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By the Numbers © Milos Podmanik Page 5

Chapter 1 Linear Functions

1.1 Functions

Functions are all around us. Every person is linked to a distinct set of characteristics. Every freezer (that is properly functioning) takes water and converts it into ice cubes. Every line of computer code takes a set of inputs and converts them into unique outputs. The applications are endless. But, we must first understand what we mean when we talk about functions and, to do that, we must consider some basic terminology. An input variable is a variable whose value determines the value of another variable, called the output variable.

For example, since the time of day determines how warm it is outside in Chandler, AZ on a summer day, we would say that:

Input: Time of day

Output: Temperature It would not make too much sense to say that the temperature determines what time it is outside.


Example 1: Identify the input and the output variables given that:

Variable 1: Temperature of water (°F) Variable 2: Time that water has been exposed to heat for (seconds)

SOLUTION: Since the time that water has been exposed to heat for determines the temperature of the water, we would say that

Variable 2 is the input Variable 1 is the output

Example 2: Identify the input and the output variables given

that:

Variable 1: Number of people in a family Variable 2: Combined family income

SOLUTION: Since the number of people in a family is likely to determine the combined family income (as having more people in a family likely means a larger overall income), we conclude that

Variable 1 is the input Variable 2 is the output

Example 3: A marketing analyst will study two variables related to its potential clients: age

and income. Identify the input and output variables and give justification for this choice.


SOLUTION: Since a person’s age helps to determine that person’s income, we would say that age is the input variable and income is the output variable.

Fairly often, it is not completely clear which variable is the input and which is the output. For example, if

Variable 1: Number of bars in City X Variable 2: Number of shoe stores in City X

we are unable to tell if it should be the case that number of bars determines number of shoe stores or if number of shoes stores determines the number of bars. In this case, researchers would choose one of the variables, in hopes that it will help them determine values of the other variable. For instance, suppose the researcher wants to predict the number of bars in a city, provided he knows how many shoe stores a city has. Then, he has effectively identified number of shoe stores as the input and number of bars as the output. He could just as well have decided to reverse the two! So, why then, is this not all a process of guessing? Because most of the time we want to better understand something we have little (or no) information on by using a piece of information that we do have. Researchers infrequently use the terms “input” and “output.” Instead, they call them independent and dependent variables, respectively. This was chose to suggest that the independent variable (input) is used to determine the dependent variable (output), since the dependent variable depends on the independent variable. Independent and Dependent Variables

Independent Variable = Input Variable Dependent Variable = Output Variable

When two variables are being studied together we call their relationship a relation. Every input/output variable relationship is a relation. However, not every relation is a function. A function is a special type of relation that we define below:


Function A function is a relation in which every input variable value has exactly one output variable value. NOTE: The opposite does not have to be true! This definition implies that we have a function only when we are able to look at each input value and find exactly one output value. This means that we cannot have more than one output value for a specific input value, nor can we have an input that has no output values. 1.1.1 Functions in Table Form Let’s look at a few tables of input and output pairings. Example 4: Determine which of the following sets of relations represent functions.

Input 1 2 4 5 6 9 Output 2 -1 3 5 7 10

Table 1

Input 1 2 4 5 2 9 Output 2 -1 3 5 7 10

Table 2

Input 1 2 4 5 6 9 Output 2 2 2 2 2 2

Table 3

SOLUTION: From Table 1we see that each input is only listed once. Each of these inputs has exactly one output variable, so there the relation between the two variables is a function. From Table 2 we see that the input value 2 has two different output values: -1 and 7. This violates the definition of a function, since it is not the case that every input has exactly one output. This relation is not a function. From Table 3 we see that every input is displayed only once and each has exactly one output value. Some people conclude that this relation isn’t a function on first glance;; most see the output value of 2 repeated and jump the gun. However, keep in mind the definition:


For every input there is exactly one output.

This definition does not specify that every input value has a unique output not used by the other inputs! Thus, this relation is a function.

We can just as easily discuss a function in a number of other ways (in addition to the comparison in tables): ordered pairs and graphs. 1.1.2 Functions in Ordered Pair Form Recall that an ordered pair is formed as a set of parenthesis that contains an input value and its corresponding output value, separated by a comma. For example, (2, 7) is an ordered pair that relates an input value of 2 with an output value of 7. This idea is no different than the tabular representation provided above. An ordered pair is simply a more compact form to represent an input with its output. To represent a collection, or set, of ordered pairs, we typically will enclose the ordered pairs in curly braces. For example:

(2,1), (4,3), (5,2) Is the set of three ordered pairs. Example 5: Determine whether the following sets of ordered pairs represent functions:

a.) (4,1), (6,1), (7,2), (9,3) b.) (2,2), (4,3), (2,3), (7,9) c.) (1,1), (5,1), (3,3), (4,3)

SOLUTION:

a.) Since each input value has exactly one output value, this relation is a function. b.) This relation is not a function since the input value 2 has outputs of 2 and 3. This violates

the definition of a function. c.) Each input value has exactly one value. This is a function.


Example 6: Using either a table or ordered pairs, explain why the relation considering a

person’s shoe size (input) and his/her hair color (output) does not represent a function. SOLUTION: Let’s say we consider people who are 60 inches tall (5 feet). Then it is very likely that not all of these people have the same hair color. For example, we might encounter the following ordered pairs for three different people of the same shoe size:

(60, brown), (60, black), (60, blonde) For the single input of 60 inches, there is a number of different outputs. This does not live up to the definition for a function.

1.1.3 Functions in Graphical Form There is really not much of a distinction between the forms of functions discussed so far: functions represented by tables and functions represented by ordered pairs. Even less so is there a difference between these two and functions in graphical form. Why? Since graphs are simply sets of (sometimes many) ordered pairs represented visually, we should be able to determine relatively quickly (even faster than by looking at tables or ordered pairs, in fact) whether a graphical relation represents a function. Recall that a graph has two axes, the horizontal axis and the vertical axis. The horizontal axis runs left-and-right and the vertical axis runs up-and-down. An easy way to remember this fact is by thinking about the horizon. If you look out to the horizon, you are looking out ahead, not up or down.


Table 4: A graph labeled with axis names

The major question is, what is a graph but a collection of ordered pairs? If we were to plot a set of points, say

(2,3), (5,9), (4,11), (1,6) We would immediately see that this data represents a function; each input is displayed only once. We would see:

Suppose that we add one more point:

(2,3), (5,9), (4,11), (1,6), (𝟐, 𝟖) We see that this set of ordered pairs no longer represents a function, since the input value 2 has two different outputs: 3 and 8. Graphically,

Horizontal Axis

Vertical Axis

10 5 5 10

10

5

5

10

1 2 3 4 5

2

4

6

8

10


We notice that a function cannot exist when, for one particular input value, there are two (or more) output values. Visually, if we draw a vertical line set at any input, and two or more points are connected by this vertical line, then a function does not exist. This must only be the case for one input value in order for us to determine that a function does not exist. This is referred to as the vertical line test. Vertical Line Test If a vertical line drawn anywhere on a graph touches more than one point, then the graph does not represent a function. Example 7: Over a one-year span of time, the U.S. Geological Survey reported the following

graph, which gives the water level of Horseshoe Reservoir at Horseshoe Dam in Arizona (SOURCE: http://water.usgs.gov/watuse/). Suppose that water levels are recorded once each day.

1 2 3 4 5

2

4

6

8

10


Does the relation represent a function? Explain in the context of the problem. SOLUTION: Yes, if we draw a vertical line through the graph on any day, we get only one water level reading. This makes sense, since the level is recorded once a day. It is not possible for the water to be at two different levels at one time.

Function Terminology We often say,

Output is a function of input When a functional relationship exists between the two variables To give an example based on the Horseshow Dam graph above, we would say that

Water level is a function of day due to the fact that graph represents a function. Example 8: A friend claims that the following graph represents his distance from home as a

function of time of day:


Does his assertion make sense? SOLUTION: His answer does not make sense. Since for multiple fixed times of day (input), there are two distances on the graph (output), he cannot claim that this is a function. For example, at 4pm he claims to be both 4 feet and 12 feet away from home. This is physically impossible. Thus, distance from home is not a function of time of day.

