.
14. For the functions below, please do the following.
I. Find the domain of each function.
II. Determine all the vertical asymptotes.
III. Determine the end behavior of the function.
IV. Use all the information found above to sketch the functions.
a. π¦ =βπ₯β1
π₯+1
b. π¦ = βπ₯ β 1 +1
π₯+1
c. π¦ = 2π₯2 β 4 +1
π₯β1
d. π¦ = 2π₯2 + 1 β1
π₯2β1
15. Create a rational function with vertical asymptotes at π₯ = β2, π₯ = 1 and slant asymptotes at
π¦ = β3π₯ + 5.
16. Create a quotient of function π¦ =π(π₯)
π(π₯) with a vertical asymptote at π₯ = 2, and so that as
π₯ β Β±β, π¦ β 3π₯2 + 5π₯ β 1.
Solving Equations and Inequalities Page 222
Chapter 3 Solving Equations and Inequalities We saw in chapter two when graphing functions that finding π₯-values where a function output or the π¦-
coordinate on the graph was equal to zero provided the location of π₯-intercepts of the graph of the
function graph. When we know the input to a function, it is often straightforward to just plug in the π₯-
value to obtain the output. It turns out that the reverse question comes up in many modeling
situations. That is, we might have some function formula that we have constructed to model a situation
and we want to know what input is required so that the output will be equal to some stated value.
Algebraically this corresponds to solving an equation π(π₯) = πΆ where we seek π₯ so the output is πΆ.
When finding π₯-intercepts, this πΆ = 0 and we use the solutions to plot polynomial and rational
functions. In this chapter we investigate and develop tools for solving equations of the form π(π₯) = πΆ
for all of the types of functions π we have worked with.
Most of you have been exposed to pre-requisite materials before getting into College Algebra. So we
will assume you are familiar with solving basic linear and some non-linear equations and inequalities. If
you feel you need a refresher, we suggest studying Module 3 in the Developmental and Intermediate
Algebra e-text (Go to page 361).
3.1 Quadratic Equations Recall that
An equation is a statement asserting the equality of two mathematical objects.
An inequality is statement asserting that one mathematical object is larger than or equal to, or
less than or equal to another mathematical object.
So we can create equations as indicated above by setting π(π₯) = πΆ or we could have both objects be
outputs of different functions e.g., set π(π₯) = π(π₯).
Examples of equations and inequalities
1. 2π₯ = 3
2. 2π₯+3 = 24π₯β1
3. π₯2 β 3π₯ = 5(π₯ + 1) β 3
4. 2π₯ < 3.5
5. 2π₯+3 β₯ 24π₯β1
6. π₯2 β 3π₯ β€ 5(π₯ + 1) β 3
Solving Equations and Inequalities Page 223
The symbols above can be used as follows:
1 The equation π = π signifies that the mathematical object π is equal to the mathematical
object π.
2 The inequality π < π signifies that the mathematical object π is smaller than or less than the
mathematical object π.
3 The inequality π > π signifies that the mathematical object π is bigger than or greater than the
mathematical object π.
4 The inequality π β€ π signifies that the mathematical object π is smaller than or equal to, or less
than or equal to, the mathematical object π.
5 The inequality π β₯ π signifies that the mathematical object π is bigger than or equal to, or
greater than or equal to, the mathematical object π.
The symbols above are our shorthand way of writing comparisons between two mathematical objects.
This is another example of the symbolic nature of writing mathematics. It is much easier to write π₯ β€ 2
than to have to say βa number that is less than or equal to 2β, or a number that is not larger than 2β.
Any number(s) when substituted for the variable(s) in the original equation or inequality that
results in a true statement is called a solution to that equation or inequality.
The process in which we use mathematical properties of equality or inequality respectively to
isolate the variable by itself is called solving the equation or inequality.
All real number solutions to an equation or inequality in one variable can be represented on a real
number line. When we equate two non-zero functions like π(π₯) = π₯2 β 3π₯ = 3π₯ β 1 = π(π₯) we are
trying to see at what π₯ values can the two functions have the same output. Thus solving π(π₯) = π(π₯)
corresponds to finding where the two graphs intersect each other. Equations in one variable can also be
rearranged so that the non-zero terms are on the left side and zero is on the right side. Thus if we set
β(π₯) = βthe left hand sideβ, solving this equation reduces to finding the zeroes of β, i.e., solving the
equation β(π₯) = 0. You can see that equations of the type π(π₯) = π(π₯) can be also be then written as
π(π₯) β π(π₯) = 0, or π(π₯) β π(π₯) = 0.
Real solutions to an equation or an inequality in two variables can be represented as points or regions in
a 2-dimensional Cartesian coordinate system. Solutions to equations of the type π(π₯, π¦) = π(π₯, π¦) can
be thought of as points of intersection of the surfaces π§ = π(π₯, π¦), and π§ = π(π₯, π¦) represented in the
3-dimensional Cartesian coordinate system.
An equation that is true for all values of the variable is called an identity.
The equation π₯ Γ π₯ = π₯2 is an identity. Both sides of the equal sign yield the same number for any real
number value of π₯.
An equation in which you end up with a false statement for all values of the variable is said to
have βNo Solutionβ.
The equation π₯ = π₯ + 2 has no solution since a quantity is never equal to one more than itself.
Solving Equations and Inequalities Page 224
Solving an equation or an inequality is like untying a knot or undoing what was done to the variable to
get it into its current state. We operate on an equation with certain mathematical tools to convert the
equation to a simpler equation, that has the same solutions as the original. (These are called equivalent
equations.) Thus solving an equation involves using tools to work the original equation through a
sequence of equivalent equations to eventually get to a statement like π₯ = βanswerβ.
Certain tools used in the process of isolating the variable can sometimes lead us to a value of
the variable that makes the original equation false. Such a solution is called an extraneous
solution and we will need to watch out for these βfalseβ solutions whenever we use those tools.
Two examples of such tools are:
1. Raising both sides of an equation to a power, e.g., square both sides of an equation.
2. Multiplying or dividing both sides of an equation by an expression that has the potential to be equal
to zero.
When making use of the tools mentioned above, it is best to check our final answers by substituting
their values back into the original equation to see if they really are the solutions.
You can see why these tools may cause an equation to have an extraneous solution by working with a
linear equation say, π₯ = 3. The only solution to this equation is π₯ = 3
a. Look what happens when we square both sides, we get π₯2 = 9.
This equation has two solutions π₯ = 3, π = βπ.
b. Look what happens if we multiply both sides by (π₯ + 2).
π₯(π₯ + 2) = 3(π₯ + 2)
π₯2 + 2π₯ = 3π₯ + 6
π₯2 β π₯ β 6 = 0
(π₯ β 3)(π₯ + 2) = 0
π₯ = 3 and π = βπ.
Even though we did not really need to square or multiply both sides of the equation π₯ = 3 to get the
solutions to it, the process demonstrates how additional solutions got inserted by doing these
operations.
Sometimes when solving polynomial equations like π₯2 = β4 we get solutions that are not real. These
solutions are referred to as complex numbers. In case you are not familiar with complex numbers below
a review of complex numbers.
Review of Complex Numbers
For a long time people believed that the set of real numbers accounted for all the numbers that there
are. However in 1545, Girolamo Cardano and others made progress on solving cubic equations such as
π₯3 β 15π₯ = 4. Their technique gave a solution to this problem that involved ββ121 . The value of this
square root cannot be any real number since any real number when squared will be positive. However
their solution method simplified to a final result of π₯ = 4 which is easily seen to make the above
equation true. This lead to a deeper study of the square roots of negative numbers. Today these kinds of
numbers are absolutely essential in higher mathematics, engineering and physics. A simpler problem
that shows the deficiency of the real numbers is to try to find a real number π₯ such that π₯2 = β1. Play
Solving Equations and Inequalities Page 225
with this for a while to convince yourself that no real number when multiplied by itself will produce the
result of negative one!
So a new kind of number system evolved where π = ββ1 is designated to represent the unit imaginary
(non-real) number that is a solution to the equation π₯2 = β1. Even though it is hard to imagine what π
is, we know that its square is β1 or π2 = π Γ π = β1. With this definition of π we expand the set of real
numbers to the set of complex numbers.
Set of complex numbers is a collection of all numbers of the form π€ = (π + ππ), where π, π
are real numbers and are called the real part and imaginary part respectively of π€. Another
way to represent this sentence in mathematical notation is
πΆ = {π + ππ |π πππ π πππ πππ¦ ππππ ππ’πππππ πππ π2 = β1}.
Note: The set of all real numbers is a subset of the set of complex numbers because every real number can be written as π + 0π, e.g., 2.4 = 2.4 + 0π. In set notation it will look like π β πΆ.
Any time mathematicians find new objects a natural questions to ask is can we can do all the arithmetic
operations we do with real numbers with complex numbers, and do the properties of arithmetic work
the same way. The most natural way to add or subtract would be to combine like terms. To multiply we
use distributive property of multiplication over addition/subtraction. The division is a little more
complex but uses properties of radicals. See examples below to see how the different operations work
with the complex numbers.
Practice Problems
1. Simply the following and write all your answers in the π + ππ form where π, and π are real
numbers.
a) (3 + 5π) + (2 β 3π)
Adding like terms we get (3 + 5π) + (2 β 3π) = (3 + 2) + (5 β 3)π = 5 + 2π
b) (3 + 5π) β (2 β 3π)
(3 + 5π) β (2 β 3π) = (3 β 2) + (5 β (β3))π = 1 + 8π
c) (3 + 5π)(2 β 3π)
Using distributive property and adding like terms we get
(3 + 5π)(2 β 3π) = 3(2) + 3(β3π) + 5π(2) + 5π(β3π)
= 6 β 9π + 10π β 15π2 (recall π2 = β1) giving us
= 6 + π + 15
= 21 + π
d) (2 + 3π)(2 β 3π)
= 4 β 6π + 6π β 9π2
= 4 + 9
= 13
Such pairs of numbers 2 + 3π and 2 β 3π are called complex conjugates of each other.
Solving Equations and Inequalities Page 226
Complex Conjugates: The pair of complex numbers π + ππ and π β ππ are called complex
conjugates of each other and their product is always a real number.
(π + ππ)(π β ππ) = π2 + π2
This result allows us to simplify fractions (representing division) of complex numbers.
e) 3+5π
2β3π Simplify to the form π + ππ.
Since we need to write the division in π + ππ form we multiply numerator and denominator
by the conjugate of 2 β 3π which is 2 + 3π.
3 + 5π
2 β 3π=
(3 + 5π)(2 + 3π)
(2 β 3π)(2 + 3π)=
6 + 9π + 10π + 15π2
4 + 6π β 6π β 9π2
=6 + 19π β 15
4 + 9=
β9 + 19π
13= β
9
13+
19
13π
f) Simplify the powers of π below. Do you notice any patterns
π1 = π π2 = β1 π3 = βπ π4 = 1
π5 = π π6 = β1 π7 = βπ π8 = 1
π9 = π π10 = β1 π11 = βπ π12 = 1
Note that the powers ππ start to repeat as π increases by 4. So we have in general
ππ = 1 if π is a multiple of 4.
ππ = π if π divided by 4 leaves a remainder of 1.
ππ = β1 if π divided by 4 leaves a remainder of 2.
ππ = βi if π divided by 4 leaves a remainder of 3.
g) π401 = π400 β π
Since 401 divided by 4 leaves a remainder of 1 we get π401 = π, because π401 = π400 β π =
(π4)100π = π because π4 = 1.
h) π39 = π36 β π2 β π = βπ, since π36 = (π4)9 = 1 and π2 = β1.
i) ββ81
ββ81 = ββ1β81 = 9π
j) ββ8
ββ8 = β8 π = 2β2 π
Note that if you had β81 and we rewrote it as 9 = β81 = β(β9)(β9) = ββ9ββ9 = (3π)(3π) =
9π2 = β9. You can see that in splitting a root you have be careful.
Solving Equations and Inequalities Page 227
Before we go into solving non-linear equations let us review the zero product property.
Zero Product Property of Complex Numbers: Let π΄, and π΅ be two complex numbers. Then π΄π΅ = 0, if
and only if either π΄ = 0 or π΅ = 0.
