5.1 equation of lines using slope-intercept · 5.1 equation of lines using slope-intercept mr....
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5.1 Equation of Lines Using Slope-Intercept
Mr. Noyes, Akimel A-al Middle School 1 Heath Algebra 1 - An Integrated Approach
Objectives:
• Learn how to use slope intercept form to write an equation of a line
• Learn how to model a real-life situation with a linear equation
Slope Intercept Form: y = mx + b Where m = slope b = y-intercept Write an equation when given the slope and y-intercept:
m = –2 m = 5
b = 4 b = 12−
y = –2x + 4 y = 5x – 12
5.1 Equation of Lines Using Slope-Intercept
Mr. Noyes, Akimel A-al Middle School 2 Heath Algebra 1 - An Integrated Approach
Write an equation from the graph
y = mx + b
y = 32 x +(–3)
y = mx + b
y = –x –1
y = mx + b
y = 3 Real-life: the Phone Company charges a flat fee of $0.75 for the first minute of long
distance plus $0.10 per minute after that. Write an equation to use to figure out each call.
1. Verbal Model:
Total cost = first minute cost + rateminute · number of minutes
2. Labels:
y = total cost
c = cost for first minute ($0.75)
x = # of additional minutes
r = rate per additional minute ($0.10)
5.1 Equation of Lines Using Slope-Intercept
Mr. Noyes, Akimel A-al Middle School 3 Heath Algebra 1 - An Integrated Approach
3. Algebraic Model:
Fill in what you know and write in slope-intercept form (y = mx + b)
y = .75 + .10x 4. Solve/Give Answer:
Find values for extra minutes: 0, 5, 10 and graph
x y
0 .75
5 1.25
10 1.75
1 2 3 4 5 6 7 8 9 10 11 x
1
2
3
y
minutes
Cost ($)
5.2 Equations of Lines Given the Slope and a Point
Mr. Noyes, Akimel A-al Middle School 1 Heath Algebra 1 - An Integrated Approach
Objectives:
• Learn how to use slope and any point to write an equation of the line
• Learn how to model a real-life situation with a linear equation
Slope intercept form: y = mx + b Need to know m and b Given the slope, m = –2 and the point (6, –3), find equation of the line:
y = mx + b
–3 = –2(6) + b substitute what you know into the slope-intercept form of the equation, and then solve for b
–3 = –12 + b +12 +12 9 = b
y = –2x + 9 graph check:
1 2 3 4 5 6 7 8 910–1–2–3–4–5–6–7–8–9–10 x
123456789
10
–1–2–3–4–5–6–7–8–9
–10
y
5.2 Equations of Lines Given the Slope and a Point
Mr. Noyes, Akimel A-al Middle School 2 Heath Algebra 1 - An Integrated Approach
Write the equation for a line that passes through point (6, 7) and has a slope 23
y = mx + b graph:
7 = 6(32 ) + b
7 = 4 + b –4 –4 3 = b
y = 23 x + 3
Real-life: Find an equation for vacation trips y (in millions) in terms of the
year, t. Let t = 0 correspond to 1980. From 1980 – 1990, vacation trips increased by about 15 million per year. In 1985, Americans went on 340 million vacation trips.
Because change is constant, you can model this as a linear equation y = mt + b Constant is 15 trips per year, so you know slope. In 1985, the year would be t = 0 + 5 and the number of vacation trips (y) would be 340 million.
y = 15t + b
340 = 15(5) + b
340 = 75 + b –75 –75 265 = b
1 2 3 4 5 6 7 8 9–1–2–3–4–5–6–7–8–9 x
123456789
–1–2–3–4–5–6–7–8–9
y
5.2 Equations of Lines Given the Slope and a Point
Mr. Noyes, Akimel A-al Middle School 3 Heath Algebra 1 - An Integrated Approach
y = 15t + 265
In 1998, the value of t = 1998 - 1980 = 18
So the number of vacations taken in 1998 would be:
y = 15(18) + 265
y = 270 + 265
y = 535
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 t
100
200
300
400
500
600
y
Year ( 0 ↔ 1980)
5.3 Equations of Lines Given 2 Points
Mr. Noyes, Akimel A-al Middle School 1 Heath Algebra 1 - An Integrated Approach
Objectives:
• Learn how to write an equation of a line given 2 points on the line
• Learn how to model a real-life problem with a linear equation
Equation of a line: y = mx + b Need slope and y-intercept Given 2 points, can you find the slope and y-intercept?