1.1.4 Function Notation As we have seen in the previous examples, we typically write

Output is a function of input Rather than to have to write this out each time, mathematicians have developed some notation to allow an abbreviation to take place. To show that the output (𝑜) is a function of the input (𝑖), we write

𝑜 = 𝑓(𝑖) Reading this from left-to-right we have:

12

34

56 7

89

1011

1213

1415

0

2

4

6

8

10

12

14

16

0 2 4 6 8 10

Dist

ance

(ft)

Time of Day (hour in the afternoon)

Distance from Home


𝑜 = 𝑓

( 𝑖) To many students, this is often confusing notation, since we usually use parenthesis to represent multiplication. Here, however, 𝑓(𝑖) does not mean to multiply a variable 𝑓 by a variable 𝑖! In fact, 𝑓 is not even a variable! It is simply a shorthand abbreviation for the word “function.” Above, we illustrated that water level, 𝑤, is a function of day, 𝑑. Thus, using function notation, we would write:

𝑤 = 𝑓(𝑑) to represent the sentence:

water level is a function of day

Function Notation To represent that the output, 𝑜, is a function of the input, 𝑖, we write:

𝑜 = 𝑓(𝑖) This reads,

Output is a function of input

CAUTION: 𝑓(𝑖) does not mean multiply!! Example 9: A biologist studies the relationship between the height of a cricket’s jump (in

inches), ℎ, and the length of its chirp (in seconds), 𝑙. He concludes that:

ℎ = 𝑓(𝑙) What does this mean? SOLUTION: This means that he has found height of a cricket’s jump to be a function of its chirp length. That is to say, if has a known chirp length, then we can determine exactly how high it will jump.


Example 10: Two variables are of interest to a data analyst for the National Basketball Association (NBA): free throw accuracy (as a % of baskets made) and age.

a.) Does it seem reasonable to conclude that free throw accuracy is a function of age?

Explain. b.) Does it seem reasonable to conclude that age is a function of free throw accuracy?

SOLUTION:

a.) In this statement, age is the input and free throw accuracy is the output. Since we would expect that two people of the same age to have two different free throw accuracies, this does not represent a function.

b.) In this statement, free throw accuracy is the input and age is the output. If one finds two people that have the same free throw accuracy, it cannot be concluded that they are both of the same age. This is not a function.

Homework Problems – 1.1

1. For each of the following identify what you believe are the input (independent) and

output (dependent) variables. Justify your answer. a. Variable 1: number of friends having dinner together; Variable 2: total cost of bill b. Variable 1: number of cars on the freeway; Variable 2: time of day c. Variable 1: size of a house; Variable 2: location of house d. Variable 1: number of eggs produced by a hen; Variable 2: temperature outside

2. The weights and heights of six mathematics students are given in the following table.

Weight (in pounds)

Height (in centimeters)

165 172 123 157 212 183 175 178 165 163 147 167

a. In the statement “Height is a function of weight,” which variable is the input and

which is the output? b. Is height a function of weight for the six students? Explain using the definition of

a function.


c. In the statement “Weight is a function of height,” which variable is the input and which is the output?

d. Is weight a function of height for the six students? Explain using the definition of a function.

For 3 – 7 below, determine whether or not each of the situations describes a function. Give a reason for your answer.

3. a. (2,5), (−3,5), (10,5), (𝜋, 5) b. (5,2), (5, −3), (5,10), (5, 𝜋)

4.

a. In the following table, elevation is the input and amount of snowfall is the output.

Elevation (in feet)

Snowfall (in inches)

2000 4 3000 6 4000 9 5000 12

b. In the preceding table, snowfall is the input and elevation is the output.

5.

a. The letter grade in this course is a function of your numerical grade. b. The numerical grade in this course is a function of the letter grade.

6. Number of hours using the Internet is the input, and the monthly cost for the Internet

service is $19.95.

7. a. On a given night, your blood-alcohol level is a function of the number of beers

you drink in a two-hour time period. b. The number of beers you drink in a two-hour period is a function of your blood-

alcohol level.

8. For each of problems 6 and 7, represent each variable with a letter, and then write the relationship in function notation (assume that for a particular group of people, all are functions).


1.2 Models and Function Notation One of the primary tasks we’re faced with in mathematics is to build models that express the relationship between input and output variables. Much like creating a Styrofoam model of a planet, building mathematical models is important in order to better help us understand an underlying “thing” of importance.

Suppose our goal is to better understand the relationship between the number of minutes a person talks on his cell phone and the total monthly cost. This person in particular decides to go with a pay-as-you-go plan, and so pays by the minute. If the cost per minute is $0.25, what would be an equation that would give us the total monthly cost? Since our total monthly cost depends on the number of minutes talked, we identify the total monthly cost as the output and number of minutes talked as the input. If we first consider a table to help us understand this relationship, we find:

Minutes Cost ($) 1 0.25 2 2(0.25) = 0.50 10 10(0.25) = 2.50 100 100(0.25) = 25.00

We quickly see that to get cost, we simply multiply the number of minutes by the 0.25, the cost per minute. Symbolically, Cost = Minutes ∙ 0.25


Of course, we know that in math our goal is to reduce the amount of wording we use, so we let

𝐶 = cost 𝑚 = minutes

Giving us,

𝐶 = 0.25𝑚 (we move the coefficient out front for simplicity sake) From the previous section, we know that cost is a function of minutes talked, so we could write:

𝐶 = 𝑓(𝑚) Why? Because for and value of the input, minutes, there can only be one output, cost. This precisely meets the requirements for a function. (REMEMBER: this does not mean multiply 𝑓 and 𝑚!) While this is a nice way of saying that a function exists, it would be even better if it could somehow incorporate the actual equation. To make this notation more useful, we usually condense it down a bit more. Instead of writing,

𝐶 = 𝑓(𝑚) We will simply write,

𝐶(𝑚) This means the exact same thing. It can still be read as “𝐶 is a function of 𝑚.” We can now complete it by writing:

𝐶(𝑚) = 0.25𝑚 IMPORTANT: 𝐶(𝑚) does not mean multiply 𝐶 and 𝑚! Condensed Function Notation When a function exists between two variables, say 𝑥 and 𝑦,


𝑦 = 𝑓(𝑥)

Then we can simply write this functional relationship in a more condensed form,

𝑦(𝑥) = whatever the equation is Now that we have 𝐶(𝑚), we can answer questions, such as, “What is the cost of 75 minutes?” Rather than to write questions like this every time, we can simply write:

𝐶(75) This can be read as, “Cost as a function of 75 minutes.” To evaluate this, we now determine what the total cost is:

𝐶(75) = 0.25(75) = 18.75

In short,

𝐶(75) = 18.75 which means, “the cost as a function of 75 minutes is $18.75.” Example 1: A chair company ships its new chair model in large

boxes. Each chair weighs 20 lbs. Come up with a model (equation) that gives the weight of a shipment, 𝑊, as a function of the number of chairs, 𝑛, that it ships. Then, find and interpret the meaning of 𝑊(10).

SOLUTION: To get the total weight of the shipment, we must multiply the weight of 20 lbs. by the number of chairs shipped. That is, Weight = 20 ∙ Number.


Symbolically,

𝑊 = 20𝑛 In condensed function notation form:

𝑊(𝑛) = 20𝑛 This means: “Weight as a function of the number of chairs shipped is given by taking 20 times the number of chairs shipped.” To find 𝑊(10), we take 20 times 10:

𝑊(10) = 20 ∙ 10 = 200

Or,

𝑊(10) = 200 This means that the weight as a function of shipping 10 chairs is 200 lbs.

Example 2: The following table is given to represent height of a tree as a function of years passed:

Years Passed, 𝑦 1 3 4 6 9 12 Height of a Tree, 𝐻, (feet)

3 5 7 12 13 14

Determine 𝐻(6) and then explain the meaning of this result. SOLUTION: Since the condensed function notation would be 𝐻(𝑦), we are being asked to find the height of the tree after 6 years have passed. From the table, we see that the answer is 12 feet. Thus,

𝐻(6) = 12


Evaluating an Equation When we are given an input value and are looking for the output value, we are evaluating the equation. 1.2.1 Solving Equations with Function Notation Function notation allows us to answer many types of questions. We know that 𝐻(6) demands to know what the output is when the input is 6. Given the table in Example 2:,

Years Passed, 𝑦 1 3 4 6 9 12 Height of a Tree, 𝐻, (feet)

3 5 7 12 13 14

Suppose we are asked to solve 𝐻(𝑦) = 13. What does this mean? It appears that we know the output value: 13. We are reading that: “The height of the three as a function of 𝑦 years passing is 13 feet.” It makes sense that our goal would be to determine 𝑦. As we scan the outputs, we find that a height of 13 goes with 9 years. Thus, we could say that:

𝐻(9) = 13 Solving an Equation When we are given an output value and seek the corresponding input value, we are said to be solving the equation. Example 3: The following graph gives the total U.S. oil

consumption (in millions of barrels) from 2003 to 2008 (SOURCE: Earth Policy Institute):


Let 𝑃 be the total production (in millions of barrels) and 𝑦 be the year.

a. Approximate 𝑃(2003). b. Solve 𝑃(𝑦) = 85.