Quadratic Equations
Lectures
Quadratic Equations (15 min)
https://www.youtube.com/watch?v=29_SBzxChMw
Playing
We have seen how to solve quadratic equations that can be factored. We will next look at a series of
examples (some of which donβt factor) that we have been able to solve using square roots. We will look
carefully at what makes these solvable and build from there.
Examples:
Solving Quadratic Equations using the square root operation.
Find all solutions of the following quadratic equations.
Equations With Real Zeros Equations with Complex Zeros
1. π₯2 = 4
π₯ = Β±β4
Square roots undo squares (remember even
root property)
π₯ = 2, or π₯ = β2
π₯2 = β4
Square roots undo squares (remember even root
property)
π₯ = Β±ββ4 (recall ββ1 = π)
π₯ = 2π, or π₯ = β2π
2. π₯2 = 5
π₯ = Β±β5
Square roots undo squares (remember even
root property)
π₯ = β5, or π₯ = ββ5
π₯2 = β5
π₯ = Β±ββ5
Square roots undo squares (remember even root
property)
π₯ = β5π, or π₯ = ββ5π
3. (π₯ + 3)2 = 5
π₯ + 3 = Β±β5
Square roots undo squares (remember even
root property)
π₯ = β3 + β5, or π₯ = β3 β β5
(π₯ + 3)2 = β5
π₯ + 3 = Β±ββ5
Square roots undo squares (remember even root
property)
π₯ = β3 + β5π, or π₯ = β3 β β5π
Solving Equations and Inequalities Page 228
Completing the Square
In all the previous examples we undid squares by taking square roots. A question we may want to ask
then is what makes a quadratic polynomial a perfect square. Also, can we modify any quadratic
polynomial so it becomes a perfect square?
Examples: (Assume all variables take on positive real values to make sense of the polynomials
as areas of the rectangles shown)
1. π₯2 + 6π₯
You can visualize the polynomial above as shown in figure 1 below. In order to change the rectangle into a perfect square using the pieces we have, we will have to move some of the 1 by π₯ rectangular pieces to see what is needed for this polynomial to be part of a perfect square. Since π₯2 already is a square, it would make sense to move the 1 by π₯ rectangles. Since a square has the same length and width, it would make sense to take half of the 6 rectangles, turn them and put them under the blue square which would look as in figure 2. You can see that a corner square is missing. To complete the square then we need to add nine 1 by 1 squares as seen in figure 3. This process of adding 9 squares to the existing quadratic polynomial is called completing the square.
π₯2 + 6π₯ + ____ = (π₯+? )2, we can see the process is to add (6
2)
2= 32 = 9
π₯2 + 6π₯ + (6
2)
2
= π₯2 + 6π₯ + 9 = (π₯ + 3)2
From the above diagram, we see that the 9 red squares needed to complete the square are
obtained by taking half of the number of π₯ pieces and squaring that, i.e., (6
2)
2= 9.
4. 4(π₯ + 3)2 = 5
(π₯ + 3)2 =5
4
π₯ + 3 = Β±β5
4
π₯ = β3 +β5
2, or π₯ = β3 β
β5
2
Square roots undo squares (remember even root property)
4(π₯ + 3)2 = β5
(π₯ + 3)2 = β5
4
π₯ + 3 = Β±ββ5
4
π₯ = β3 +β5
2π, or π₯ = β3 β
β5
2π
Square roots undo squares (remember even root property)
π₯
π₯ 6
Figure 1
π₯
π₯
6
2= 3
6
2= 3
Figure 2
π₯
π₯
6
2= 3
6
2= 3 v
vvv
vv
3
3
Figure 3
Solving Equations and Inequalities Page 229
2. π₯2 β 10π₯
π₯2 β 10π₯ + (β10
2)
2
= π₯2 β 10π₯+25 = (π₯ β 5)2
3. π₯2 + 5π₯
π₯2 + 5π₯ + (5
2)
2
= π₯2 + 5π₯+25
4= (π₯ +
5
2)
2
We can now use this new completing the square process to help us solve any quadratic equation. We
can make all quadratic equations look like the examples 1-3 by making them perfect squares as shown
below.
Examples
Find solutions to the quadratic equations below.
Equations With Real Zeros Equations with Complex Zeros
1. π₯2 + 6π₯ β 14 = 0
π₯2 + 6π₯ = 14
π₯2 + 6π₯ + 9 = 14 + 9 (π₯ + 3)2 = 23
π₯ + 3 = Β±β23
π₯ = β3 + β23, or π₯ = 3 β β23
We start by keeping just the π₯2 and π₯ terms on one side.
Now, to complete the square on the left hand side
we have to add (6
2)
2= 9 to both sides giving us
Solving an equation in this manner is called solving
the equation using the completing the square
method.
π₯2 + 6π₯ + 14 = 0
π₯2 + 6π₯ = β14
π₯2 + 6π₯ + 9 = β14 + 9 (π₯ + 3)2 = β5
π₯ + 3 = Β±ββ5
π₯ = β3 + β5π, or π₯ = β3 β β5π
We get just the variable terms on one side.
and then to complete the square on the left
hand side we have to add (6
2)
2= 9 to both
sides giving us
2. 2π₯2 β 10π₯ β 10 = 0
2π₯2 β 10π₯ = 10
π₯2 β 5π₯ = 5
π₯2 β 5π₯ + (β5
2)
2
= 5 + (β5
2)
2
(π₯ β5
2)
2
= 5 +25
4
π₯ β5
2= Β±β
20
4+
25
4
Completing the squares is easiest with the coefficient of the π₯2 as 1. Rewriting the equation so that the coefficient of the square term is 1 and the constant term is on the right hand side we get
(divide both sides by 2)
2π₯2 β 10π₯ + 15 = 0
2π₯2 β 10π₯ = β15
π₯2 β 5π₯ = β15
2
π₯2 β 5π₯ + (β5
2)
2
= β15
2+ (β
5
2)
2
(π₯ β5
2)
2
= β15
2+
25
4
π₯ β5
2= Β±β
β30
4+
25
4
Rewriting the equation so that the coefficient of the square term is 1 and the constant term is on the right hand side we get
(divide both sides by 2)
Solving Equations and Inequalities Page 230
Quadratic Formula
Using the method of completing the square, we can solve any quadratic equation. Since the problem of
solving quadratic equations comes up frequently and to save us some time later, we will generalize the
completing the squares process to a generic quadratic equation ππ₯2 + ππ₯ + π = 0 to arrive at the
βQuadratic Formula.β We recommend knowing the process of completing the square as well as
memorization of the quadratic formula. Sometimes completing the square is simpler.
A generic quadratic equation can be written as ππ₯2 + ππ₯ + π = 0, where π β 0, πππ π, and π are any
real numbers. Applying the completing the square process we get
ππ₯2 + ππ₯ + π = 0
ππ₯2 + ππ₯ = βπ (Get the variable terms alone on one side.)
π₯2 +π
ππ₯ = β
π
π (Make the π₯2 term have coefficient of 1.)
π₯2 +π
ππ₯ + (
π
2π)
2
= βπ
π+ (
π
2π)
2
(Make the left side a perfect square by adding the square of half of the π₯-coefficient.)
(π₯ +π
2π)
2
= βπ
π+
π2
4π2
(π₯ +π
2π)
2
= β4π(π)
4π(π)+
π2
4π2 (Get a common denominator on the right.)
(π₯ +π
2π)
2
=β4ππ + π2
4π2
π₯ +π
2π= Β±β
π2 β 4ππ
4π2 (Take square root of both sides.)
π₯ = βπ
2πΒ±
βπ2 β 4ππ
β4π2 (Simplify the radical and combine the two terms.)
π₯ =5
2Β± β
45
4
π₯ =5
2Β±
β45
β4
π₯ =5
2Β±
3β5
2
π₯ =5
2+
3β5
2 ππ π₯ =
5
2β
3β5
2
π₯ =5
2Β± β
β5
4
π₯ =5
2Β±
β5
β4π
π₯ =5
2Β±
β5
2π
π₯ =5
2+
β5
2π ππ π₯ =
5
2β
β5
2π
Solving Equations and Inequalities Page 231
Quadratic Formula: The solution set to the quadratic equation πππ + ππ + π = π, π β π and is given by
π₯ = βπ
2πΒ±
βπ2β4ππ
2π=
βπΒ±βπ2β4ππ
2π.
Note: This means that the solutions to the quadratic equation ππ₯2 + ππ₯ + π = 0 are given by
π₯ =βπ+βππβπππ
ππ, or π₯ =
βπββππβπππ
ππ.
Terminology
The radicand under the square root ππ β πππ is called the discriminant.
Note: Depending on the discriminantβs value we can identify what kinds of solutions we will get. In fact:
1. If the discriminant π2 β 4ππ > 0, then the quadratic equation has two distinct real solutions.
2. If the discriminant π2 β 4ππ = 0, then the quadratic equation has one real solution.
3. If the discriminant π2 β 4ππ < 0, then the quadratic equation has two distinct complex solutions.
Examples
Find all solutions to the quadratic equations below.
1. 3π₯2 β π₯ + 3 = 0
Note that π = 3, π = β1, π = 3
Using the quadratic formula we get
π₯ =β(β1) Β± β(β1)2 β 4(3)(3)
2(3)
π₯ =1 Β± β1 β 36
6
π₯ =1 Β± ββ35
6=
1 Β± β35π
6
π₯ =1
6+
β35
6π ππ π₯ =
1
6β
β35
6π
Note that here π2 β 4ππ < 0.
2. 4π₯2 β 20π₯ + 25 = 0
π₯ =β(β20) Β± β(β20)2 β 4(4)(25)
2(4)
π₯ =20 Β± β400 β 400
8
π₯ =20 Β± 0
8=
20
8=
5
2
Or by factoring:
Solving Equations and Inequalities Page 232
4π₯2 β 20π₯ + 25 = 0
4π₯2 β 10π₯ β 10π₯ + 25 = 0
2π₯(2π₯ β 5) β 2π₯(2π₯ β 5) = 0
(2π₯ β 5)2 = 0
2π₯ β 5 = 0
2π₯ = 5 or π₯ =5
2
Note that here π2 β 4ππ = (β20)2 β 4 β 4 β 25 = 0 and hence only one solution.
3. 4π₯2 β π₯ β 5 = 0 Using the quadratic formula
π₯ =β(β1) Β± β(β1)2 β 4(4)(β5)
2(4)=
1 Β± β1 + 80
8
π₯ =1 Β± β81
8=
1 Β± 9
8
π₯ =1 + 9
8=
10
8=
5
4 ππ π₯ =
1 β 9
8=
β8
8= β1
Or by factoring:
4π₯2 β π₯ β 5 = 0
4π₯2 β 5π₯ + 4π₯ β 5 = 0
π₯(4π₯ β 5) + 1(4π₯ β 5) = 0
(π₯ + 1)(4π₯ β 5) = 0
π₯ + 1 = 0 or 4π₯ β 5 = 0
π₯ = β1 or 4π₯ = 5
π₯ = β1 or π₯ =5
4
Note that here π2 β 4ππ = (β1)2 β 4 β 4 β (β5) = 81 is positive, so two real solutions.
4. 3π₯4 + 14π₯2 + 8 = 0
Here we can let π’ = π₯2 and then π₯4 = π’2 so we have 3π’2 + 14π’ + 8 = 0.
Factoring we get (3π’ + 2)(π’ + 4) = 0 and with π’ = π₯2,
π₯2 = β4 or 3π₯2 = β2 or
π₯ = Β±ββ4 = Β±2π or π₯ = Β±β2
3π = Β±
β2
β3π = Β±
β2β3
β3β3π
π₯ = 2π, β2π, β6
3π, β
β6
3π
5. (π₯ + 3)2/3 β 3(π₯ + 3)1/3 β 4 = 0
We set π’ = (π₯ + 3)1/3 and then have π’2 β 3π’ β 4 = 0.
Factoring, we have (π’ β 4)(π’ + 1) = 0
Thus π’ = 4, ππ π’ = β1. Now we go back to π₯.
(π₯ + 3)1/3 = 4 or (π₯ + 3)1/3 = β1. Now cube both sides of each.
Solving Equations and Inequalities Page 233
(π₯ + 3) = 64 or (π₯ + 3) = β1. Now subtract 3.