Points (3, 5) and (–7, 2)
1. Find the slope: m = 2 1
2 1
riserun
y yx x−
=−
m = 2 57 3−
− − = 3
10
y = mx + b Substitute the value of m back into the slope-intercept form of the linear equation
y = 310 x + b
2. Substitute the x and y values from one point to find the y-intercept (the value of b):
y = mx + b (3, 5)
5 = 310 (3) + b
5 = 910 + b
– 910 – 9
10
5.3 Equations of Lines Given 2 Points
Mr. Noyes, Akimel A-al Middle School 2 Heath Algebra 1 - An Integrated Approach
4 110 = b
3. Substitute both m and b back into the slope-intercept form of the linear equation and solve:
y = 3
10 x + 4 110 This is the equation of the line passing through points
(3, 5) and (–7, 2)
(3, 5)
(–7, 2)
1 2 3 4 5 6 7 8 910–1–2–3–4–5–6–7–8–9–10 x
123456789
10
–1–2–3–4–5–6–7–8–9
–10
y
Example: Find linear equation with points (–3, 2) and (5,–2), and then graph
Let’s say (x1, y1) = (–3, 2) and (x2, y2) = (5,–2) (Though it really doesn’t matter)
1. Find the slope, m:
m = 2 25 ( 3)− −− −
= 48− = 1
2−
y = mx + b Substitute m into the slope-intercept form of the linear equation
y = 12− x + b
5.3 Equations of Lines Given 2 Points
Mr. Noyes, Akimel A-al Middle School 3 Heath Algebra 1 - An Integrated Approach
2. Substitute the x and y values from one of the two points given in the question to find the y-intercept (the value of b):
y = mx + b (–3, 2)
2 = 12− (–3) + b
2 = 32 + b
– 32 – 3
2
12 = b
3. Substitute both m and b back into the slope-intercept form of the linear equation and solve:
y = 1
2− x + 12
(–3, 2)
(5, –2)
1 2 3 4 5 6 7 8 910–1–2–3–4–5–6–7–8–9–10 x
123456789
10
–1–2–3–4–5–6–7–8–9
–10
y
5.3 Equations of Lines Given 2 Points
Mr. Noyes, Akimel A-al Middle School 4 Heath Algebra 1 - An Integrated Approach
Can we find the equation of a line if we know both the x, and y-intercepts?
Write an equation that has points whose y-intercept is –4 and x-intercept is –6
What are your 2 points? (0, –4) (–6, 0)
m = 2 1
2 1
y yx x−−
= 0 ( 4)6 0− −− −
= 46−
= – 23
y = mx + b 0 = – 2
3 (–6) + b
0 = +4 + b –4 –4
–4 = b
y = mx + b
y = – 23 x – 4
(0, –4)
(–6, 0)
1 2 3 4 5 6 7 8–1–2–3–4–5–6–7–8 x
12345678
–1–2–3–4–5–6–7–8
y
5.3 Equations of Lines Given 2 Points
Mr. Noyes, Akimel A-al Middle School 5 Heath Algebra 1 - An Integrated Approach
Real–Life: (see textbook, page 253, Question #29) The Tunnel (aka, “The Chunnel”) from Calais, France to Dover, England Write a linear equation of the line formed from A to B
Point A: (0, 60) Point B: (15, – 70)
m = 2 1
2 1
y yx x−−
= 70 6015,000 0− −
−
= 130 1315000 1500− = −
y = mx + b
60 = 131500− (0) + b
60 = b
y = mx + b
y = – 131500 x + 60
5.4 Exploring Data: Fitting a Line to Data
Mr. Noyes, Akimel A-al Middle School 1 Heath Algebra 1 - An Integrated Approach
Objectives:
• Learn how to find a linear equation that approximates a set of data points
• Learn how to use scatter plots to determine positive, negative or no correlation
The winning Olympic times for the 100-meter run from 1928 to 1988 are plotted in the graph below. Approximate the best-fitting line for these times. Let y represent the winning time and x the year (x = 0, corresponding to 1928).