SOLUTION:

a. Since production is a function of the year, 𝑃(2003) asks us to determine the total production in 2003. We estimate from the graph that this is about 79.8 million barrels. Thus,

𝑃(2003) = 79.8

b. We are given that the output is 85. This is asking us to find the year in which 85 million barrels were produced. From the graph, it appears that this occurred in 2007 (though 2006 also appears pretty close). Thus, our answer would be:

𝑃(2007) = 85

1.2.2 Domain and Range Just as every person can be described in terms of his or her characteristics, so too can a function. Domain

79.00

80.00

81.00

82.00

83.00

84.00

85.00

86.00

2002 2003 2004 2005 2006 2007 2008 2009

Prod

uctio

n (M

illio

n Ba

rrel

s per

Day

)

Year

U.S. Oil Consumption


The set of all possible inputs for a given function is called the domain. Example 4: A function is given by the set (2,1), (−4,3), (−2,−1), (6,4). What is the

domain of this function. SOLUTION: Since domain is the set of all possible inputs, here the domain would be -4, -2, 2, and 6. Using set notation, we could write:

Domain = −4,−2,2,4 \

Note that it is common practice to arrange the values in a domain from smallest to largest. Range The set of all possible outputs for a given function is called the range.

Example 5: Suppose gas is $3.90 per gallon at the local gas station. To determine the total cost of filling up your car, 𝐶, as a function of number of gallons pumped, 𝑔, is

𝐶(𝑔) = 3.90𝑔 Determine a practical domain and practical range for this function.

SOLUTION: A practical domain is very subjective. It would make sense that one can pump as few as 0 gallons (arrive at the pump and change your mind) and as many as 50 gallons (for a large truck). Thus, we can write the domain in a couple of ways: In words:

Domain: 0 to 50 gallons Or, in symbols as an interval:


Domain: 0 ≤ 𝑔 ≤ 50

So, how would we find a practical range (set of output values)? The range must jive with our selected domain. Since we know the least number of gallons we can pump is 0, the corresponding cost would be:

𝐶(0) = 3.90(0) = 0

The largest cost would be if the maximum number of gallons were pumped:

𝐶(50) = 3.90(50) = 195

Thus, our practical range would be: In words:

Range: $0 to $195

In symbols: Range: 0 ≤ 𝐶 ≤ 195

Homework Problems –1.2

1. You are given the following table that represents 5 children’s shoe sizes as a function of

their ages: (Video Solution)

Age, 𝑎 4 6 5 7 9 Shoe Size, 𝑆 4 5 5 5 6

a. Evaluate 𝑆(6) and explain the meaning of your answer. b. Solve 𝑆(𝑎) = 6 and explain the meaning of your answer.

2. According to national data, the number of alcohol-related deaths in the U.S., 𝐷, is a

function of the year, 𝑡 (SOURCE: U.S. Statistical Abstract, Table 23). The following graph is provided: (Video Solution)


a. Evaluate 𝐷(1999). Explain the meaning of your answer. b. Solve 𝐷(𝑡) = 22. Explain the meaning of your answer.

3. The distance you travel while hiking is a function of how fast you hike and how long you

hike at this rate. You usually maintain a speed of three miles per hour while hiking. (Video Solution)

a. Write a verbal statement (using words) that describes how the distance that you travel is determined.

b. Identify the input and output variables of this function. c. Write the verbal statement in part c using function notation for the input variable.

Let 𝑡 represent the input variable. Let ℎ represent the output variable. d. Use the equation from part c to determine the distance traveled in four hours. e. Evaluate ℎ(7) and write a sentence describing its meaning. Write the result as an

ordered pair. f. Determine the practical domain and the practical range of the function.

4. According to the Federal Highway Administration, it costs approximately $0.36 per mile

to own and operate a car. The total cost, 𝐶, is a function of the number of miles traveled, 𝑚. When you finally take your car to the junkyard, the odometer reads 157,200 (not good by today’s standards!). (Video Solution)

a. Write a function (equation/model), 𝐶(𝑚), to represent the situation b. What is the practical domain of this function? c. What is the practical range of this function?

5. You’re having a party, and you want to fill balloons with helium gas. The volume of

helium needed for one balloon is a function of the size of the balloon. You can write an equation 𝑉 = 𝑓(𝑟) to represent this function. If you assume the balloon is approximately the shape of a sphere, the volume 𝑉, of a spherical balloon is given by the formula 𝑉 = 𝜋𝑟 , where r is the radius of the balloon. The balloon will pop when the radius is approximately 25 centimeters. (Video Solution)

a. Complete the following table.

19

19.5

20

20.5

21

21.5

22

22.5

23

23.5

1998 2000 2002 2004 2006 2008

Deat

hs (i

n th

ousa

nds)

Year

Alcohol-Related Deaths


Radius of the Balloon (cm) 12 14 16 18 Helium Needed (cubic cm)

b. Identify the DV and IV. c. What would be a practical domain and practical range for this function? d. How much helium would you need to fill 100 balloons if each has a radius of 15

centimeters? e. Rewrite the volume formula using function notation. Use 𝑟 to represent the input

and 𝐻(𝑟) to represent the output. f. Is the graph increasing, decreasing, or constant? Explain. g. How can you determine, just by looking at the graph, that it is a function?


1.3 Graphs and Average Rate of Change Mathematicians often describe data as beautiful. Data are capable of telling incredible stories about the world around us. For example, over time we might see a trend in U.S. oil consumption. When extraordinary world events take place, we often see the effect it has on oil consumption by observing what occurs with the data values themselves.

Let’s start with a simpler example. Suppose you set the global positioning system (GPS) in your car to keep track of your trip from Gilbert, AZ to San Diego. You record your distance traveled as a function of hours since you left your house. Parts of your recordings are illustrated in the graph below:

What story does this graph tell us?

0

20

40

60

80

100

120

140

160

0 0.5 1 1.5 2 2.5 3 3.5

Dist

ance

Tra

vele

d (M

iles)

Time Traveled (hours)

Trip Distance Graph


First off, we can see that at time 0 hours, no distance had been traveled. The trip was just beginning. In the first 1.5 hours, it appears you traveled a total of 75 miles. At that time, you decided to pull over at a diner for a tasty meal. This took about ½ an hour, after which time, you proceeded to pick up your speed in order to make up for lost time. Now, one thing that might appear questionable is: how do we know that the driver was traveling at a faster rate in hours 2 to 3 as compared to hours 0 to 1.5? To answer this question, we first observe that the vehicle traveled 60 miles in the first 1.5 hours. If we compare the two as a ratio, we get:

60 miles1.5 hrs

Note that the division of the two units is , which can be read as “miles per hour.” If we divide 60 by the amount of time, 1.5 hours, we get:

40mihr

We can conclude that, on average, the vehicle traveled 40 miles per hour for the first 1.5 hours. Similarly, over the second period of driving, the vehicle traveled from 60 miles to 140 miles in the span of one hour. The difference is 80 miles. Thus, we can conclude the vehicle traveled:

80 mi1 hr

Or, more simply

80mihr

This proves the car was traveling, on average, faster after the rest stop than before the rest stop. The computations that we performed, taking the increase in distance and dividing it by the increase in time, is called average rate of change. Average Rate of Change The average rate of change is defined as


change in outputchange in input

The units for this ratio are:

units of outputunits of input

The most important things to remember when computing an average rate of change is to keep track of whether the change represents an increase or a decrease. When the change represents an increase, it is positive. When the change represents a decrease, it is negative.

Example 1: The graph below looks at the total amount of water (in gallons) used by an Salt River Project (SRP) customer over the course of a 1-day period (in hours).

a.) Tell a story that describes what the graph is telling us. b.) What would be the units that would describe an average rate of change for this situation? c.) Did the user use water at a faster rate between hours 4 am and 6 am, or between 8 pm and

10 pm? SOLUTION:

0

5

10

15

20

25

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

Wat

er U

sed

(Gal

lons

)

Hours after Midnight

Water Usage


a.) Between midnight and 4 am, no water was used. Between 4 am and 6 am, the user’s drip watering system ran and used a total of 10 gallons of water. Between 6 am and 8 am, the user did some laundry. From 8 am to 4 pm, the user was at work and so used no water. Upon returning at 4 pm, the user ran the dishwasher and did some more laundry. Between 8 pm and 10 pm, the user watered his lawn and took a shower. He then went to bed and no more water was used.

b.) The average rate of change units are

= , or gallons per hour.

c.) We determine the average rate of change over each time period.