π₯ = 61 or π₯ = β4 . Both solutions work in the original Equation.
6. Find the points of intersection the linear graph π¦ = π(π₯) =1
2π₯ + 2 and the parabola
graph of π¦ = π(π₯) = β1
3π₯2 + 3π₯ + 4.
We need to find all π₯-values so that the output from both functions is the same π¦-value. In other words, we find all π₯, for which π(π₯) = π(π₯), i.e., find any π₯ such that:
1
2π₯ + 2 = β
1
3π₯2 + 3π₯ + 4
Multiply both sides by 6 to clear fractions 3π₯ + 12 = β2π₯2 + 18π₯ + 24
Then bring all terms to one side 2π₯2 β 15π₯ β 12 = 0
Substituting into the quadratic formula, the π₯-coordinates
are: π₯ =15Β±β321
4
π₯ =15+β321
4 or π₯ =
15ββ321
4 are the exact coordinates
or π₯1 β 8.23 ππ π₯2 β β0.73 are the approximations. We can obtain the π¦-coordinates by plugging into π or π,
Thus π¦1 = π(π₯1) =1
2(
15+β321
4) + 2 =
31
8+
β321
8β 6.11,
and
π¦2 = π(π₯2) =1
2(
15ββ321
4) + 2 =
31
8β
β321
8β 1.635. The
graph shows the two graphs and their intersection points.
7. Find the points of intersection intersection of the curves with equations π¦ = 3π₯ β 2
and π₯2
9+
π¦2
25= 1.
We expect two solutions if the line actually crosses the ellipse. Since the π¦ from the first equation must be the same as the π¦ in the second equation, at a solution value for π₯, we can substitute in the expression 3π₯ β 2 into the second equation for π¦ to get: π₯2
9+
(3π₯β2)2
25= 1. Expand, multiply by 9 β 25 = 225 and
collecting terms, we get: 25π₯2 + 9(9π₯2 β 12π₯ + 4) = 225 and 106π₯2 β 108π₯ β 189 =
0. Thus π₯ =(108Β±β91800)
212β β0.920 ππ 1.949. Substituting
into π¦ = 3π₯ β 2, π¦ β β4.76 ππ π¦ = 3.816. The graph illustrates the curves and the intersection points.
Remember, the graph of a quadratic function π¦ = π(π₯) = ππ₯2 + ππ₯ + π is a parabola. Finding real
solutions to quadratic equationsπ(π₯) = 0 locates the π₯-intercepts of the parabola if they exist.
Solving Equations and Inequalities Page 234
Section 3.1 Worksheet Date:______________ Name:__________________________
Concept Meaning in words and/or examples as required
1. Equation
2. Inequality
3. Solutions, or zeros of an equation
4. Solutions of an inequality
5. Quadratic Formula
6. Discriminant
7. Complex numbers
8. Maximum or Minimum Value of a quadratic function
Difficulties encountered in the section:
Solving Equations and Inequalities Page 235
Exercises 3.1
1. What is the difference between π₯-intercepts, π¦-intercepts of a function, and solutions of an
equation?
2. How would you find the maximum or minimum of a parabola?
3. What is the difference between solutions of an equation and that of an inequality?
4. Can you use your knowledge of solving quadratic equations, exponential and logarithmic
functions to solve the following problems?
a. 22π₯ β 2π₯ β 2 = 0
b. π₯4 β 3π₯2 β 4 = 0
Solving Equations and Inequalities Page 236
5. Simplify the following and write your answer in standard π + ππ form.
a. ββ72 b. ββ66
ββ6 c. ββ121β144
d. (3 + 5π)(β5 + 6π) e. 4β2π
β2β5π f. π35
6. Solve the following equations for the given variable. If there is more than one solution, separate them with commas.
a. (5π¦ + 4)(2π¦ β 3) = 0 b. π’2 β 10π’ + 21 = 0 c. 5π€2 = 17π€ β 6
d. π₯2 β 10π₯ + 10 = 0 (by completing the square) Form:
o (π₯ + _____ )2 = ________ o (π₯ β _____ )2 = ________
Solution π₯ = __________
e. 2π₯2 + 5π₯ β 1 = 0 f. 2π₯2 β 3π₯ + 6 = 0
Solving Equations and Inequalities Page 237
7. Find the discriminant and determine the number of real solutions of the quadratic equation. Then find the actual solutions (real or complex).
a. 4π₯2 β 12π₯ + 9 = 0 Discriminant: Number of solutions: Actual solutions:
b. β2π₯2 β 6π₯ + 8 = 0 Discriminant: Number of solutions: Actual solutions:
c. 2π₯2 β 5π₯ + 8 = 0
Discriminant: Number of solutions: Actual solutions:
8. Determine all the solutions to the equations below. If there is more than one solution, separate them with commas.
a. π₯4 β 5π₯2 β 6 = 0 b. π₯4/3 β 5π₯2/3 β 6 = 0 c. π¦6 β 5π¦3 β 6 = 0
d. 1
π₯2 β37
π₯+ 36 = 0 e. (π₯ β 1)2 β 37(π₯ β 1) + 36 = 0 f. π’4 + 2π’2 + 1 = 0
Solving Equations and Inequalities Page 238
g. 2π₯(π₯ β 1)3(2π₯ + 3)4(5 β π₯)2(π₯2 + 4) = 0
Solve the following problems. If there is no solution, please state so.
9. The length of a rectangle is 5 yd less than twice the width, and the area of the rectangle is 33 π¦π2. Find the dimensions of the rectangle.
10. A model rocket is launched with an initial velocity of 235 ft/s. The rocketβs height β (in feet ) π‘ seconds after it is launched is given by the following. β = 235π‘ β 16π‘2
a. Find all the values of π‘ for which the rocketβs height is 151 feet. Round your answer(s) to the nearest hundredth.
b. Also locate the time when the rocket is at its maximum height.
c. The rocket deploys a parachute just as it reaches maximum height and after that, it falls at a constant
rate of 20 ππ‘
π . Determine how long after the launch
the rocket will hit the ground.
Solving Equations and Inequalities Page 239
11. The cost πΆ in (dollars) of manufacturing π₯ wheels at Raviβs Bicycle Supply is given by the function
πΆ(π₯) = 0.5π₯2 β 170π₯ + 25,850. What is the minimum cost of manufacturing wheels? Do not
round your answer.
12. Consider the parabola shaped graph of the quadratic function π(π₯) = β2π₯2 + 16π₯ β 34.
a. Complete the square to locate the vertex. Then find the maximum or minimum of the function
π(π₯) and indicate which it is a maximum, or a minimum.
b. Locate the focus and π₯- and π¦-axis intercepts and plot the graph of π¦ = π(π₯) with these features
labled.
Solving Equations and Inequalities Page 240
13. Find the intersection points of the graphs of functions below. Then sketch graph of both functions, and label the intersection points clearly. π¦ = π(π₯) = π₯2 β 6π₯ + 4 and π¦ = π(π₯) = 1 β 2π₯
14. Find the intersection points of the graphs of both relations below. Then the sketch graph of both the relations, and label the intersection points clearly. π¦ = π(π₯) = π₯2 β 5 and the circle given by π₯2 + π¦2 = 5.
Solving Equations and Inequalities Page 241
3.2 Polynomial Equations of Degree Three or Higher When trying to solve higher degree polynomial equations, we resort to the zero-product property of real
numbers just as we did when solving quadratic equations by factoring. The word βtryingβ is intentional
here because solving polynomial equations can be a very difficult problem and sometimes exact
solutions cannot be found. By taking all terms to one side (with the other side = 0) and factoring (when
possible) into two lower degree polynomials, we reduce the problem of solving a given polynomial
equation to solving two polynomial equations of lower degree. In this chapter, we build on our
observations from chapter 2 where we saw that π₯-intercepts of a graph at π₯ = π correspond to (π₯ β π)
being a factor of the polynomial. The key idea is to somehow find a number π₯ = π where π(π) = 0.
We call π a zero of the function π and for polynomials, weβll see that (π₯ β π) is a factor. Weβll use this
idea extensively in what follows. We focus mainly on polynomials with integer coefficients.
We start by considering some observed patterns and rigorously explaining these patterns to establish
Theorems or Laws that will always hold true. One observation concerns the nature of rational solutions
to polynomials with integer coefficients. Consider the polynomial equation (7π₯ β 5)(3π₯ + 4) = 0. It
is clear that π₯ =5
7 and π₯ = β
4
3 are the only solutions by the zero-product property of real numbers. If
we look at the unfactored form of the equation 21π₯2 + 13π₯ β 20 = 0, we notice that the numerators
of the zeros are factors of the constant term β20 and the denominators of the zeros are factors of the
leading coefficient 21. Formally stated this pattern is called the Rational Zeros Theorem.
Rational Zeros Theorem:
Let π(π₯) = π0 + π1π₯ + π2π₯2 + π3π₯3 + β― + πππ₯π be a non-constant polynomial function of
degree π with no common factor between all the integer coefficients π0, π1, π2, π3, β¦ ππ.
Any rational zero of π, π =π
π in reduced form must have π as factor of the constant term π0
and π as a factor of the leading coefficient ππ.
Thus for π(π₯) = 3π₯3 + 4π₯ β 5 = 0, this theorem says any rational number zero must be of the form
π = Β±1,5
1,3 . It is easy to check that none of π = Β±1,
1
3, 5,
5
3 make π(π) = 0, so all zeros of π are
irrational or complex numbers.
Proof: To prove this statement we what is given, i.e., that π(π₯) = π0 + π1π₯ + π2π₯2 + π3π₯3 + β― +
πππ₯π, and that π₯ =π
π is a rational zero of this polynomial.
That means π (π
π) = 0 or that π0 + π1 (
π
π) + π2 (
π
π)
2+ π3 (
π
π)
3+ β― ππβ1π₯πβ1 + ππ (
π
π)
π= 0 .
Multiply both sides ππ. We get
π0ππ + π1πππβ1 + π2π2ππβ2 + π3π3ππβ3 + β― ππβ1π β ππβ1 + ππππ = 0.
Now we can take the first or the last term to the right side to get the equations below.
Solving Equations and Inequalities Page 242
π0ππ + π1πππβ1 + π2π2ππβ2 + π3π3ππβ3 + β― ππβ1π β ππβ1 = βππππ .
And,
π1πππβ1 + π2π2ππβ2 + π3π3ππβ3 + β― ππβ1π β ππβ1 + ππππ = βπ0ππ.
In the first equation each term on the left is divisible by q and hence the right side must be too. Since π
π
is in simplest form, π and π donβt share any common factors, and all the factors in π must be in ππ or π
must be a divisor of ππ .
Likewise, in the second equation above each term on the left is divisible by p and hence the right side
must be too. Since π
π is in simplest form, π and π donβt share any common factors, and all the factors in
π must be in π0 or π must be a divisor of π0. This proves our theorem.
Division and Factor Theorems
Recall that when doing long division of numbers or polynomials we can use the division algorithm to
write the dividend in terms of the divisor, the quotient, and the remainder.
Division Algorithm: Let π(π₯) be a polynomial of degree π. On long division of π(π₯) Γ· (π₯ β π)
for any fixed real number π, we obtain a remainder of π that is a constant since it must be
lower degree than the divisor (π₯ β π) and a quotient π(π₯) of degree one less than π. Thus we
can write
π(π₯) = π(π₯)(π₯ β π) + π
We can see by this algorithm then that π(π) = π(π)(π β π) + π = π . This observation is stated as:
Remainder Theorem: Let π(π₯) be a polynomial with real coefficients. Then the remainder when
dividing π(π₯) by (π₯ β π) is equal to the value of the function at π₯ = π.
π = π(π)
Note that when the remainder is zero, then from the division algorithm above, π(π₯) = π(π₯)(π₯ β π) + 0
or π(π₯) = π(π₯)(π₯ β π) and (π₯ β π) is a factor of the polynomial. This important result is stated as:
Factor Theorem: Let π(π₯) be a polynomial. Then (π₯ β π) is a factor of π(π₯) if and only if
π(π) = 0.
Complex Zeros of Real Polynomials
We have already seen that not all polynomials have real zeros, e.g., show that π(π₯) = π₯2 + 4 is never
zero for any real value of π₯. Even when polynomials do have real zeros, they may not be rational zeros.