20 40 60 x
10
11
12
13
y
Looking at the graph, do you see the trend in the info given?
Can we write a linear equation to match?
Often in life the data collected (no matter how carefully done) to analyze whether a relationship exists between two variables will not appear as a nice and neat straight line. However, while all the data may not fall on one line, they may still exhibit a trend that can best be described as linear. When this happens we draw a single line that best represents (or approximates) the set of data points. This line is called the line of best fit.
5.4 Exploring Data: Fitting a Line to Data
Mr. Noyes, Akimel A-al Middle School 2 Heath Algebra 1 - An Integrated Approach
Draw a line of best fit and pick any 2 points along the line (they don’t have to be actual points plotted on the scatter plot).
Let’s say points (0, 12) and (50, 11) Use the method described in Chapter 5.3
m= 2 1
2 1
riserun
y yx x−
=−
=11 1250 0−−
m = 150−
y = mx + b
11 = 150− (50) + b
11 = –1 + b +1 +1
12 = b
y = 150− x + 12
Correlation: A quantitative assessment of whether a relationship exists between
two variables. Describes the slope of the line of best fit.
x
y
x
y
x
y
Positive correlation Negative correlation No correlation When a line cannot be drawn to represent the set of data points, we say there is no correlation.
5.4 Exploring Data: Fitting a Line to Data
Mr. Noyes, Akimel A-al Middle School 3 Heath Algebra 1 - An Integrated Approach
Real-life: Make a graph, find the equation of the line of best fit (if possible), and make a prediction.
Find approximate wing area of a 400 g bird
Bird Weight (g) Wing Area (cm2)
Sparrow 25 87 Martin 47 186 Blackbird 78 245 Starling 93 190 Dove 143 357 Crow 607 1344 Gull 840 2006 Blue Heron 2090 4436 Let x represent weight and y wing area
200 400 600 800 1000 1200 1400 1600 1800 2000 2200 x
400800
1200160020002400280032003600400044004800
y
Pick any two points: Let’s say (x1, y1) is (140, 360) and (x2, y2) is (1400, 3200)
5.4 Exploring Data: Fitting a Line to Data
Mr. Noyes, Akimel A-al Middle School 4 Heath Algebra 1 - An Integrated Approach
y = mx + b
m= 2 1
2 1
y yx x−−
= 3200 360 28401400 140 1260
− =−
= 14263 ≈ 2.25
y = mx + b
360 = 2.25 (140) + b
360 = 315 + b
45 = b y ≈ 2.25x + 45
5.5 Standard Form of a Linear Equation
Mr. Noyes, Akimel A-al Middle School 1 Heath Algebra 1 - An Integrated Approach
Objectives:
• Learn how to transform a linear equation into standard form
• Learn how to model a real-life situation using the standard form of a linear equation
So far, we have focused on the Slope-Intercept Form of the Linear Equation, though
the Standard Form of the Linear Equation is also commonly used.
Slope-Intercept Form Standard Form y = mx + b Ax + By = C With the Standard Form of the Linear Equation, notice that the two variables are shown
on the left side of the equation, while a constant is found on the right side. One
advantage of the Standard Form is that it can be used for any type of line, including
vertical lines.
Changing/Transforming Slope-Intercept Form to Standard Form:
y = 34 x + 2 1) Multiply by 4 to clear the fraction
4y = 3x + 8 –3x –3x 2) Subtract 3x from both sides
–1 · (–3x + 4y) = 8 · (–1) 3) Standard Form
3x – 4y = –8 4) Multiply by –1 to make x positive
Same Standard Form -- acceptable in either form
5.5 Standard Form of a Linear Equation
Mr. Noyes, Akimel A-al Middle School 2 Heath Algebra 1 - An Integrated Approach
Example:
y = –.23x + 5.2 1) multiply by 100 to clear the decimal
100y = –23x + 520 2) move the term containing x to the left side +23x +23x
23x + 100y = 520 Standard Form - it’s now easy to find both the x- and y-intercepts by setting each to zero.