From 4 am to 6 am, a 2-hr period, 10 gallons of water were used. Thus,

10 gal2 hr = 5

galhr

So, on average, the user consumes 5 gallons per hour between 4 am and 6 am. Between 8 pm and 10 pm, a 2-hours period, the user consumed from 13 to 23 gallons of water. This represents a usage of 10 gallons. Thus,

10 gal2 hr = 5

galhr

This allows us to conclude that his average usage was the same over each of the two time periods.

1.3.1 Describing Graphs with Non-Constant Rates of Change We began this section by describing graphs with segments straight lines. From our early years of learning math, we clearly learned about the difference between a straight line and curves. By definition, a line is the shortest path between two points. There are no bends. First off, however, we must understand how the idea of rate of change and straight lines are linked. Suppose that you go to the gas pump and pay $3.50 per gallon. We understand this to mean that, for every additional gallon you purchase, you are to pay $3.50 more. Imagine that we begin plotting cost of filling up as a function of number of gallons purchased:


We begin by plotting one obvious point: purchasing 0 gallons requires paying $0 and that purchasing 1 gallon requires paying $1.

Purchasing 2 gallons will cost us $3.50 more, for a total of $7. Viewing this another way, the cost of 2 gallons will be found by:

3.50$gal ∙ 2 gal = $7.00

0

0.2

0.4

0.6

0.8

1

1.2

0 0.2 0.4 0.6 0.8 1 1.2

Cost

($)

Gallons

Cost to Fill Up Gasoline

0

0.5

1

1.5

2

2.5

3

3.5

4

0 0.2 0.4 0.6 0.8 1 1.2

Cost

($)

Gallons



For 3 gallons, we would pay:

3.50$gal ∙ 3 gal = $10.50

We begin to see the pattern… the points begin to form a line-like pattern. That is, if we were to connect them, we would get a straight line. Only one important question begs to be answered: why?! First off, anytime we have a rate of change that is the same for all inputs, such as paying $3.50 per gallon, we say that we have a constant rate of change. We can still refer to it as an average rate of change, but constant rate of change is more specific. Constant Rate of Change

0

1

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8

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Cost

($)

Gallons


0

2

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12

0 0.5 1 1.5 2 2.5 3 3.5

Cost

($)

Gallons



An average rate of change that remains the same for all inputs is called a constant rate of change. Notice the set-up of our calculation of new outputs, given the constant rate of change:

3.50$gal ∙ (number of gallons)

In other words, if we summarize constant rate of change as AROC, we generically have: Finding Change in Output Given a constant rate of change, the change in output can be found by

Change in output = AROC ∙ (Change in input) Now, although it appears to be the case that the connection of points would form a line, how can we be certain that this will always be the case? Let’s test a couple of decimals. For instance, if we purchase 0.5, 1.2 and 2.4 gallons of gas, we pay:

Increase in cost = 3.50$gal ∙ 0.5 gal = $1.75



Plotting these points gives:


Our hypothesis that the points could be connected by one straight line is still consistent with what we are observing. Connecting the points gives:

If you’re still not convinced, you might continue to plot ordered pairs;; However, you would quickly be disappointed that none differ from the linear trend we are observing. Example 2: In 2009, the U.S. used 6,035 million pounds of

domestically caught fish for food. This represented a 598 million pound per year decrease from 2008. Assuming this average rate of change is valid for the foreseeable future, what is the expected amount of domestically caught fish that should be used for food in 2015?

SOLUTION: If we assume that the rate of change is constant over the next 6 years (2015-2009=6), then

0

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Cost

($)

Gallons


0

2

4

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8

10

12

0 0.5 1 1.5 2 2.5 3 3.5

Cost

($)

Gallons



change in output = −598mil lbsyr ∙ 6 yr

= −3588 mil lbs fewer Thus, we would expect there to be a food usage of

6,035 mil lbs − 3,588 mil lbs = 2,447 mil lbs

Many times, it is not the case that we have constant rate of change. Life is far more complex than we often let on. For this reason we cannot always assume a constant rate of change. Consider the following graph, which shows enrollment at U.S. colleges:

17.8

18.2

19.1

20.4 20.6

17.0

18.0

19.0

20.0

21.0

2006 2007 2008 2009 2010

Colle

ge E

nrol

lmen

t (m

il.)

Year

U.S. College Enrollment


Suppose our goal is to describe the trend in U.S. College Enrollment. Certainly we could say that enrollment has been on the rise since 2006, and clearly we see that it hasn’t been constant in the amount of growth. Between 2006 and 2009, the growth was quite large, but between 2009 and 2010, that growth largely slowed down considerably. What if we were asked to summarize the growth over the whole period of time? On the news you would likely hear something along the lines of, “the average…” So, we will describe the overall growth by describing it in terms of the average rate of change between 2006 and 2010. We see that in 2006, 17.8 million students were enrolled, and that in 2010, 20.6 million students were enrolled. The change in output is 20.6 − 17.8 = 2.8 million students. We expect this to be a positive number, since there is growth taking place. The change in input is 2010 – 2006 = 4 years. Thus,

𝐴𝑅𝑜𝐶 =2.8 mil students

4 years = 0.7mil students

yr

We can describe the average growth as 0.7 million students per year between 2006 and 2010. BUT, one might argue that this does not accurately describe the extraordinary amount of growth that took place initially and the miniscule growth that took place recently! If the news were to report the above number, you might be misled to believe that the trend is continuing. Let’s illustrate by finding the average rate of change over the “fast” growth period ,and then another average rate of change over the “slow” growth period. Average Rate of Change between 2006 and 2009:

Change in Output = 20.4 − 17.8 = 2.6 mil. students

Change in Input = 3 years

𝐴𝑅𝑜𝐶 =2.63mil. students

yr = 0.87mil. students

yr


Thus, the period of fast growth had an average rate of change of 0.87 million students per year, rather than the original 0.7 million students per year. This is a difference of 0.17 million (170,000) students per year! Average Rate of Change between 2009 and 2010:

Change in Output = 20.6 − 20.4 = 0.2 mil. students

Change in Input = 1 year

𝐴𝑅𝑜𝐶 =0.21mil. students

yr = 0.2mil. students

yr

This is 0.5 million (500,000) students per year smaller than the original. Thus, if the news wishes to be consistent with respect to the whole trend, it might wish to report 0.2 million students per year as the most recent average rate of change. This will emphasize that enrollment is growing, but not as quickly.

Example 3: The price of soybeans has been on the rise in the U.S. over the last several years. Soybeans represent a large source of the food we eat every day. In fact, reading the ingredients on your favorite snack will likely reveal the use of soybean oil. The graph below shows price of one bushel (60 lbs.) of soybeans as a function of year (SOURCE: U.S. Statistical Abstract, Table 860):

`

a. Find the average rate of change between 2005 and 2010 by using the graph. Explain the real-world meaning of your result.

5.00

6.00

7.00

8.00

9.00

10.00

11.00

12.00

2005 2006 2007 2008 2009 2010

Pric

e pe

r Bus

hel (

$)

Year

Price of Soybeans


b. Find the average rate of change between 2007 and 2009 by using the graph. Explain the real-world meaning of your result.

c. Explain why the two rates of change you calculated tell completely different stories. SOLUTION:

a. In 2005, the price per bushel was about $5.75. In 2010 the price per bushel was about $11.75. We find the change in output and the change in input:

Change in Output = $11.75 − $5.75 = $6

Change in Input = 2010 − 2005 = 5 years

𝐴𝑅𝑜𝐶 =65$yr = 1.20

$yr

On average, the price of one bushel of soybeans went up by $1.20 per year between 2005 and 2010.

b. In 2007, the price per bushel was about $10.25. In 2009 the price was about $9.50.

Change in Output = $9.50 − $10.25 = −$0.75

Change in Input = 2009 − 2007 = 2 years

𝐴𝑅𝑜𝐶 =−0.752

$yr ≈ −0.38

$yr

On average, the price of one bushel of soybeans decreased by $0.38 per year between 2007 and 2009.

c. While the practical reasons for this difference aren’t clear, we see that the first rate of change represented an interval over which the overall trend showed growth (from beginning to the end). We can show this trend by representing it with a straight (red) line below:


Between 2007 and 2009, this line goes down from left-to right, since this is the single time interval over which the price per bushel is decreasing:


1. The following graphs show various real-world scenarios. For each, write a possible story

describing the scenario. a.

5.00

6.00

7.00

8.00

9.00

10.00

11.00

12.00

2005 2006 2007 2008 2009 2010

Pric

e pe

r Bus

hel (

$)

Year

Price of Soybeans

5.00

6.00

7.00

8.00

9.00

10.00

11.00

12.00

2005 2006 2007 2008 2009 2010

Pric

e pe

r Bus

hel (

$)

Year

Price of Soybeans


b.

c.