For example, π₯2 β 2 = 0 has zeros π = β2 and π = ββ2 both real, but irrational. The quadratic
formula provides complex zeros of any quadratic function when the discriminant is negative. For
Solving Equations and Inequalities Page 243
example π(π₯) = π₯2 β 6π₯ + 13 has zeros π₯ =6Β±ββ16
2= 3 Β± 2π. When a polynomial π has real
coefficients, it turns out that the outputs π(π + ππ) and π(π β ππ) are complex conjugates of each
other. Thus if either one of these outputs is equal to 0 + 0π, the other must be 0 β 0π which are both
zero.
Complex Conjugate Zeros Theorem: If π is a polynomial with real coefficients then all non-real
zeros come in pairs π = πΌ + π½π and οΏ½Μ οΏ½ = πΌ β π½π.
Fundamental Theorem of Algebra (FTA): Let π(π₯) = π0 + π1π₯ + π2π₯2 + π3π₯3 + β― + πππ₯π, be
a polynomial with coefficients that are real or complex and is of degree π β₯ 1. Then π(π₯) has
at least one zero in the complex number system. It may be that the zero is actually real.
The proof of this theorem was established by Carl Friedrich Gauss about 200 years ago and is not
obvious. If we look at a selection of polynomials and their zeros, we can see that even for innocent
looking polynomials, the set of real numbers does not always contain a zero of a polynomial. Consider
the polynomials below with real coefficients.
i. π₯2 β π₯ β 6 = 0 β This factors as (π₯ β 3)(π₯ + 2) and the zeros are both integers.
ii. 4π₯2 β 9 = 0 β This also factors as (2π₯ β 3)(2π₯ + 3) and the zeros are rational
numbers π₯ = Β±3
2.
iii. π₯2 β 5 = 0 β Using the square root property, we have π₯ = Β±β5 . Here we have to
go beyond rational numbers to irrational real numbers to find the zeros.
iv. π₯2 β π₯ + 6 = 0 β The quadratic formula provides the solutions π₯ =1Β±πβ23
2. Here we
needed to go beyond the real numbers to the complex numbers to find a zero of this
innocent looking polynomial.
So it would seem plausible that with a little tinkering, we might come up with a polynomial where we
have to go beyond the complex numbers to find a zero. Well, the Fundamental Theorem of Algebra
says that canβt be done and that every polynomial will have at least one zero somewhere in the complex
plane.
An easy consequence of the FTA is that any polynomial of degree π can be factored into linear
factors and written as π(π₯) = ππ(π₯ β π§1)(π₯ β π§2) β¦ (π₯ β π§π) where the π§π are the zeros of the
polynomial.
Some of the π§π may be repeated. Some may be real and some may be complex.
The proof of this result follows from repeated use of the Factor Theorem and the FTA. Letβs say we
start with an ππ‘β degree polynomial. Then the FTA says there is some zero, say π₯ = π§1. Then the Factor
Theorem says that we can write π(π₯) = π1(π₯)(π₯ β π§1) with the degree of π1 being π β 1. Now if π >
1, then the FTA insures that π1 has a zero say π₯ = π§2 . Now the Factor Theorem on π1 says π1(π₯) =
π2(π₯)(π₯ β π§2) with the degree of π2 being π β 2. Thus we have π(π₯) = π2(π₯)(π₯ β π§2)(π₯ β π§1). We
Solving Equations and Inequalities Page 244
can keep doing this π time until we get to ππ which will be of degree zero and will be equal to the
leading coefficient of π(π₯).
In the case where the coefficients or π are real numbers, then the Conjugate Zeros Theorem says that all
the non-real complex zeros appear in conjugate pairs of the form π§π = πΌ + π½π and π§π+1 = πΌ β π½π. Thus
if 2 + 3π is a zero, then 2 β 3π is a zero too. One immediate consequence of this is that real
polynomials must always have an even number of non-real complex zeros. If one takes all the complex
pair factors of the form (π₯ β (πΌ + π½π))(π₯ β (πΌ β π½π)) and expands them, you get real quadratic factors
of the form (π₯2 β 2πΌπ₯ + πΌ2 + π½2). Thus any real polynomial can be written in terms of the product of
real linear factors (π₯ β π§π) for each real zero and real quadratic factors of the form (π₯2 β 2πΌππ₯ + πΌπ2 +
π½π2) for each pair of non-real complex zeros.
Practice Problems
1. Find the factored form of each polynomial below
a. π₯2 β 4π₯ + 1
We find the zeros as π₯ =(4Β±β16β4)
2=
4Β±2β3
2= 1 Β± β3.
Thus the factored form is (π₯ β (1 + β3)) (π₯ β (1 β β3)) ππ (π₯ β 1 β β3)(π₯ β 1 + β3)
b. π₯3 β 8
We see that π₯ = 2 is a real zero. This factors as a difference of cubes as (π₯ β 2)(π₯2 +
2π₯ + 4).
Now find the zeros of π₯2 + 2π₯ + 4 as π₯ =β2Β±β4β16
2=
β2Β±2β3 π
2= β1 Β± β3π.
Finally, we have:π₯3 β 8 = (π₯ β 2)(π₯ + 1 β πβ3)(π₯ + 1 + πβ3)
c. π₯4 β 3π₯2 β 4
We can factor this as (π₯2 β 4)(π₯2 + 1) and the zeros of the first are Β±2 and of the second
are Β±π. Thus the factored form is π₯4 β 3π₯2 β 4 = (π₯ β 2)(π₯ + 2)(π₯ β π)(π₯ + π)
2. Find real polynomials given the following information about their zeros.
a. π(π₯) is degree 3 and has zeros 2, 3, and 3
5.
With these zeros, π must be factored as π(π₯ β 1)(π₯ β 3) (π₯ β3
5) where π is the leading
coefficient that is not known. We might multiply the last factor by 5 and make the π out
front 1
5 as big to have π(π₯) = π(π₯ β 1)(π₯ β 3)(5π₯ β 3). Weβd need to know the value of π
at some other point to figure out what the leading coefficient should be. For example if we
were given that π(2) = 5, we substitute 2 and 5 for the input and output to get:
5 = π(1)(β1)(7) so that π = β5
7 and π(π₯) = β
5
7(π₯ β 1)(π₯ β 3)(5π₯ β 3).
b. π(π₯) is degree 5 with zeros -4, 2 β 3π and β1 + 4π.
The conjugates 2 + 3π and β1 β 4π are also zeros. The factored form of π(π₯) is
π(π₯) = π(π₯ + 4)(π₯ β (2 β 3π))(π₯ β (2 + 3π))(π₯ β (β1 + 4π))(π₯ β (β1 β 4π))
Solving Equations and Inequalities Page 245
If we multiply the conjugate pair factors we get two quadratic factors and a linear factor and
π(π₯) = π(π₯ + 4)(π₯2 β 4π₯ + 13)(π₯2 + 2π₯ + 17).
c. Suppose π(π₯) is a real polynomial of degree 11 and its zeros include: 3, 5π, β2 β 4π.
i. List another zero of π.
For sure, the conjugates β5π and β2 + 4π must also be zeros of π.
ii. What is the maximum number of real zeros π could have?
Since the total number of zeros can be no more than 11 and at least 4 of them are
non-real, we could have at most 7 real zeros.
iii. What is the maximum number of non-real zeros that π could have?
With 11 zeros total and one known real zero at π = 3, the other 10 zeros could
potentially all be non-real, (5 pairs of conjugates).
iv. If the leading coefficient of π is β8, give several possible expressions for π(π₯).
With the 5 known zeros, there are 6 that could be chosen, so several examples of
what π might look like in factored form are:
π(π₯) = β8(π₯ β 3)(π₯ β 5π)(π₯ + 5π)(π₯ + 2 + 4π)(π₯ + 2 β 4π)π₯6
π(π₯) = β8(π₯ β 3)7(π₯ β 5π)(π₯ + 5π)(π₯ + 2 + 4π)(π₯ + 2 β 4π)
π(π₯) = β8(π₯ β 3)(π₯ β 5π)(π₯ + 5π)(π₯ + 2 + 4π)(π₯ + 2 β 4π)(π₯ + 6)3(π₯ β 9)3
Using Division to Find Zeros
When presented with finding zeros of a complicated polynomial function, often one or more zeros are
known or can be found to a high degree of accuracy. In this scenario the Factor Theorem says our
polynomial can be written as π(π₯) = (π₯ β ππππ€π π§πππ)π(π₯) and further zeros of π must be zeros of
π(π₯) which is one lower in degree than π. We can find this function π by long division of π(π₯) by
(π₯ β ππππ€π π§πππ). If we know a zero of π we can divide out the corresponding factor and reduce the
problem to one lower in degree again. If we can get down to a quadratic, we can always resort to the
quadratic formula to find the last two zeros. Weβll take a brief look at long division and develop a
streamlined version called βSynthetic Divisionβ which is helpful for our present purpose in dividing by
expressions of the form (π₯ β π).
We used long division of polynomials to find oblique asymptotes of rational functions. See example
below for (6π₯2 + 7π₯ + 4) Γ· (π₯ β 2). As you can see in long division, the first term of the quotient will
always have the leading coefficient of the dividend but π₯ will be to one degree less. Subtracting out the
second term amounts subtract (6) Γ (β2) or adding 6 Γ 2. When dividing by π₯ β π these second
terms will always be subtracting (#) Γ (βπ). The steps under Synthetic Division column convert all
these (subtracting negatives) to adding and we end up with the synthetic division algorithm described
just after the diagram.
Solving Equations and Inequalities Page 246
Long Division
Synthetic Division
Synthetic Division Steps for dividing a polynomial by a divisor of the form (π₯ β π).
Step 1: The coefficients of the dividend polynomial go in the boxes in the top row in descending powers
with zeros inserted when powers are missing. In the example above we put 6 7 4 to indicate the
dividend.
Step 2: When dividing by π₯ β π put π ( Here π = 2. ) in the second row goes in the second row to Left
of the vertical bar.
Step 3: Then bring the leading coefficient down to row three and this is the first coefficient of the
quotient.
Step 4: Multiply π by the first number in the bottom row and put the result in the second row under the
coefficient of the next term in the dividend. Add the two terms in that column and put that resulting
number in the last row in that column. Then multiply that number by π = 2 and put that number in the
second row just below the third coefficient of the dividend and add. Continue the process and the final
entry in the bottom row is the remainder. The last number in the last row is the remainder. The
remaining numbers are the coefficients of the quotient polynomial which is one degree lower than the
dividend.
This method is very fast and the addition avoids the common mistakes of subtracting negatives. Now
we will utilize this to solve higher degree polynomial equations.
6π₯ + 19
(π₯ β 2) 6π₯2 + 7π₯ + 4
β(6π₯2 β 12π₯)
19π₯ + 4
β(19π₯ β 38)
42
Quotient 6π₯ + 19
Remainder is 42
6 7 4 2 12 38
6
19 42
Quotient 6π₯ + 19
Remainder is 42
Solving Equations and Inequalities Page 247
Practice Example:
1. Find all the zeros of 3π₯4 + 2π₯3 β 4π₯2 β 29π₯ + 10 = 0
Solution:
The potential rational zeros are Β±ππππ‘ππ ππ 10
ππππ‘ππ ππ 3.
Thus the possible rational zeros could be Β±1, Β±1
3, Β±2, Β±
2
3, Β±5, Β±
5
3, Β±10, Β±
10
3,
We can now use synthetic division to see if any of these are zeros. We could also just substitute
each of these into the function to see if we get zero for any of the outputs. The advantage of doing the
division is that it gives you the quotient which can be used to find additional zeros.