Real-life: You have $6 to buy bananas and apples. Bananas cost $0.49 per pound,
and apples cost $0.34 per pound. Write a linear equation that represents
the different amounts of fruit you can buy.
Let x represent bananas and y represent apples
.49x + .34y = 6
2 4 6 8 10 12 14 16 x
2468
10121416182022
y
5.6 Point-Slope Form of the Equation of a Line
Mr. Noyes, Akimel A-al Middle School 1 Heath Algebra 1 - An Integrated Approach
Objectives:
• Learn how to use the point-slope form to write an equation of a line
• Learn how to model a real-life situation using point-slope form
Consider a line containing the point (x1, y1) = (2, 5) with a slope of 23 . Let (x, y) be
any other point on the line.
Because you know (2, 5) and (x, y) are 2 points on the line, you can find the slope of
the line m= 5riserun 2
yx−=−
and since we know the slope was given as 23 , we can
write:
m= 52
yx−−
= 23 by cross-multiplying and rewriting we can say
y – 5 = 23 (x – 2) This is called point-slope form
Point-Slope Form: y – y1 = m(x – x1)
When using point-slope form, be sure you see the difference between
(x1, y1) and (x, y).
(x1, y1) = the given point
(x, y) = any point on the line
5.7 Problem Solving Using Linear Models
Mr. Noyes, Akimel A-al Middle School 1 Heath Algebra 1 - An Integrated Approach
Objectives:
• Learn how to create and use Linear Models to solve problems
• Learn how to make linear models that are accurate but simple to use
Linear Models are used to Solve 2 Basic Types of Real-Life Problems: 1. Constant rate of change
2. Two variables such that the sum Ax + By = C is a constant
Rate of Change Example: The cost of renting a lawn mower is $10 for the first hour and $5 for each additional
hour. Write a linear model that gives the cost of renting a lawn mower in terms of the
number of hours rented, then determine how much it would cost to rent a lawn mower
for 3 hours
1. Verbal Model:
Total cost = # of additional hours · (additional hourly cost) + 1st hour cost
2. Label: identify what you know and assign variables for what you don’t know
y = total cost
1st hour cost = $10
Additional hourly cost = $5/hour
x = the total number of hours needed (including the 1st hour), so . . .
x – 1 = number of additional hours needed
3. Algebraic Model:
y = 5(x – 1) + 10
5.7 Problem Solving Using Linear Models
Mr. Noyes, Akimel A-al Middle School 2 Heath Algebra 1 - An Integrated Approach
4. Solve:
y = 5(x – 1) + 10 when x = 3 hours
y = 5(3 – 1) + 10
y = 5(2) + 10
y = 20 It would cost $20 to rent a lawn mower for 3 hours
Constant Addition Example: The total income from the sale of raffle tickets was $202.50. If student tickets sold for
$0.75 and adult tickets sold for $1.25, write a linear equation that describes the
relationship between the number of student and adult tickets and construct a graph to
provide a visual model
Let: s = the number of student tickets sold
a = the number of adult tickets sold
Linear Equation: .75s + 1.25a = 202.50
Graph:
If only adult tickets were sold, s = 0
.75s + 1.25a = 202.50
.75(0) + 1.25a = 202.50
1.25a = 202.50 1.25 1.25
a = 162
If only student tickets were sold, a = 0
.75s + 1.25a = 202.50
.75s + 1.25(0) = 202.50
.75s = 202.50 .75 .75
s = 270
5.7 Problem Solving Using Linear Models
Mr. Noyes, Akimel A-al Middle School 3 Heath Algebra 1 - An Integrated Approach
(270, 0)
(0, 162)
25 50 75 100 125 150 175 200 225 250 275 s
25
50
75
100
125
150
175 a
Student Tickets
Adult Tickets