135

136

137

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144

145

0 1 2 3 4 5 6

Wei

ght (

lbs)

Week Since Diet Began

0

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180

0 0.5 1 1.5 2 2.5 3

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ance

to F

lags

taff

Hours

0

0.5

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0 1 2 3 4 5 6 7

Tree

Hei

ght (

ft)

Years of Age


2. The following table of data follows the value of a home in Chandler from 2000 until 2007. (Video Solution)

Year, 𝑡 2000 2001 2002 2003 2004 2005 2006 2007 Value (in thousands of dollars), 𝑉(𝑡)

200 215 220 224 240 234 270 265

a. Determine the average rate of change in value from 2000 to 2004 b. Describe what the average rate of change in part a represents in this situation. c. Determine the average rate of change in value from 2004 to 2005. Interpret. d. Determine the average rate of change in value across the entire 7-year period

(from 2000 to 2007). Interpret. e. Between what two years did the home’s value increase the most? f. What does it mean to have a negative average rate of change when dealing with

home values across various years? g. Hypothetically, what would it mean if the average rate of change were zero for

the above scenario? h. Plot the above scenario as a scatterplot using pairs of the form (𝑡, 𝑓(𝑡)). Then,

connect each pair of points with a line, so as to “smooth” out the trend. Describe the trend in home prices.

3. In a new scenario looking at a company’s profit as a function of the number of Tickle-Me

Elmo’s sold, let 𝑛 represent number of units sold and 𝑓(𝑛) represent the company’s profit. Suppose the following ordered pairs are found. (Video Solution)

(1000, 25000) (1500, 26500) (2300, 29000)

a. Express (1500, 26500) in function notation. What does this mean? b. Find the average rate of change between the first two ordered pairs, being sure to

include the units. What does this mean? c. Find the average rate of change between the second two ordered pairs, being sure

to include the units. What does this mean? d. In the real-world, why might there be a difference in dollars per Tickle-Me Elmo

across two ordered pairs?

4. U.S. college enrollment in 2010 was 20.6 million. The average rate of change between 2009 and 2010 was 0.2 million enrolled per year. Assuming a constant rate of change, how many enrollees will there be in 2015? (Video Solution)

5. In 2009, one bushel (60 lbs.) of soybeans cost $9.59. In 2010 it cost $11.70. Assuming a

constant rate of change, estimate the price of one bushel of soybeans in 2014. (Video Solution)

6. The following graph shows the number of iPhones sold as a function of quarter (NOTE: there are 4 quarters in one year). (Video Solution)


a. What will be the units for the rate of change? b. Find the average rate of change between the 2nd quarter of 2007 and the 4th quarter

of 2011. Explain the real-world meaning of your answer. c. Find the average rate of change between the 3rd quarter of 2008 and the 1st quarter

of 2009. Explain the real-world meaning of your answer. d. Assuming the average rate of change between the 2nd quarter of 2007 and the 4th

quarter of 2011 is valid, estimate iPhone sales in the 4th quarter of 2012.


1.4 Slope and the Linear Function

It is definitely true that not all changing things in the world do so with a constant rate of change, but from day-to-day we encounter many situations in which they do. For instance, when you pay your electric bill, you are paying a constant price per kilowatt-hour (a measurement of energy usage). When you pump gas, you pay a fixed price per gallon. When you apply fertilizer to your roses, you are using a fixed amount per gallon of water. Of course, you can probably think of a handful of others fairly quickly. Our goal now is to use the idea of constant rate of change to generate an equation that will allow us to easily come up with estimates for different time points. 1.4.1 When One Point Represents the Initial Value


For example, consider Arizona State University (ASU) tuition rates. As with much of the nation, they have been on the rise. For example, in the 2010-2011 academic school year, undergraduate fulltime tuition was $2,960.74. In the 2011-2012 academic school year, tuition was $3,152.74 (SOURCE: www.asu.edu). Suppose our goal is to write an equation that allows us to make tuition predictions for the future. We would like to do so with ease: plug-in a number corresponding to the correct year as an input, and output the tuition. Let’s begin by modeling tuition as a function of years since the 2010-2011school year. That is, we will allow an input of 0 represent the 2010-2011 school year and 1 to represent the 2011-2012 school year (1 year later). This simplifies our problem considerably. Why? Because 0 is a smaller number than 2010! We will begin by plotting these two points:

Let’s begin by representing tuition with the variable 𝑇, and years since 2010 with the variable 𝑦. We know that, initially, the tuition was $2,960.74. Thus

𝑇(𝑦) = 2960.74 This does not completely solve our problem. After all, it is only a constant value that should adjust each year. We need to know by how much, so we find the average rate of change:

Change in Output = $3,152.74 − $2,960.74 = $192

Change in Input = 1 − 0 = 1 year

2950

3000

3050

3100

3150

3200

0 0.2 0.4 0.6 0.8 1 1.2

Tuiti

on ($

)

Years since 2010

ASU Tuition


So,

𝐴𝑅𝑜𝐶 =1921

$yr = 192

$yr

We know that, for every year that passes, the tuition should go up by $192. From the previous section:

change in output = ARoC ∙ (change in input) = 192𝑦

This will only give us how much more will be required for tuition for each year after 2010. So, if you find 192(10)=1,920 is the dollar amount by which tuition will be higher in 2020. This must then be added back to the initial amount of $2,960.72 to get the total in 2020, which would be

$2,960.72 + $1,920 = $4,880.72 in 2020 That is:

𝑇(𝑦) = initial value + change in output = 2960.72 + 192𝑦

To verify the tuition we found for 2020, we find:

𝑇(10) = 2960.72 + 192(10) = 4,880.72

This verifies our intuition! Per our discoveries in the previous section, this is no doubt a situation that embodies constant rate of change, if we assume that tuition will increase according to the $192 per year increase from 2010-2011 to 2011-2012. Our graph , if points were connected, would look like a straight line:


For obvious reasons, this is called a linear function. Linear Function A linear function is a function that has a constant rate of change. If we let 𝑖 represent the input and 𝑂 represent the output, then an equation modeling the linear relationship would be:

𝑂 = initial value + change in output Or,

𝑂 = initial value + (rate of change) ∙ 𝑖 We say that the output is a linear function of the input.

Example 1: An pest control company is interested in creating a function that represents the total yearly cost of it service as a function of the number of visits. Each year, the company charges $75 as a fixed annual membersh ip cost. The company then charges $20 per visit.

Create a function that will input the number of visits

and will output the total cost of services. SOLUTION: If we let 𝑛 = number of visits and 𝐶 = total cost, then we know

𝐶 = 75

2900

2950

3000

3050

3100

3150

3200

3250

3300

3350

3400

0 0.5 1 1.5 2 2.5

Tuiti

on ($

)

Years since 2010

ASU Tuition


is the minimum yearly cost, if only to be a member. The change in output will be given to us by taking the initial value and adding to it the change in output. That is:

𝐶 = 75 + (change in output) Since the cost for each visit is $20, the cost for 𝑚 visits will be

change in output = 20$

visit ∙ 𝑚 visits = 20𝑚 $

So,

𝐶 = 75 + 20𝑚 In function notation, we would write:

𝐶(𝑚) = 75 + 20𝑚

Example 2: The median income of a family in 2005 was $65,

906 and in 2009 was $71,627 (SOURCE: U.S. Statistical Abstract, Table 699).

a. Assuming constant rate of change, create a function that gives income as a function of years after 2005.

b. Predict the median family income in 2015 by using your function.

SOLUTION:

a. Since the input is in years after 2005, we represent this year with an input of 0. Since 2009 is 4 years later, we represent it with a 4.

Since 2005 is the initial year, we know the initial value will be $65,906. Thus, out output (income), 𝐼, and our input (years), 𝑦, are related:

𝐼 = 65906 + total change in output We calculate the change in output (income) and the change in input (years):

Change in Output = $71,627 − $65,906 = $5,721


Change in Input = 4 years

The rate of change is:

57214

$yr = 1430.25

$yr

After 𝑦 years, the median income is predicted to be

𝐼 = 65906 + 1430.25𝑦

b. 2015 is represented by 𝑦 = 10. We evaluate 𝐼(10):

𝐼(10) = 65906 + 1430.25(10) = $80,208.50

The median income in 2015 is estimated to be $80,208.50.