3 2 β4 β29 10
β2 β6 8 β8 74
3 β4 4 β37 84
3 2 β4 β29 10
2 6 16 24 β10
3 8 12 β5 0
π₯ = 2 is a zero so we can work with the degree-3 polynomial 3π₯3 + 8π₯2 + 12π₯ β 5 which is our quotient
3 8 12 β5 1
3
1 3 5
3 9 15 0
π₯ =1
3 is a zero also and now the quotient is
3π₯2 + 9π₯ + 15 = 0. Note that we can factor 3 out of the quotient before plugging into the quadratic formula. Thus we seek zeros of π₯2 + 3π₯ + 5. We get the remaining zeros to be
π₯ =β3Β±β9β4(1)(5)
2 after simplifying you will get π₯ = β
3
2+
β11
2π, β
3
2β
β11
2π
So all our solutions of 3π₯4 + 2π₯3 β 4π₯2 β 29π₯ + 10 = 0 are
π₯ =1
3, 2,
3
2+
β11
2π, β
3
2β
β11
2π. We could also use this to write the polynomial in factored form as:
3π₯4 + 2π₯3 β 4π₯2 β 29π₯ + 10 = 3(π₯ β1
3)(π₯ β 2)(π₯ β
3 + β11π
2)(π₯ β
3 β β11π
2)
2. Find the zeros of π(π₯) = 4π₯5 β 13π₯3 β 8π₯2 + 3π₯ + 2 and write it in factored form. Solution:
The potential rational zeros are: π = Β±1 ππ 2
1 ππ 2 ππ 4= Β± {1,
1
2,
1
4, 2} .
Synthetic division shows that π₯ = 2, β1
2 and
1
2 are zeros and we get:
4 0 -13 -8 3 2 We are left with the quadratic 4(π₯2 + 2π₯ + 1) = 0
which factors as 4(π₯ + 1)2 = 0. Thus β1 is a zero of multiplicity 2.
The zeros are thus π₯ = β1, β1
2,
1
2, πππ 2
The factored form of π is:
π(π₯) = 4(π₯ + 1)2 (π₯ +1
2) (π₯ β
1
2) (π₯ β 2).
π = 2 8 16 6 -4 -2
4 8 3 -2 -1 0
π =1
2
2 5 4 1
4 10 8 2 0
π = β1
2
-2 -4 -2
4 8 4 0
Solving Equations and Inequalities Page 248
3. Given the graph of the polynomial π(π₯) = 2π₯3 + π₯2 + 4π₯ β 15. Find all the zeros of π and the complete factored form of π.
Solution:
From the graph we see that π₯ =3
2 is a zero
and by synthetic division we get: 2 1 4 -15 3/2 3 6 15
2 4 10 0 Thus the quadratic quotient is: 2(π₯2 + 2π₯ + 5) which by the quadratic formula has complex zeros π₯ = β1 Β± 2π
The zeros are: π₯ =3
2, β1 + 2π, β1 β 2π
π(π₯) = 2 (π₯ β3
2) (π₯ + 1 β 2π)(π₯ + 1 + 2π)
Solving Equations and Inequalities Page 249
Section 3.2 Worksheet Date:______________ Name:__________________________
Concept Meaning in words and/or examples as required
Rational Zeros Theorem
Factor Theorem
Remainder Theorem
Complex Conjugates Zeros Theorem
Real Zeros
Complex Zeros
Synthetic Division
Fundamental Theorem of Algebra
Difficulties encountered in the section:
Solving Equations and Inequalities Page 250
Exercises 3.2
1. Let π(π₯) be a polynomial.
a. What can you say about the non-real zeros of the equation π(π₯) = 0, if the coefficients
of π(π₯) are real?
b. What is the maximum number of non-real zeros a polynomial of odd degree can have?
c. Can a polynomial of odd degree have all non-real zeros?
d. Why must an odd degree polynomial with real coefficients have at least one real zero?
e. Can you have a polynomial with non-real coefficients of odd degree with all non-real
zeros?
Solving Equations and Inequalities Page 251
2. Suppose π (π₯) is a polynomial of degree 13 whose coefficients are real numbers. Also, suppose that π (π₯) has the following zeros: 7, β8 , 5π, β2 β 4π.
a) Find another zero of π (π₯).
b) What is the maximum number of real zeros that π (π₯) can have?
c) What is the maximum number of non-real zeros that π (π₯) can have?
d) If the leading coefficient of the polynomial is β3, list three possible formulas for π (π₯) with the degree and zeros given above.
3. Find a polynomial π(π₯) of degree 4 that has the following zeros β2, 1, β6, πππ 0. Leave your
answer in factored form.
4. Find the real polynomial of degree 4 with zeros π = 3, β1
2, πππ 2 β 3π and leading coefficient
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5. Perform the following division to find the quotient and the remainder. Write your answer in the
Quotient +π ππππππππ
π₯β4 form. When possible use synthetic division.
a. (6π₯2 + 37π₯ + 39) Γ· (π₯ + 5)
b. (3π₯2 β π₯3 + 6π₯ β 8) Γ· (π₯ β 4)
c. (24π₯3 + 4π₯2 + 14π₯ + 3) Γ· (6π₯ β 2)
Solving Equations and Inequalities Page 253
6. Use the rational zeros theorem to list all possible zeros of the following. Be sure that no value in
your list appears more than once.
a. π(π₯) = β5π₯4 β π₯3 β 3π₯2 β 3π₯ + 3
b. π(π₯) = β6π₯4 β π₯3 β 3π₯2 β 3π₯ + 9
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7. The function below has at least one rational zero. Use this fact to find all the zeros of the function.
π(π₯) = 4π₯3 + 12π₯2 β π₯ β 3
If there is more than one zero, separate them with commas. Write exact values, not decimal
approximations.
8. The function below has a rational zero. Use this fact to find all the zeros of the function.
π(π₯) = 2π₯3 β 3π₯2 + π₯ β 6
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9. The function below has at least one rational zero. Use this fact to find all the zeros of the function.
Write exact values, not decimal approximations.
π(π₯) = 7π₯4 + 20π₯3 + 10π₯2 β 5π₯ β 2
10. For the polynomial below β2 is a zero. Express β(π₯) as a product of linear factors.
β(π₯) = π₯3 + 8π₯2 + 30π₯ + 36.
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11. The function below has at least one rational zero. Use this fact to find all the zeros of the function.
β(π₯) = 5π₯4 β 29π₯3 β 40π₯2 β 13π₯ β 7
12. Find all the other zeros of π(π₯) given that 3β3π is a zero. (Hint: find a quadratic factor with real
zeros, and then use long division to obtain a second quadratic factor.)
π(π₯) = π₯4 β 7π₯3 + 22π₯2 β 6π₯ β 36
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13. Find all the zeros of π(π₯) = 12π₯4 + 5π₯3 + 46π₯2 + 20π₯ β 8 whose graph is given below.
14. Find the exact values of the points of intersection of the two functions below.
π(π₯) = 3π₯5 β 2π₯4 β 4π₯3 β 5π₯2 + 6π₯ β 10 and π(π₯) = π₯5 + π₯4 β 2π₯3 β 8π₯2 + 10π₯ β 16
15. Create two polynomial functions that intersect each other at π₯ = 2, π₯ = β3, π₯ = 2β3 , and
π₯ = β2β3.
Solving Equations and Inequalities Page 258
3.3 Exponential and Logarithmic Equations
In logarithmic and exponential equations, the unknown appears in the equation as an input to a
logarithmic or exponential function. In this section we deal with equations where the variable appears
only in a logarithm or only in an exponent and not both in the same equation. The standard method of
solving these equations is to isolate the logarithmic or exponential function expression on one side of
the equation and then operate on both sides of the equation with the inverse function. Thus for the
logarithmic equation log2(π₯ β 4) = 7 we operate on both sides with the 2π₯ function to get (π₯ β 4) =
27. This is referred to as exponentiating both sides of the equation. Likewise, weβd operate on both
sides of an exponential equation with a logarithmic function to undo the exponential function. This is
much like what you do with radical or power equations, e.g., for π₯3 = 7, you undo the cube function by
operating on both sides with the cube root function to get π₯ = β73
.
Logarithmic Equations and Inequalities
The basic idea for solving logarithmic equations is to isolate the logarithm expression on one side and
then exponentiate both sides. This is equivalent to converting the logarithm statement into its
exponential form as shown below:
loga (π₯-expression) = # β πloga (π₯βexpression) = π# β (π₯-expression)= π#
Once the logarithm is undone, further steps may be needed to isolate the variable π₯.
Practice Example
1. log(2π₯ β 1) = 5
Solution:
The common logarithm here is base-10 and we exponentiate by taking 10πππβ π πππ to get.
2π₯ β 1 = 105 2π₯ = 100,001 π₯ = 50,000.5 Check log(2 β 50,000.5 β 1) = log(100,001 β 1) = log 100,000 = 5 It is important to check the answer since the domain of a logarithmic function is restricted.
2. log3(π₯ β 1) = β2
Solution:
Here we operate on both sides with the 3( ) function to undo the log3( ).
π₯ β 1 = 3β2
π₯ = 1 +1
9=
10
9
Check that the answer works.
Convert from logarithmic to exponential form.
Solving Equations and Inequalities Page 259
When more than one logarithmic term appears in an equation, we can often use properties of
logarithms to combine them into a single logarithm and then proceed as above. We list these properties
below:
ππππ(π’π£) = ππππ(π’) + ππππ(π’)
ππππ (π’
π£) = ππππ(π’) β ππππ(π’)
ππππ(π’π) = ππππππ’
3. Solve log2(π₯ + 7) + log2 π₯ = 3
Solution:
Combine the log terms to get: log2((π₯ + 7)π₯) = 3
Exponentiate taking 2πππβ π πππ: π₯2 + 7π₯ = 23 = 8.
Take to one side and factor: π₯2 + 7π₯ β 8 = (π₯ + 8)(π₯ β 1) = 0
Answers: π₯ = β8 ππ π₯ = 1.
Checking the answers:
π₯ = β8 β log2(β1) + log2(β8) these are undefined and hence this solution is extraneous!
π₯ = 1 β log2(8) + log2 1 = 3 + 0 = 3 so, The solution is just π₯ = 1.
4. Solve log4(π₯Β² β 15π₯ β 16) = log4(π₯ + 1) + 3
Solution:
Taking the log terms to the left side and combining, we get: log4π₯Β²β15π₯β16
π₯+1= 3
Exponentiating we get π₯Β²β15π₯β16
π₯+1= 64 .
The left side reduces to (π₯ β 16) and solving we get π₯ β 16 = 64 β π₯ = 80
Check this answer in the original equation to get: log4(5184) = log4(81) + 3.
Subtract log4 81 and combine to get log45184
81= log4 64 = 3 . So π₯ = 80 is the solution.
Solving logarithmic inequalities is done much like for equalities except we must be careful when
undoing the logarithm, i.e, taking π πππβ π πππ on both sides of an inequality. The question is whether
the new inequality will be true exactly when the original inequality was true. The answer comes
down to whether the undoing function ππ₯ is increasing for all π₯. Note that the exponential function
ππ₯ is increasing when π > 1, but decreasing when 0 < π < 1 . Consider operating with the ππ₯
function to undo a natural logarithm in an inequality. If we start with π < π will it be true that
ππ < ππ ? The answer is clearly yes since the π¦ = ππ₯ graph increases as we move from π₯ = π to the
right to π₯ = π. Note that if we were to operate with (1
2)
π₯ on both sides, this is a decreasing function
and would thus reverse the order of the inequality. For example 2 < 5, but (1
2)
2=
1
4> (
1
2)
5=
1
32.
Luckily mostly the logarithms we deal with are either the common logarithm with base 10 or the
Solving Equations and Inequalities Page 260
natural logarithm with base π and thus we can exponentiate both sides with these bases and not
worry about switching the direction of the inequality. One must also check that the expression
contained within the logarithm is in the domain of the logarithm.
Aside: We saw this switching of inequalities before when we multiplied or divided both sides of an
inequality by (-1). That operation of Γ· (β1) can be thought of as operating with the decreasing
function π(π₯) = βπ₯ function on both sides.
5. log3(π₯ β 1) β€ β2
Solution:
First off, we must have π₯ β 1 > 0 (or π₯ > 1) in order for the logarithm to be defined.
Next operate on both sides with 3π₯ (which preserves the direction of the inequality since π¦ = 3π₯ is
an an increasing function).
π₯ β 1 β€ 3β2
π₯ β€ 1 +1
9 or π₯ β€
10
9 and also π₯ > 1
Thus the solutions is the interval 1 < π₯ β€ 11
9.
Exponential Equations and Inequalities
In exponential equations, the unknown appears within the exponent on some base. These equations
show up in financial problems and growth and decay problems where exponential models are often
used.
Consider the annual compounding problem of determining how long it would take an investment to
double in value given that it grows by 5.6% each year. This is annual compounding of interest and is
describe by π΄ = π(1.056)π‘ where π is the initial value and π΄ is the value π‘ years later. The doubling
problem is to find what π‘ should be so that π΄ = 2π, or 2π = π(1.056)π‘. We could divide out the π so
that our doubling problem comes down to solving 2 = (1.056)π‘ which is a typical exponential
equation. To undo the exponential, one might operate on both sides with the base 1.056 logarithm.
The drawback to doing this is that it is not readily apparent how to evaluate log1.056 2 = π‘. A more
practical method is to just take the natural or common logarithm on both sides since these are available
on any scientific calculator. The right side is then simplified using the third property of logarithms
above.
Thus weβd solve 2 = (1.056)π‘ as: ln 2 = ln 1.056π‘ = π‘ ln 1.056 and π‘ =ln 2
ln 1.056β 12.72 .
Or, using the common logarithm, log 2 = log 1.056π‘ = π‘ and π‘ =log 2
log 1.056β 12.72 .
The basic tool for solving exponential equations is to isolate the exponential expression on one side and
take either the natural or common logarithm of both sides. For inequalities, we isolate the exponential
so both sides are a positive quantity and then operate on both sides with either the common or natural
logarithm functions. Since both of these are increasing functions, we donβt need to alter the direction
of the inequality in this step of taking logs on both sides.
Solving Equations and Inequalities Page 261
Sample Problems:
1. Solve the equations and inequalities below.
a) 2π₯ = 3 Solution: We take the common (or could use natural log) of both sides to get: log 2π₯ = log 3 π₯ log(2) = log (3) and therefore
π₯ =log 3
log 2
This is the exact solution. An approximate solution by a scientific calculator is
π₯ =πππ3
πππ2β 1.585
Graphically that means the two functions π¦ = 2π₯ and
π¦ = 3 intersect at π₯ =πππ3
πππ2β 1.585. Or that the π₯-
intercept of the function π(π₯) = 2π₯ β 3 is at
π₯ =πππ3
πππ2β 1.585
a) 2π₯ < 3 Solution: We take ln (or could use log ) on both sides.
π₯ ln 2 < ln 3 As with the equation, we divide by the positive number ln 2 to get:
π₯ <ln 3
ln 2β 1.585
Solution: (ββ,ln 3
ln 2)
b) The half-life of penicillin (a form of antibiotic) is about 35 minutes for an adult with normal renal function (or kidneys). A typical dose for an adult is between 250-3000mg 2 to 4 times a day to treat different kinds of bacterial infections. An adult was accidentally given an overdose of 6000ππ. How long will it take for the amount in this person to come down to 200mg? Assume this person has normal kidney function and a half-life of 35 minutes.
Solution:
The model is: 6000 (1
2)
π‘
35= 200.
To undo here we first divide both sides by 6000 and then take natural logarithms on both sides.
(1
2)
π‘
35=
200
6000 or ππ (
1
2)
π‘
35= ππ (
200
6000)
π‘
35ln (
1
2) = ππ (
200
6000)
π‘ =35ππ (
2006000)
ln (12)
β 171.74 ππππ’π‘ππ β 2 βππ’ππ 51.6 ππππ’π‘ππ
So assuming you drink a lot of fluids to flush things out of the body it will take just a little under 3 hours.
(1.585,3)
Solving Equations and Inequalities Page 262
c) 2π₯β5 = 45βπ₯ Solution:
Method 1: We might notice that 4 = 22, so we can rewrite the right side as:
2π₯β5 = (22)5βπ₯
2π₯β5 = 210β2π₯ That means the exponents must be the same. π₯ β 5 = 10 β 2π₯ 3π₯ = 15
Solution: π₯ = 5
Method 2: We can take natural logarithm of both sides to get: (π₯ β 5) ln 2 = (5 β π₯) ln 4 Now expand and collect the π₯-terms so:
π₯ ln 2 β 5 ln 2 = 5 ln 4 β π₯ ln 4 β
π₯(ln 2 + ln 4) = 5(ln 4 + ln 2) β Solution: π₯ = 5
d) 2π₯β5 = 34βπ₯ Solution: Here we have to use logarithmic functions.
ln (2π₯β5) = ππ(34βπ₯) (π₯ β 5) ln 2 = (4 β π₯) ln 3 Since ππ2 and ππ3 are just numbers we treat them like it and solve the equation above as a linear equation can be solved. π₯ ln 2 β 5 ln 2 = 4 ln 3 β π₯ ln 3 π₯ ln 2 + π₯ ln 3 = 4 ln 3 + 5 ln 2 π₯(ln 2 + ln 3) = 4 ln 3 + 5 ln 2
π₯ =4 ln 3 + 5 ln 2
ln 2 + ln 3
This is the exact solution to the equation and approximate solution can be found by using your calculators which would give you
π₯ =4ππ3+5ππ2
ππ2+ππ3β 4.387
e) Amy invested $4000 at 5% interest compounded quarterly. How many years will she will have
to wait for her money to grow to $5000. Solution:
The quarterly compounding formula is: 4000 (1 +0.05
4)
4π‘= 5000.
We use the natural logarithmic function to isolate the unknown in the exponent.
(1 +0.05
4)
4π‘
=5000
4000
4π‘ β ln (1 +0.05
4) = ln (
5
4)
π‘ =ln(
5
4)
4 ln(1+0.05
4)
β 4.49 yrs
She would have to wait about 4 years and approximately 6 months.
Solving Equations and Inequalities Page 263
f) log3(π₯ β 1) = log3(5π₯ β 7)
Since the logarithmic function is one-to-one we get: π₯ β 1 = 5π₯ β 7 Or β4π₯ = β6
π₯ =β6
β4=
3
2
Check your answer pleaseβ¦.
g) log3(π₯ β 1) β log3(5π₯ β 7) = 2 We can use properties of logarithms to solve this problem.
log3 (π₯ β 1
5π₯ β 7) = 2
Now using the fact that logarithmic function is the inverse function of the exponential function we get
π₯ β 1
5π₯ β 7= 32
π₯ β 1 = 9(5π₯ β 7) π₯ β 1 = 45π₯ β 63
62 = 44π₯
π₯ =62
44=
31
22
We need to check this solution, so go ahead do that.
h) log 2 + log(π₯ β 1) = log(3π₯ β 1) We can use properties of logarithms to solve this problem
log(2(π₯ β 1)) = log(3π₯ β 1)
Using the fact that log functions are one-to-one we get 2π₯ β 2 = 3π₯ β 1
π₯ = β1 Note that the domain of log (π₯ β 1) is π₯ > 1 so we have an extraneous solution There is no solution to this problem.
Solving Equations and Inequalities Page 264
Application Problems 1. A sample from a spruce tree that was discovered buried under 40 feet of clay while digging a
well in eastern Wisconsin was found to contain 20% of its original carbon-14. This tree was
buried thousands of years ago by glaciers advancing out of Lake Michigan and pushing the clay
lake-bottom ahead of them. It is known that the half-life of carbon-14 is 5730 years. Estimate
how long ago the tree lived.
Solution:
The amount of a radioactive compound remaining is given in terms of the initial amount and
half-life by: π΄(π‘) = π΄0 (1
2)
π‘
5730.
It is known that we now have only 0.2π΄0 remaining, so: 0.2 π΄0 = π΄0 (1
2)
π‘
5730.
We divide out the π΄0 and take natural logarithms on both sides to get:
ln 0.2 =π‘
5730ln
1
2
Isolate π‘, by multiplying by 5730 and dividing by ln1
2 to get: π‘ =
5730 ln 0.2
ln1
2
β 13,300 π¦ππππ .
2. How long will it take for $5000 earning 6% interest compounded quarterly to have the same
value as $400 which earns 11% interest compounded continuously?
Solution:
We set the value for each account equal to get:
5000 (1 +0.06
4)
4π‘
= 400π0.11π‘
We can divide by 400 and then take logarithms of both sides. On the left, we have a product
and we will use the product property of logarithms to separate the two factors since the
exponent 4π‘ is not applied to the factor 12.5.
Γ· 400 β 12.5 (1.015)4π‘ = π0.11π‘
Take logarithms β ln 12.5 + ln 1.0154π‘ = 0.11π‘
Bring down the 4π‘ and collect the π‘-terms on the right β
β ln 12.5 + 4π‘ ln 1.015 = 0.11π‘
β ln 12.5 = π‘(0.11 β 4 ln 1.015)
β π‘ =ln 12.5
0.11 β 4 ln 1.015β 50 π¦ππππ
Solving Equations and Inequalities Page 265
3. Determine the π»+ concentration range if the ππ» is to be kept between 4.5 and 6.8 .
Solution:
The definition of ππ» is: ππ» = β log[π»+] with [π»+] being the π»+concentration, we require that:
4.5 < β log[π»+] < 6.8
We first divide by (-1) and then operate with the 10π₯ function to undo the log .
β4.5 > log[π»+] > β6.8
10β4.5 > [π»+] > 10β6.8
Approximating: 0.000000158< [π»+] < 0.0000316
4. A cooked turkeyβs temperature is modeled by the function π(π‘) = 74 + 115πβ0.02π‘ where the
input π‘ is the number of minutes after it was taken out of the oven and the temperature is in
degrees Fahrenheit. Determine the time range when the bird temperature will be between 140
and 160 degrees.
Solution:
We require: 140 β€ 74 + 115πβ0.02π‘ β€ 160
Subtract 74, then Γ· by 115, then take ln to get:
66 β€ 115πβ0.02π‘ β€ 86 β 66
115β€ πβ0.02π‘ β€
86
115 β ln
66
115β€ β0.02π‘ β€ ln
86
115
Dividing by β0.02 and approximating we get: 14.5 πππ β€ π‘ β€ 27.88 πππ.
Solving Equations and Inequalities Page 266
Section 3.3 Worksheet Date:______________ Name:__________________________
Concept Meaning in words and/or examples as required
List different methods of solving exponential equations
List different methods of solving exponential equations
Difficulties encountered in the section:
Solving Equations and Inequalities Page 267
Exercises 3.3
1. How are the product, quotient and power properties of logarithmic functions used to solve
logarithmic or exponential equations?
2. List at least three applications of being able to solve exponential and logarithmic equations not
listed in the section.
Find all the solutions to the following equations.
3. πππ3(π₯ + 1) = 2
4. πππ3(π₯ + 1) β πππ3(π₯) = 2
5. πππ3(2π₯ β 1) + πππ3(π₯ + 1) = 2
Solving Equations and Inequalities Page 268
6. πππ4(β1 β 2π₯) = β1
7. 4 + log(2π₯ β 1) = 5
8. πππ5(π₯ β 3) = 1 β πππ5(π₯ β 7)
9. ln(π₯ + 4) β ln 18 = ln 5
Solving Equations and Inequalities Page 269
10. 125 = 25βπ₯β2
11. 2π₯2β61π₯ = 643β9π₯
12. 15β8π¦ = 6
(Round your answer to the nearest hundredth. Do not round any intermediate computations.)
Solving Equations and Inequalities Page 270
13. πβ8π’ = 6
(Round your answer to the nearest hundredth. Do not round any intermediate computations.)
14. 17βπ₯β3 = 16β8π₯
(Write the exact answer using base-10 logarithms)
15. 3π₯β1 = 52π₯β1
(Write the exact answer using natural logarithms)
Solving Equations and Inequalities Page 271
16. 200π0.05π‘ = 50
17. 3π2π₯ β 5ππ₯ + 2 = 0 (Hint a substitution π’ = ππ₯ will be useful.)
18. 9π₯ β 3π₯ β 2 = 0 (Hint a substitution π’ = 3π₯ will be useful
19. 1500 (1 +0.05
4)
4π‘= 3000
Solving Equations and Inequalities Page 272
20. Write a word problem that would give rise to the exponential equation below and solve the
equation
2000(1.02 )4π‘ = 5000
Solving Equations and Inequalities Page 273
21. Write a word problem that would give rise to the exponential equation below and solve the
equation.
20 = 250 (1
2)
π‘
5730
22. A car is purchased for $28,500. After each year the resale value decreased by 35%. What will
be the resale value be after 4 years? Also determine when the resale value will be less than
$5000. Round your answers to the nearest whole number.