1.4.2 When Neither Point Represents the Initial Value In the previous problems, we noticed that one of the two data points given represented the initial value. Unfortunately, this is not always the case. Suppose you purchase a 2012 Toyota Camry. According to www.toyota.com, it is expected that you will get about 35 miles per gallon (mpg) on the highway. To test this hypothesis, you begin driving the car. When you begin keeping track of your mileage counter, you find that you have used 2 gallons and have traveled a total of 60 miles. After having consumed 5 gallons you have traveled a total of 145 miles. Since you did not reset the counter after the last gas fill-up, you are not sure what your current gas mileage is. Assuming a constant rate of gas use, you model distance traveled (miles) as a function of gallons used. You begin by plotting the two points on a graph:


There is only one problem – we don’t know the initial value. This would be the total distance traveled after having consumed 0 gallons. We cannot assume this value is 0 miles, since the counter was not reset after the last fill-up (per the given information). We can first begin by finding the rate of change:

change in distancechange in gallons =

145 − 605 − 2

migal =

853migal ≈ 28

migal

Perfect! We know the average gas mileage! We can actually use this value quite easily to figure out the counter reading at 0 gallons. To reach the initial value, we geometrically need to move to the left 2 gallons:

Assuming the same rate of change, if we “undo” 2 gallons, then we must “undo” 28 miles twice: once for the first gallon and once for the second. Thus, we are moving downwards:

0

20

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120

140

160

0 1 2 3 4 5 6

Dist

ance

Tra

vele

d (M

iles)

Gallons Used

Distance vs. Gallons


Remember, change in output is the rate of change multiplied by the change in input. Thus, if the change in input is -2 and the rate of change is 28 mpg, then we must change our output -56 mi. From this value of 60 miles, we must decrease it by 56 in order to figure out where on the vertical axis our point will rest:

60 mi − 56 mi = 4 mi Thus, the initial value is 4 mi. We can proceed to write our equation to calculate output values:

𝐷 = initial value + change in output = 4 + 28𝑔

Where 𝑔 is the number of gallons consumed and 𝐷 is the total distance, in miles, traveled. Another Way If you favor solving equations, then we can begin with one fact:

𝐷 = initial value + (rate of change)g This is the basic form of a linear equation. We don’t know the initial value, but we do know the rate of change:

𝐷 = initial value + 28g


To solve for the initial value, we must know values for 𝑔 and 𝐷. Otherwise we have too many unknowns. We know, for instance, that after 3 gallons, 60 miles have been driven. So,

60 = initial value + 28(2) 60 = initial value + 56

Solving for initial value gives:

initial value = 60 − 56 = 4 This gives us:

𝐷 = 4 + 28𝑔 We have thus effectively confirmed our answer Example 3: A researcher is working to model life expectancy in the U.S.

as a function of years since 2000. He knows that 2005 the life expectancy was 77.4 years and 77.8 years in 2008 (SOURCE: U.S. Statistical Abstract, Table 104).

a. Determine a function that gives life expectancy, 𝐿, as a function of years after 2000, 𝑦. b. What was the life expectancy in 2002?

SOLUTION:

a. We are given that in year 5, life expectancy was 77.4 years of age and in year 8, life expectancy was 77.8 years. We do not have the initial value for year 0.

We begin by finding the average rate of change:

change in outputchange in input =

77.8 − 77.48 − 5

yrs of ageyr

=0.43

yrs of ageyr

= 0.13 yrs of age per year Our equation must have the following structure:


𝐿 = inital value + 0.13y

Using the fact that we know in year 5 the life expectancy was 77.4 years, we find the initial value:

77.4 = initial value + 0.13(5) 77.4 = initial value + 0.65

Solving for initial value gives:

77.4 − 0.65 = initial value Before blindly doing this calculation, we think about what it represents:

77.4 years of age − 5 years ∙ 0.13years of age

yr

In order to figure out the life expectancy 5 years ago, we are subtracting away the life expectancy increase 5 times, once for each year that has passed since the initial year. We get:

initial value = 77.4 − 0.65 = 76.75 years of age Our final function is:

𝐷 = 76.75 + 0.13𝑦

b. 2002 is represented by 𝑦 = 2. Finding 𝐿(2) gives

𝐿(2) = 76.75 + 0.13(2) = 77.01 years of age



1. A classmate of yours missed class on the day we discussed this section. Explain to him how to create a linear function provided an initial value and another ordered pair.

2. To purchase a smartphone through Verizon, it would have cost you $69.99 each month for a 900-minute voice plan (texting not included), with a $0.40 per minute overage charge (SOURCE: www.verizon.com on June 3, 2012). (Video Solution)

a. Create a linear function that models monthly cost to a customer as a function of

number of minutes of overage. b. What is the rate of change and what does it mean? c. How much would the monthly bill be for a customer that went over by 100

minutes?

3. A concert hall planner keeps track of cars in the parking lot of a local venue in order to determine if cars are efficiently leaving after a concert. As he begins, he finds that there are 5,353 cars in the parking lot (from ticket receipts). Twenty minutes later, there are 2,154 cars in the parking lot. (Video Solution)

a. Find the rate of change. Then, explain the real-world meaning of this value. b. Create a linear function that models the number of cars in the lot as a function of

time (minutes) after the concert. c. Using your function, estimate the number of cars after 25 minutes.

4. An educational administrator is interested in considering the percentage of the population that has at least graduated high school. He would like to model this as a function of years after 2001. In 2006, 85.5% of people were high school graduates. In 2010 this number jumped to 87.1% (SOURCE: U.S. Statistical Abstract, Table 229). (Video Solution)

a. Find the rate of change. Then, explain the real-world meaning of this value. b. Create a linear function that models the percentage of high school graduates as a

function of years after 2001. c. What percentage of the U.S. public are expected to graduate high school in the

year 2015?

5. An outdoor waterpark is building a new slide. The slide spans 100 feet (horizontally) and is a straight slide that uses water to accelerate the rider to high speeds. When you walk from the bottom of the stairs of the slide to the middle of the slide, its height is 60 feet. When you walk a total of 100 feet, the height of the slide is at ground level (0 feet). (Video Solution)


a. An architect is modeling the height of the slide as a function of its horizontal distance from the stairs. Find and interpret the rate of change in this problem.

b. Create a linear function that models the height of the slide as a function of horizontal distance.

c. How tall is the slide?


1.5 Intercepts and Intersections 1.5.1 Horizontal and Vertical Intercepts

Imagine that you work for a consulting firm and are working on constructing a model that gives a company’s stock share value (an investment you can make to own a part of the company, and from which you benefit only if the company does well) as a function of years since the company opened. You have been recruited for this job due to the fact that the company has been experiencing a bad several years. You find that an appropriate model assumes a roughly constant rate of change. Letting 𝑉 = value of one share of stock and 𝑦 = years since the company started, you find they are related in the following way:

𝑉(𝑡) = 40 − 1.25𝑡 Two questions come to mind: according to this model,

x What was the stock share worth initially? x When (if ever) will the share value tank completely?

When we think about what these two questions are actually asking, we come up with the following mathematical equations, respectively:

x Find 𝑉(0); that is, what is the share worth at time 0? x Solve 𝑉(𝑡) = 0;; that is, when will the stock’s value be 0?


The first one we can answer quickly, and the second one requires our algebra skills:

x 𝑉(0) = 40 − 1.25(0) = $40; the stock was initially worth $40. x We must substitute $0 in for the actual value, giving:

0 = 40 − 1.25𝑡

We must solve for 𝑡:

0 = 40 − 1.25𝑡 −40 = −1.25𝑡 (subtract 40 from both sides) −40−1.25 = 𝑡 (divide both sides by − 1.25)

32 = 𝑡

This means that the stock is estimated to reach $0, 32 years after the company has opened. If we plot these points on a graph, we notice that each falls at an interesting place:

The initial value falls on the vertical axis and the 0-output value falls on the horizontal axis. These two points have special names: Horizontal and Vertical Intercept The point of a function falling on the vertical axis is called a vertical intercept. It is the point whose input value is always 0. That is, it is of the form (0, 𝑦).

0

5

10

15

20

25

30

35

40

45

0 5 10 15 20 25 30 35

Shar

e Va

lue

($)

Years Since Company Opened

Company Share Value


To find the vertical intercept, replace the input variable with 0 and evaluate for the output.

The point(s) of a function falling on the horizontal axis is/are called a horizontal intercept(s). It is the point whose output value is always 0. That is, it is of the form (𝑥, 0). To find the horizontal intercept, replace the output variable with 0 and solve for the input.

Horizontal and vertical intercepts are often important in determining points that help us graph a function. While they do not always provide real-world significance, they are the “anchors” of the graph. If nothing else, they allow us to generate two points through which we can then sketch a graph.

Example 1: The Basic plan for electrical service with Salt River Project (SRP) in the month of November are $.078 (almost 9 cents) per kilowatt-hour. Suppose that SRP also charges a $10 account fee just for keeping your electrical lines open to service. Let 𝑃 be the total monthly price for electrical service and let 𝑘 be the number of kilowatt-hours used.

a. Write an equation to model price as a function of number of kilowatt-hours used. b. Find the vertical intercept. Explain the real-world meaning of this value. c. Find the horizontal intercept. Explain the real-world meaning of this value. d. Graph the equation by plotting both intercepts. Explain what quadrant of the graph is

important.