23. A loan of $39,000 is made at 5% interest, compounded annually. Assuming no repayment is
made, after how many years will the amount due reach $63,000 or more? (Use a calculator if
necessary.) Write the smallest possible whole number answer.
Solving Equations and Inequalities Page 274
24. The number of bacteria in a certain population increases according to a continuous exponential
growth model, with a growth rate parameter of 4.1% per hour. How many hours will it take for
the sample to double?
Note: This is a continuous growth model. (π΄(π‘) = π΄0πππ‘)
Do not round any intermediate computations, and round your answer to the nearest hundredth of an
hour.
25. An initial amount of $1800 is invested in an account at an interest rate of 2% per year
compounded continuously. Find the amount in the account after 6 years. Round your answer to
nearest cent. Also how many years are required for the value to reach $10,000.
26. The light intensity in Lake Superior decreases due to absorption by 12% for each meter of depth.
If the intensity just below surface is at 150π€
π2. Write a formula for the intensity at dept π₯
meters. Also determine how deep one must go for the light intensity to be 0.05π
π2.
27. A model for world population in billions π‘ years after the year 2000 is given by (π‘) =24
1+3πβ0.04π‘ .
Determine in what year the population is projected to be 12 billion by this model.
Also, what does the model predict as π‘ gets very large?
Solving Equations and Inequalities Page 275
28. The value of an exponential growth stock fund increased from $10,000 to $18,000 over a 12
year period.
a. Find the continuous growth rate and model π΄(π‘) = ππππ‘ that describes this growth.
b. Find the quarterly compounding growth rate and model π΄(π‘) = π (1 +π
4)
ππ‘that describes this
growth.
29. A sample of radioactive iodine decreased from 50 micrograms to 10 micrograms over a 19 day
period.
a. Find a decay model π΄(π‘) = π΄0πππ‘ that models this decay.
b. Determine the half-life of this radioactive iodine.
30. Extra Credit: You want to buy a house that might be in range of $180,000 to $200,000 in 10
years. You know you will need to save at least a 20% down payment cost. How much should you
invest per month starting now in an account that pays 5% interest so that in 10 years you will
have enough for your down payment at that time? Please explain carefully how you computed
this amount. You might start with $100/month and see how that does and adjust this amount
accordingly.
Solving Equations and Inequalities Page 276
3.4 Systems of Equations We know how to solve some types of equations in one variable. But very often there is a need to solve a collection of more than one equation in
two or more variables. So by now you are familiar on how a mathematician uses prior knowledge to play with the objects at hand to deliver the
solutions. So let us look at certain types of systems of equations which is a collection of equations in two or more variables and see how we
could find the solutions if they exist.
A system of linear equations in two or more variables is a set of two or more linear equations. A system of nonlinear equations in two or more
variables is a set of two or more nonlinear equations. A solution to a system of equations is a point in the plane that satisfies all the equations in
the system.
Recall that working with a system of linear equations in two variables means working with lines. If two lines intersect each other they are said to
form a consistent independent system of equations with a unique solution, which is the point of intersection. If two lines overlap each other, or
are the same line, then all points on the line are solutions to the system of equations. Such a system of equations is called a consistent
dependent system of equations with infinitely many solutions. If two lines are parallel to each other they are said to form an inconsistent system
of equations with no solution. Graphically, the three scenarios are depicted below.
Consistent Independent Consistent Dependent Inconsistent
Unique Solution Infinitely many solutions No Solutions
Example
Example
Example
The two lines are not parallel and therefore intersect in one point. The point of intersection will be the unique solution to the system of equations.
The lines are parallel and have the same π¦-intercept or the lines overlap. In this situation all points on the line are solutions to the system of equations and therefore has infinitely many solutions.
The lines are parallel and have different π¦-intercepts. In this situation no point in the plane can be the solution and therefore the system has no solutions.
Solving Equations and Inequalities Page 277
When working with a system of nonlinear equations, knowing what shape the graphs of each of the
equations look like may also help in understanding the nature of the solutions or how many to expect.
Methods to solve a system of equations and inequalities
Graphical Method: Plot the graphs of all the equations or inequalities in the system on the same
coordinate system and then from the graph locate the solution or the solution set if one exists.
Depending on the graph though you may at times only be able to get an approximate value of the π₯ and
π¦ coordinates of the solution or the point that satisfies all the equations, or inequalities in the system.
Graphing method is not always the best but can give you an insight into the solutions.
Substitution Method: Solve one of the equations for either the π₯ or π¦ and substitute this value into the
other equation making it then an equation in one variable. This will allow you to solve the new equation
in one variable. Then replace that value in one of the original equations to find the other coordinate.
Elimination Method: Align the equations one above the other so that all the variables and degrees of
each of the equations are lined up. Then you multiply one of both of the equations by the required
constants so that when you add the two equations it results in a one variable equation and you can then
proceed as in the substitution method.
When dealing with more than two variables the first two methods will work, the elimination method
becomes a little more complicated but can still come in handy. In this case we will see what our other
options are later.
Practice examples
Solve the systems of equations if possible.
1. {3π₯ β π¦ = 22π₯ + π¦ = 3
Substitution Method Solve the first equation for π¦, giving us π¦ = 3π₯ β 2 Substitute this value of π¦ in the second equation 2π₯ + (3π₯ β 2) = 3 or 5π₯ = 5, or π₯ = 1
π¦ = 3(1) β 2 = 3 β 2 = 1 Solution: (1,1) Elimination Method Add the two equations and we get
5π₯ = 5 π₯ = 1
3(1) β π¦ = 2 π¦ = 1
Solution: (1,1)
Graphical Method
Solving Equations and Inequalities Page 278
2. {2π₯ + π¦ = 24π₯ + 2π¦ = 3
Elimination Method Multiply the first equation by β2
{β4π₯ β 2π¦ = β44π₯ + 2π¦ = 3
add the two equations and we get
0 = β1 a false statement therefore this system of equations has no solution or is an inconsistent system.
A question to ask is why this method works. Remember that when working with equations, you can create equivalent equations (equations with same solutions as the original equation) by multiplying/dividing both sides by a constant or adding/subtracting equal quantities from both sides. So essentially we are manipulating the two equations to get what we want.
3. {π₯2 β π¦2 = 4π₯ + 3π¦ = 5
We can see from our knowledge of graphs that the first equation is a hyperbola and the second a line so we expect two solutions. As you can see it is not easy to read the solutions from their graphs but we got an insight into how many solutions to expect. So here substitution method will come in handy. Solve the second equation for π₯ and we get π₯ = β3π¦ + 5. Now substitute this value in the first equation we get
(β3π¦ + 5)2 β π¦2 = 4 9π¦2 β 30π¦ + 25 β π¦2 = 4 8π¦2 β 30π¦ + 21 = 0
Using quadratic formula we get
π¦ =30 Β± β(β30)2 β 4(8)(21)
16=15 Β± β57
8
π¦ =15
8+β57
8 will give us π₯ = β3(
15
8+β57
8) + 5 Or π₯ = β
5
8β3β57
8
π¦ =15
8ββ57
8 will give us π₯ = β3(
15
8ββ57
8) + 5 Or π₯ = β
5
8+3β57
8
Solutions are
(β5
8β3β57
8,15
8+β57
8) and (β
5
8+3β57
8,15
8ββ57
8)
Solving Equations and Inequalities Page 279
4. {π₯2 + 2π¦ = 4π₯ β 2π¦ = β2
Elimination Method Adding the two equations we get π₯2 + π₯ = 2 or π₯2 + π₯ β 2 = 0
(π₯ β 1)(π₯ + 2) = 0 π₯ = 1 or π₯ = β2
π₯ = 1 gives us 1 β 2π¦ = β2 or π¦ =3
2
π₯ = β2 gives us β2 β 2π¦ = β2 or π¦ = 0
Solutions are (β2,0) and (1,3
2).
Graphing does give us solutions here as
(β2,0) and (1,3
2).
Solutions to system inequalities are all points in a region that satisfy both inequalities. What this means
is that if we have a strict inequality the points on that curve are automatically not part of the solution
and we draw the curve as dotted lines to indicate this. When solving inequalities in one variable the
solutions are regions on a number line, now when working with inequalities in two variables we will
work with regions in the plane. Finding solutions to inequalities in two variables is really asking for a
particular relationship be satisfied between the (π₯, π¦) coordinates. For example, if we say we want a
solution to the inequality 3π₯ β π¦ > 5 that is the same as saying find all coordinates (π₯, π¦), so that
3π₯ β 5 > π¦ or that the π¦-coordinate is always smaller than five less than three times the π₯-coodinate.
That means all points in the solution sets are coordinates whose π¦-coordinate satisfies that relationship.
See below for an example of how (1,1) is not a solution to the inequality, but (1, β5) is a solution.
All points in the region below the line shown by the shaded blue region are the solutions as all the π¦-coordinates in this region satisfy the required condition.
To extend this to a system of inequalities would mean finding solutions to each of the inequalities in the system and seeing where all the solutions sets intersect. That is pretty much how you would solve system of inequalities and find the points in the regions that are solutions to all the inequalities in the system.
Solving Equations and Inequalities Page 280
5. {3π₯ β π¦ > 22π₯ + π¦ < 3
Step 1: Plot each of the lines in the system. Determine whether each of the lines is dotted or solid based (dotted if working with a inequality, and solid if working with either β€, or β₯). Step 2: Take a test point on the lines to see if it is part of the solution or not, then shade the appropriate region. In this case each of the lines are dotted lines since we have strict inequalities. Let us pick (0,0) as our test point. In the first inequality we see 0 > 2 is a false statement so all the points in the region under the line 3π₯ β π¦ = 2 are solutions to this inequality. For the other inequality (0,0) as our test point we see 0 < 3 which is true so all points in the place above the line 2π₯ + π¦.
6. {π₯2 + 2π¦ < 4π₯ β 2π¦ β₯ β2
Plot each inequality and the overlap is the solution as you see in the picture to the right.
Solving Equations and Inequalities Page 281
Section 3.4 Worksheet Date:______________ Name:__________________________
Concept Meaning in words and/or examples as required
1. System of Equations
2. System of Inequalities
3. Solutions to system of equations
4. Solutions to system of inequalities
5. Graphing Method
6. Elimination Method
7. Substitution Method
Difficulties encountered in the section:
Solving Equations and Inequalities Page 282
Exercises 3.4
1. What is a system of equations?
2. When would you use the graphing method?
3. When would you use the elimination method?
4. When would you use the substitution method?
5. What is the difference between solving system of equations versus system of inequalities?
6. What is most number of solutions you expect from of a system of equations where both
equations are second degree polynomials?
7. What is least number of solutions you expect from of a system of equations where both
equations are second degree polynomials?
Solving Equations and Inequalities Page 283
8. Find the solutions to system of equations and inequalities below.
a. {5π₯ β π¦ = 4π₯ β 2π¦ = 3
b. {1
2π₯ β π¦ = 4
π₯ β 2π¦ = 3
c. {π₯ β 2π¦ = 4
β4π₯ + 16 = β8π¦
d. {1.5π₯ β 2π¦ = 1π₯ β 5π¦ = β3
e. {2π¦ = 1 β π₯4π₯ β 5π¦ = β3
f. {π¦ = 1 β 3π₯6π₯ + 2π¦ = 5
g. {π₯ = 4 β 3π¦2π₯ + 8 = 5π¦
h. {
2
3π¦ β 3π₯ =
1
21
3π₯ β
2
5π¦ =
2
5
i. {0.55π₯ β 1.01π¦ = 4.131.2 + 3.24π¦ = β7.32
j. {π₯2 + π¦2 = 4
π₯ β π¦2 = β2
k. {π₯2 + π¦2 = 4
π₯2 β π¦2 = 1
l. {π₯2 + π¦2 = 4
π₯ β π¦2 = β6
m. {9π₯2 + 4π¦2 = 36π₯ = 3
n. {9π₯2 + 4π¦2 = 36π¦ = 3π₯ β 1
o. {9π₯2 + 4π¦2 = 363π₯ β π¦ = β12
p. {π₯ β π¦ < 4π₯ β 2π¦ β₯ 3
q. {π₯2 + π¦2 β€ 4
π₯ β π¦2 < 1
r. {π₯2 + π¦2 > 4
π₯2 β π¦2 < 1
s. {π₯2 + π¦2 β₯ 4
π₯ β π¦2 β₯ β6
9. Use systems of equations or inequalities to solve the following.
a. Paul invests $100,000 in three stocks that pay dividends of 6%, 8%, and 10%. The
amount invested in the 10% is twice the amount at 6% and the return on the whole
investment is $8600 per year. How much is invested in each stock type?
b. Tickets at a play were sold for either $5 for children, $7 for college students and $9 for
general adult admission. The total number of tickets sold was 400 and the total ticket
revenue was $2500. How many tickets of each type were sold if the number of general
adult tickets was 1/7th of the total of the child and student tickets sold.
Solving Equations and Inequalities Page 284
3.5 Introduction to Matrices and Gauss Elimination Method We saw in the previous section how to solve a system of equations. The first thing to notice in the
examples we solved was for a system of linear equations we created equivalent systems of equations by
performing what are called elementary row operations which is basically allowing you to multiply or
divide one of the equations in the system by a constant, or add two or more equations together creating
a new equivalent system of equations. Two equivalent systems of equations have the same set of
solutions.
Matrix: A matrix is a collection of numbers organized into rows and columns.
The size of a matrix is determined by the number of rows and columns it has.
For counting numbers π, and π, a π Γ π matrix is said to have π rows and π columns.
For example,
1. [3 β1β2 4
] is a 2 by 2matrix (with 2 rows and 2 columns).
2. [1 β2 30 1 47 β1 2
] is a 3 by 3 matrix (with 3 rows and 3 columns)
3. [β1 0 43 β2 5
] is a 2 by 3 matrix (with 2 rows and 3 columns).
4. [2 3 4β1 3 5β2 1 5
|6β21] this matrix is called an augmented matrix and is a 3 by 4 matrix. Such matrices
can be used to represent system of equations. This particular matrix could represent the system of
equations given by {
2π₯ + 3π¦ + 4π§ = 6βπ₯ + 3π¦ + 5π§ = β2β2π₯ + π¦ + 5π§ = 1
as you can see the coefficients of π₯, π¦, and π§ are in the
first, second, third columns respectively and the constant term are in the last column. The equality
symbol is replaced with the bar between the third and the fourth column.
Playing
As with any new object mathematicians define, we may wonder how to do arithmetic with these new
objects. Can you think of a way we could add and subtract two matrices?
Just like any objects we can add and subtract like objects which in this case would mean two matrices
must have the same size to be able to add and subtract. And then a natural way to perform these
operations would be to do it term by term.
[3 β1β2 4
] + [2 53 β6
] = [3 + 2 β1 + 5β2 + 3 4 + (β6)
] = [5 41 β2
]
Subtraction would be similar.
What do you think would the value of 2 [3 β1β2 4
]. Trust your intuition before peeking on the next
page.
Solving Equations and Inequalities Page 285
2 [3 β1β2 4
] = [2 Γ 3 2 Γ β12 Γ β2 2 Γ 4
] = [6 β2β4 8
]
This is referred to as multiplying the matrix by a scalar.
We can also multiply two matrices but in this case it is not a very intuitive how the multiplication
works. See if you can come up with a clever way to multiply matrices and decide on which matrices we
could multiply together and how.
Matrices as such may have been used earlier but became mostly popular around 1850βs. Matrix theory
is a subject in mathematics that deals with intricacies of matrices. They are used in many fields like
quantum mechanism, chemistry to describe atomic structures, computer graphics, and many other
interesting fields.
We can use matrices to solve very complex systems of equations. For now we will only focus on how to
use the elementary row operations to solve system of equations of three or more variables.
The Gauss Elimination Method is basically taking the augmented matrix of a system of linear equations
and using elementary row operations bring the matrix into a triangular form that has ones in the
diagonals and zeros under them. For example, for a system of linear equations in three variables to get
it in the form [ 1 β1 20 1 β30 0 1
|4β12] this will allow us to get the solutions to the system of equations
easily since here for example we will get {π₯ β π¦ + 2π§ = 4π¦ β 3π§ = β1π§ = 2
and then we can proceed to get the solutions
to system of equations.
Elementary Row Operations
Let us see why this works. We will use a two by two system of linear equations and then show you the
beauty of this approach as we can then solve systems of linear equations that has three or more
variables. We will represent rows by writing π π or π π to mean nth row. Elementary row operations will
be denoted as π 2 = β3π 1 + 2π 2 means the new row 2 is found by adding β3 times row1 to 2 times
old row2.
Pay careful attention to the solving of the system of linear equations in two variables and the
equivalent step by step use of matrices to solve the same system. This careful attention will pay
dividends later to solve more complex system of linear equations.
Solving Equations and Inequalities Page 286
Practice Examples
Solve the system of equations below.
Solving by elimination method Solving using matrices
1. {ππ β ππ = πππ β ππ = π
So first multiply the first equation by π and second equation by βπ creating the equivalent system of
{ππ β ππ = ππ
βππ + πππ = βππ
Add the two equations πππ = βπ or
π =βπ
ππ
Use this value of π into one of the equations to get the value of π. So we get
ππ β π(βπ
ππ) = π or
ππ = π βππ
ππ=ππβππ
ππ=ππ
ππ
Or π =π
π(ππ
ππ) =
ππ
ππ.
The solution of the system is
π =ππ
ππ, π = β
π
ππ
1. {ππ β ππ = πππ β ππ = π
The augmented matrix to represent this system of equations will be
[ 3 β22 β5
| 6 7]
π 1=2π 1π 2=β3π 2β [
6 β4β6 15
| 12β21
]
π 2=π 2+π 1β [
6 β40 11
| 12β9]
π 2=
1
11π 2
β [6 β40 1
| 12
β9
11
]
π 1=4π 2+π 1β [
6 00 1
| 12 β
36
11
β9
11
]
π 1=1
6π 1
β [1 00 1
| 2 β
6
11
β9
11
] = [1 00 1
|
16
11
β9
11
]
π₯ =16
11 and π¦ = β
9
11
It may look like a lot of work compared to use matrices but when working with more than two variables
you will see the efficiency of it. So letβs attempt to solve a three by three system of linear equations.
Solving Equations and Inequalities Page 287
2. {
2π₯ β 3π¦ + π§ = 73π₯ + 2π¦ β 2π§ = β3βπ₯ + π¦ + 3π§ = 4
Our goal is to get ones in the diagonals and zeros under it so we can start with interchanging
rows one and three represented as below. All that means is we are writing the third equation
first and the first equation third. Interchanging equations creates an equivalent system of
equations. Then continue with the elementary row operations.
[2 β3 13 2 β2β1 1 3
|7β34]π 1βπ 3β [
β1 1 33 2 β22 β3 1
|4β37](β1)π 1βπ 1β [
1 β1 β33 2 β22 β3 1
|β4β37]
π 2β3π 1βπ 2π 3β2π 1βπ 3β [
1 β1 β3
0 5 7
0 β1 7
|β4
9
15
] π 2βπ 3β [
1 β1 β3
0 β1 7
0 5 7
|β4
15
9
]
β1(π 2)βπ 2β [
1 β1 β3
0 1 β7
0 5 7
|β4
β15
9
]π 3β5π 2βπ 3π 1+π 2βπ 1β [
1 0 β10
0 1 β7
0 0 42
|β19
β15
84
]
1
42π 3βπ 3
β [1 0 β10
0 1 β7
0 0 1
|β19
β15
2
]π 1+10π 3βπ 1π 2+7π 3βπ 2β [
1 0 0
0 1 0
0 0 1
|1
β1
2
]
Solution: π₯ = 1, π¦ = β1, π§ = 2
Interchange R1 and R3
+(β3 3 9 12)
0 5 7 9
π 2 β 3π 1β π 2
3 2 β2 β3
+(β2 2 6 8)
0 β1 7 15
π 3 β 2π 1 β π 3
2 β3 1 7
Interchange R2 and R3
+(0 β5 35 75)
0 0 42 84
π 3 β 5π 2 β π 3
0 5 7 9
+( 0 0 10 20)
0 0 0 1
π 1 + 10π 3 β π 1
1 0 β10 β19
Solving Equations and Inequalities Page 288
3. {
π₯ + π¦ β π§ = 22π₯ + 2π¦ β 2π§ = 43π₯ β 2π¦ + π§ = 1
Our augmented matrix will be
[1 1 β12 2 β23 β2 1
|241]
π 2β2π 1βπ 2π 3β3π 1βπ 3β [
1 1 β1
0 0 0
0 β5 4
|2
0
β5
]π 2βπ 3β [
1 1 β1
0 β5 4
0 0 0
|2
β5
0
]
1
β5π 2βπ 2
β [1 1 β1
0 1 β4/5
0 0 0
| 2
1
0
]π 1βπ 2βπ 1β [
β1 0 β1/5
0 1 β4/5
0 0 0
| 1
1
0
]
Since the bottow row is always true we can set π§ = π‘ where π‘ = πππ¦ ππππ ππ’ππππ
That makes π¦ =4
5π‘ + 1 and π₯ =
1
5π‘ + 1 .
There are infinitely many solutions to this sytem.
4. {
π₯ + π¦ β π§ = 22π₯ + 2π¦ β 2π§ = 6βπ₯ β 2π¦ + π§ = β3
Our augmented matrix will be
[1 1 β12 2 β2β1 β2 1
|26β3]
π 2β2π 1βπ 2π 3+π 1βπ 3β [
1 1 β1
0 0 0
0 β1 0
|2
2
β1
]
In row 2 we notice the equation 0 = 2 which is never true so this system of equations has no solutions.
Solving Equations and Inequalities Page 289
Section 3.5 Worksheet Date:______________ Name:__________________________
Concept Meaning in words and/or examples as required
1. Matrix
2. Elementary Row Operations
3. Gauss Elimination Method
Difficulties encountered in the section:
Solving Equations and Inequalities Page 290
Exercises 3.5
1. Please answer the following without looking at other resources.
a. Is it possible to add two different matrices? If yes, explain when and how.
b. Is it possible to subtract two different matrices? If yes, explain when and how.
c. Is it possible to multiply two different matrices? If yes, explain when and how.
d. Is it possible to divide two different matrices? If yes, explain when and how.
2. Why should we bother learning about the Gauss Elimination method?
3. What are the advantages of writing the system of equations using matrices?
Solving Equations and Inequalities Page 291
4. Do you think we use matrices for a two by two system of linear equations?
5. What about using matrices for two by two nonlinear system of equations? If you think we can,
explain how?
6. Solve the following systems using Gauss Elimination Method.
a. {
π₯ β 2π¦ + π§ = 43π₯ β π¦ + 6π§ = β14π₯ + 3π¦ β 7π§ = 5
b. {4π₯ + 5π¦ + 2π§ = β5π₯ + π¦ β 3π§ = β4
β3π₯ + 2π¦ β π§ = β13
c. {
π₯ + π¦ + 2π§ = 42π₯ + 2π¦ + 4π§ = β4β3π₯ + 2π¦ β π§ = β13
d. {
π₯ β π¦ + 2π§ = β5π₯ + π¦ β 3π§ = β4
β2π₯ + 2π¦ β 4π§ = 10
e.
{
1
2π₯ β
1
3π¦ +
1
5π§ =
11
30
π₯ β2
3π¦ + π§ =
4
3
4π₯ β π¦ +1
2π§ =
7
2
f. {
0.21π₯ β 0.2π¦ β 1.4π§ = 2.01π₯ β 0.33π¦ + 1.4π§ = 0.26
β0.4π₯ + 3.1π¦ + 2.1π§ = β8.7
g. {
3π₯ β 2π§ = 7βπ₯ + 4π¦ + 5π§ = β7β2π₯ β 4π¦ β 3π§ = 0
h. {
π₯ + π¦ + π§ + π€ = 12π₯ β π¦ + 3π§ β π¦ = β9π₯ β π¦ β π§ + π€ = 1
π₯ β 2π¦ + π§ β 2π€ = β8
i. {
π₯ β 2π¦ + 3π§ β π€ = 12βπ₯ + 3π¦ β 2π§ β 3π€ = 1β4π₯ + 6π¦ + π§ β 3π€ = 1π₯ β 2π§ + π€ = β6