SOLUTION:

a. The initial value is $10, since this is the beginning price. The rate of change is a constant $.078. So,

𝑃(𝑘) = 10 + .078𝑘

b. The vertical intercept is when the input is 0. So,

𝑃(0) = 10 + .078(0) = 10

This is simply the initial value. That is, when 0 kilowatt-hours are used, the total price in November will be $10.

c. The horizontal intercept occurs when the output is 0. So,

0 = 10 + .078𝑘

−10 = .078𝑘 −10. 078 = 𝑘

−128.805 = 𝑘

This says that the cost is $0 when a negative number of kilowatt-hours are used. This is impossible, so we conclude that the horizontal intercept is not of practical significance.

d. Plotting the points (0, 10) and (-128.8, 0):

0

2

4

6

8

10

12

-140 -120 -100 -80 -60 -40 -20 0

Pric

e ($

)

Kilowatt-hours used

November Electric Costs


It does not really make sense to talk about the second quadrant, since the kilowatt-hours are negative here. Thus, we extend our linear function into quadrant 1:

1.5.2 Evaluating and Solving for Other Values Consider again the problem from the value of the company’s stock as a function of years since the company opened from above:

𝑉(𝑡) = 40 − 1.25𝑡 We have found vertical and horizontal intercepts for this function. In a similar way, we could use it to determine outputs for other values. For instance, we know how to determine 𝑉(2), 𝑉(10), and 𝑉(100); we simply replace 𝑡 with the correct value. This same process holds for solving for times at which the output variable takes on certain values. For example, lets determine when the stock value will be cut in half. That is, solve 𝑉(𝑡) = 20. This statement demands we replace the value with 20 and solve for time:

20 = 40 − 1.25𝑡 −20 = −1.25𝑡 (subtract 40 from both sides) −20−1.25 = 𝑡 (divide both sides by − 1.25)

0

2

4

6

8

10

12

14

16

18

20

-150 -100 -50 0 50 100 150

Pric

e ($

)

Kilowatt-hours used

November Electric Costs


16 = 𝑡 We find that 𝑉(16) = 20. That is, 16 years after the company opened, we expect the share value to drop to half ($20) of what it started at ($40). The process for this remains the same as for finding horizontal intercepts. 1.5.3 Comparing Different Linear Functions

Many times, it is of interest to compare two different functions. For example, the population of Indiana, 𝐼, (in millions of people) as a function of years since 2000, 𝑦, is given by:

𝐼(𝑦) = 6.1 + .04𝑦 While the population of Arizona, 𝐴, (in millions of people) over the same set of inputs is:

𝐴(𝑦) = 5.1 + .17𝑦 We wish to determine, when, if ever, we can expect the populations to be the same size. First off, we see that the starting population of Indiana in 2000 was larger than Arizona’s;; however, we also see that Arizona’s population is growing at a faster rate than Indiana’s (Arizona’s 0.17 million people per year vs. Indiana’s 0.04 million people per year). Under the assumption of constant growth, we would expect Arizona to surpass the population of Indiana. If we started with a table comparing the two populations for various years, we would find:


In 2008, it that somewhere between years 7 and 8, Arizona’s population grew to be larger than Indiana’s. Is there, however, a way to determine the exact time? Yes! We want to know, when does 𝐴 = 𝐼? This brings up an important mathematical idea worth considering. Principle of Equality Suppose that two quantities are equal,

𝑎 = 𝑏 Then, anywhere an 𝑎 is encountered, it can be replaced with 𝑏. Similarly, anywhere a 𝑏 is encountered, it can be replaced with 𝑎. This is a critical idea, since anywhere we see an 𝐴, we can replace it with 5.1 + .17𝑦 and anywhere we see an 𝐼 we can replace it with 6.1 + .04𝑦. So,

𝐴 = 𝐼 5.1 + .17𝑦 = 6.1 + .04𝑦

Now what? Well, now our goal is to solve for the year, since this is the only variable present on each side of the equation. We proceed by:

1. Moving constants to one side 2. Moving variable expressions to the other

y A I0 5.1 6.1

1 5.27 6.14

2 5.44 6.18

3 5.61 6.22

4 5.78 6.26

5 5.95 6.3

6 6.12 6.34

7 6.29 6.38

8 6.46 6.42

9 6.63 6.46

10 6.8 6.5


3. Isolating 𝑦

5.1 + .17𝑦 = 6.1 + .04𝑦 . 17𝑦 = 1 + .04𝑦 (subtract 5.1 from both sides) . 13𝑦 = 1 (subtract .04𝑦 from both sides)

𝑦 =1. 13

(divide both sides by .13) 𝑦 ≈ 7.7

We thus find that the populations are both equal when 𝑦 is about 7.7, meaning that in the later part of 2007, the populations were equal. One question remains: what was the population of each state? We need to evaluate either 𝐴(7.7) or 𝐼(7.7), since the outputs should be equal (though slight rounding error might apply). We will check both:

𝐴(7.7) = 6.4

𝐼(7.7) = 6.4 Thus, the populations were both around 6.4 million people. Graphically,

We see that the Indiana population is slow-growing while Arizona’s sores past it between 2007 and 2008. The point at which these two function cross is called the intersection point.

0

1

2

3

4

5

6

7

0 2 4 6 8 10

Popu

latio

n (in

mil.

)

Years Since 2000

Arizona vs. Indiana Populations

Arizona Population (inmil.)

Indiana Population (inmil.)



1. Suppose you are given the function 𝑓(𝑥) = 1.1𝑥 − 7 (Video Solution)

a. Find the vertical intercept. b. Find the horizontal intercept.

2. The time it takes a particular jogger to run one mile (in minutes), 𝑇, is a function of the

number of weeks she has been training, 𝑤, is given by 𝑇(𝑤) = −0.2𝑤 + 8.1. (Video Solution)

a. Identify the rate of change. What is its real-world meaning? b. Find the vertical intercept. What is its real-world meaning? c. Find the horizontal intercept. What is its real-world meaning?

3. A car insurance company guarantees it will reduce a customer’s monthly insurance rate as

long as the driver is free of accidents. Suppose that after 2 years as a customer, a person is charged a monthly rate of $120. After 5 years as a customer, he is charged a monthly rate of $105. (Video Solution)

a. Assuming a constant rate of change, create a function that gives the customer’s monthly insurance premium as a function of years as a customer.

b. Identify the rate of change. What is its real-world meaning? c. Find the vertical intercept. What is its real-world meaning? d. Find the horizontal intercept. What is its real-world meaning? e. When will the customer being paying half of what he started off paying as a new

customer?

4. The average number of people in a household in the U.S. as a function of the years since 2005 is given by: 𝑁(𝑦) = 0.0029𝑦 + 2.5629 (SOURCE: U.S. Statistical Abstract, Table 59). (Video Solution)

a. At what rate are households growing each year? b. How many people were in a family in the initial year? c. Find the horizontal intercept. What is its real-world meaning? d. In what year is it expected that there will be 4 people per household? e. In what year is it expected that there will be 5 people per household? f. Why is it not okay to continue making projections with the given equation?

5. It is interesting to note how population density, or the number of people living on one square

mile of land, changes for each state. The population density of Hawaii, 𝐻, as a function of years since 2000 is given by 𝐻(𝑦) = 2.3𝑦 + 188.6. For North Carolina, it is 𝑁(𝑦) = 3.1𝑦 +165.6 (SOURCE: U.S. Statistical Abstract, Table 14). (Video Solution)


a. Find the vertical intercept for each state. Explain the real-world meaning of the values you find.

b. Why will it be the case that both states will eventually have the same population densities (look at the equations and compare them)?

c. When will the two states have the same population density? d. What will the population density be?

6. Between 2005-2009, the percentage of people who own a home has changed drastically from

state to state. The percentage of individuals who own a home in Indiana can be modeled by 𝐶(𝑡) = −0.6𝑡 + 76, where 𝑡 is the number of years since 2005. For New Hampshire, the same model is given by 𝑁(𝑡) = 0.5𝑡 + 73 (SOURCE: Modeled from U.S. Statistical Abstract, Table 988). Did or will Indiana and New Hampshire ever have the same homeownership rate? If so, what will it be? (Video Solution)


1.6 Linear Regression

Often times, we struggle selecting just two points in a data set that we will use to model a set of real-world data. For example, consider the graph below that shows national U.S. average interest rates for 30-year home loans (mortgages):

Suppose we decide that the trend in the data appears linear. If we were to choose the first and last points, we would have a rate of change of −0.4% per year. If we were to choose the first and second-to-last point, we would get a rate of change of −.5% per year. We could go on and discover other rates of change. Our question is this:

0

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3

4

5

6

7

0 1 2 3 4 5

%

Years since 2006

30-Year Mortgage Interest Rates


How do we choose the most appropriate pair of points? This is a daunting task. Wouldn’t it be great if there were a method that would take all points into consideration? The good news: there is one! The method that takes all points into consideration when finding an estimate for the constant rate of change is called linear regression. There is some tough mathematics behind this idea, so we will simply provide a mild bit of background. The basic idea is that a mathematical process provides selects a constant rate of change so that it minimizes the amount, that is, the differences between the actual points and the points that are actually in the dataset. After all, you can probably see that it is impossible to create a single straight line that goes through all points in this dataset. So, linear regression is a method that finds the “best-fitting” line that considers all points in the data set. To carry this out in a Texas Instruments graphing calculator. Calculator Clinic: Linear Regression We will use our data from above. The data points displayed in the graph above are shown in the table below:

Our first task is to enter this into our calculator. To do so:

1. Press STAT, then press ENTER to select 1:Edit. This takes us to the table lists where we can input our data. You should see something similar to the window below:

Years after 2006 Rate (%)

0 6.47

1 6.4

2 6.23

3 5.38

4 4.86


2. If you have numbers entered in, as does the window above, scroll up to highlight the L2 at the top of the screen. Press CLEAR and then ENTER. Do not press delete.

Repeat this for L1.

3. Begin entering your input values in L1. Press ENTER after each entry to advance to the next line.

Repeat this for the outputs in L2. Be sure that each input in L1 matches the output next to it in L2.

4.


a. Do this only the first time you perform a regression. You will not need to repeat step 4a) on a regular basis.

Press 2ND and the 0 key. This will take you to the catalog:

Scroll down to the DiagnosticOn. Press ENTER twice.

This doesn’t initially show anything interesting, but it turns on an important feature of our regression diagnostics.

b. Now perform the regression by going to STAT, over to the right to the CALC column and select 4:LinReg(ax+b). You will see the following window:


c. Before pressing ENTER, we will tell our calculator to put the equation it generates into our Y= screen for us. To do this, go to VARS, over to the Y-VARS column, and press ENTER twice. You should see this screen:

This tells our calculator to perform the linear regression (come up with the best-fitting equation) and to then place the equation into our Y= screen into Y .

d. Press enter to see the following screen:

Notice that it tells us our equation is of the form 𝑦 = 𝑎𝑥 + 𝑏. Here, 𝑎 is the constant rate of change and 𝑏 is the initial value. Our equation would thus be:


𝑦 = −.424𝑥 + 6.716

Letting 𝑟 represent the interest rate (output) and 𝑦 represent the number of years after 2006, we would have:

𝑟 = −.424𝑦 + 6.716 The other two values that have been provided, 𝑟 and 𝑟, are diagnostics that tell us how well the data fit. We will mention those in a minute. We now want to see the graph along with the data points. To do this:

5. Press ZOOM and scroll down to select 9:ZoomStat (or simply press the number 9 key). You should now see the data points along with the best-fitting line!

1.6.1 Analysis of the Regression We first off want to consider the graph of the data points and the linear function above. You almost instantly notice that the line doesn’t touch all of the points. In fact, it doesn’t appear that it touches any of the points! This is okay! Mathematically, this minimizes the overall error that we have between the function and the data. What does that mean? First off, notice that for example, an input of 0 actually corresponds with an output of 6.47. But, according to our equation, an input of 0 produces:

𝑟(0) = −.424(0) + 6.716 = 6.716 The error falls into play in the difference between these two values. If our model was perfect, the model value would perfectly match the actual table value. Why, then, would we go through all of this hassle if we introduce error?


The answer is that we want to make predictions. Suppose we want to estimate interest rates for the coming several years. We now have an equation that will allow us to do this. Caution, however, should be in place: Extrapolation vs. Interpolation Any time a model is used with inputs that are not part of the dataset, one is said to be extrapolating. This can be dangerous in that outputs outside of our data range are always uncertain. When using inputs that fall within the dataset to make a prediction, this process is called interpolation. Suppose we wanted to predict the interest rate in 2011. Since this year was not actually part of our data, we have extrapolated. While it would not be desirable, in this situation, to predict for a year between 2006 and 2010 since we already have this information, there are situations where one would like to do this. As long as you are predicting outputs that fall inside of your data input domain, you are interpolating. 𝒓𝟐-value The 𝑟 value given by the calculator when performing regression is an indicator of goodness-of-fit. The closer 𝑟 is to 0, the worse the fit. The closer 𝑟 is to 1, the better the fit. The value of 𝑟 often is used, though it is also important to visually inspect the regression line against the data points. The closer the line is to the data points, overall, the better the fit is. Use these two feature in conjunction to verify that a linear model is appropriate. Notice that the 𝑟 value was . 885 (the calculator gave us this value). You can think about this as a test score. If you get .885, you are getting an 88.5%. This is a fairly good grade! Thus, we can say that the fit of the line to the data points is pretty good. 1.6.2 Now What?


The equation you come up with can be used in the same way that one is used when you have to create it from scratch! Congratulations, you saved yourself the work of having to find the constant rate of change and initial value by hand! Take a break, and prepare to answer some questions.

Example 1: Answer the following questions about the interest rate equation found above (𝑟 = −.424𝑦 + 6.716)

a. Identify and explain the real-world meaning of the rate of change. b. Find the vertical intercept and explain its meaning in real-world terms. c. Find the horizontal intercept and explain its meaning in real-world terms. d. Predict interest rates in 2015 by using the model.

SOLUTION:

a. The constant rate of change is −.424. Its units are %. This means that interest rates

decreased by .424% per year between 2006 and 2010. b. The vertical intercept is the point of the form (0, output). We replace the input

variable with 0 and evaluate:

𝑟(0) = −.424(0) + 6.716 = 6.716%

In 2006, we estimate that the rate of change was 6.716%

c. The horizontal intercept is the point of the form (input, 0). We replace the output variable with 0 and evaluate:

0 = −.424𝑦 + 6.716

−6.716 = −.424𝑦 (subtract 6.716 from both sides) −6.716−.424 = 𝑦 (divide both sides by − .424) 15.8 ≈ 𝑦

15.8 years after 2006, or about the end of 2006 + 15 = 2021, the equation predicts that interest rates will drop to 0%. This is absurd, since rates can only drop so much before they begin rising again. We see that, according to our graph, this is expected to happen:


d. We evaluate 𝑟(9) = −.424(9) + 6.716 = 2.9%. The predicted rate in 2015 is 2.9%.


1. The following data table gives regular gas prices in Houston, TX as a function of years

since 2005 (SOURCE: U.S. Statistical Abstract, Table 731): (Video Solution)

a. Find a linear regression equation for this data. b. Graph the function. Does it appear fit the data points well? c. Compare your answer in b) to the value of 𝑟 . Does it say the line fits well?

Comment. d. Identify the rate of change. What does it mean in real-world terms. e. According to the model, when will gas prices reach $4.00? f. What are gas prices expected to be in 2013?

2. The following data table displays the number of births (in millions) as a function of the

number of the actual year (SOURCE: U.S. Statistical Abstract, Table 78): (Video Solution)

Years since 2005 Price per gallon ($)0 2.17

1 2.47

2 2.61

3 3.08

4 2.17

5 2.59


a. Find a linear regression equation for this data. (Use 0 to represent 2000, 1 to represent 2001, etc.)

b. Graph the function. Does it appear fit the data points well? c. Compare your answer in b) to the value of 𝑟 . Does it say the line fits well?

Comment. d. Identify the rate of change. What does it mean in real-world terms. e. According to the model, in what year can we expect 5 million births? f. Find the horizontal intercept. What is its real-world meaning?

3. Following is a data table that gives the number of divorces as a function of marriages in the United States between 2003 and 2008 (SOURCE: U.S. Statistical Abstract, Table 78): (Video Solution)

a. Find a linear regression equation for this data. b. Graph the function. Does it appear fit the data points well? c. Compare your answer in b) to the value of 𝑟 . Does it say the line fits well?

Comment.

Year Births (in mil.)2000 4.06

2001 4.03

2002 4.02

2003 4.09

2004 4.11

2005 4.14

2006 4.27

2007 4.32

2008 4.25

Marriages (in millions)

Divorces (in millions)

2.245 0.927

2.279 0.879

2.249 0.847

2.193 0.872

2.197 0.856

2.157 0.844


d. Identify the rate of change. What does it mean in real-world terms. e. How many divorces could we expect to see if there were 2.5 million marriages? f. How many marriages could we expect to see if there were 1.5 million divorces?

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