Chapter 4 · 2016. 10. 10. · Chapter 4 3. Solve a simpler problem. Suppose you drive 250 miles in...

Copyright © Big Ideas Learning, LLC Algebra 1 165All rights reserved. Worked-Out Solutions

Chapter 4

Chapter 4 Maintaining Mathematical Profi ciency (p. 173)

1. (−5, −2) corresponds to point G.

2. (2, 0) corresponds to point D.

3. Point C is located in Quadrant I.

4. Point E is located in Quadrant IV.

5. x − y = 5 x − x − y = 5 − x −y = 5 − x

−y — −1 = 5 − x

— −1

y = −5 + x The rewritten equation is y = −5 + x.

6. 6x + 3y = −1 6x − 6x + 3y = −1 − 6x 3y = −1 − 6x

3y — 3 = −1 − 6x —

3

y = − 1 — 3 − 2x

The rewritten equation is y = − 1 — 3 − 2x.

7. 0 = 2y − 8x + 10 0 − 2y = 2y − 2y − 8x + 10 −2y = −8x + 10

−2y — −2

= −8x + 10 — −2

y = 4x − 5 The rewritten equation is y = 4x − 5.

8. −x + 4y − 28 = 0 −x + 4y − 4y − 28 = 0 − 4y

−x − 28 = −4y

−x −28 — −4

= −4y — −4

1 — 4 x + 7 = y

The rewritten equation is y = 1 — 4 x + 7.

9. 2y + 1 − x = 7x 2y + 1 − x + x = 7x + x 2y + 1 = 8x 2y + 1 − 1 = 8x − 1 2y = 8x − 1

2y — 2 = 8x − 1 —

2

y = 4x − 1 — 2

The rewritten equation is y = 4x − 1 — 2 .

10. y − 4 = 3x + 5y y − 5y − 4 = 3x + 5y − 5y −4y − 4 = 3x −4y − 4 + 4 = 3x + 4 −4y = 3x + 4

−4y — −4 = 3x + 4 —

−4

y = − 3 — 4 x − 1

The rewritten equation is y = − 3 — 4 x − 1.

11. When both coordinates of a point are multiplied by a negative number, points in Quadrant I move to Quadrant III, points in Quadrant II move to Quadrant IV, points in Quadrant III move to Quadrant I, and points in Quadrant IV move to Quadrant II.

Chapter 4 Mathematical Practices (p.174)

1. Solve a simpler problem.

Suppose you work 40 hours and earn $360.

Hourly wage = $360 — 40 h

= $9 — 1 h

In the simpler problem, you earn $9 per hour. Apply the strategy to the original problem.

Hourly wage = $352.50 _______ 37 1 — 2 h

= $9.4 — 1 h

In the original problem, you earn $9.40 per hour.


Suppose you drive 1250 miles and use 50 gallons of gasoline.

Gas mileage = 1250 mi — 50 gal

= 25 mi — 1 gal

In the simpler problem, your car’s gas mileage is 25 miles per gallon.

Apply the strategy to the original problem.

Gas mileage = 1244.5 mi — 47.5 gal

= 26.2 mi — 1 gal

In the original problem, your car’s gas mileage is 26.2 miles per gallon.

166 Algebra 1 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.

Chapter 4


Suppose you drive 250 miles in 5 hours. At the same rate, how long will it take you to drive 450 miles?

Driving speed = 250 mi — 5 h

= 50 mi — 1 h

Driving time = 450 mi ÷ 50 mi — 1 h

= 450 mi ⋅ 1 h — 50 mi = 450 h

— 50

= 9 h

In the simpler problem, it will take 9 hours to drive 450 miles.

Apply the strategy to the original problem.

Driving speed = 236 mi — 4.6 h

≈ 51.3 mi — 1 h

Driving time = 450 mi ÷ 51.3 mi — 1 h

= 450 mi ⋅ 1 h — 51.3 mi = 450 h — 51.3

≈ 8.8 h

In the original problem, it will take about 8.8 hours to drive 450 miles.

4.1 Explorations (p.175)

1. a. m = y2 − y1 — x2 − x1

= 3 − (−1) — 2 − 0

= 3 + 1 — 2 − 0

= 4 — 2 = 2

Because the line crosses the y-axis at (0, −1), the y-intercept is −1.

An equation is y = 2x −1.

b. m = y2 − y1 — x2 − x1

= −2 − 2 — 4 − 0

= −4 — 4 = −1

Because the line crosses the y-axis at (0, 2), the y-intercept is 2.

An equation is y = −x + 2.

c. m = y2 − y1 — x2 − x1

= −1 − 3 — 3 − (−3)

= −1 − 3 — 3 + 3

= −4 — 6 = − 2 —

3

y = mx + b or Because the line crosses the y-axis at (0, 1), the y-intercept is 1. 3 = −

2 — 3 (−3) + b

3 = 2 + b − 2 − 2 1 = b An equation is y = − 2 —

3 x + 1.

d. m = y2 − y1 — x2 − x1

= −1 − 0 — 2 − 4

= −1 — −2

= 1 — 2

y = mx + b or Because the line crosses the y-axis at (0, −2), the y-intercept is −2.

0 = 1 — 2 (4) + b

0 = 2 + b − 2 − 2 −2 = b

An equation is y = 1 — 2 x − 2.

2. a. The y-intercept is the point where the line crosses the y-axis, or the value of y when x = 0. For this situation, the y-intercept is 20. In the context of this problem, the y-intercept is the base cost of the plan, or $20 per month.

b. Sample answer: Pick two points: (0, 20) and (1500, 65)

m = y2 − y1 — x2 − x1

= 65 − 20 — 1500 − 0

= 45 — 1500

= 0.03

So, for this smartphone plan, the charge is $0.03 per megabyte of data used.

c. Sample answer:

y = mx + b y = 0.03x + 20 So, an equation is y = 0.03x + 20.

3. Use the graph to fi nd the slope m of the line by identifying two points on the graph and calculating the change in y over change in x. The y-intercept is the point (0, b) where the line crosses the y-axis. Once you have found the values of m and b, you can write an equation of the line by substituting them into y = mx + b.

4. Sample answer:

x

y

−4

42−2−4

2

4

(2, 0)

(0, −2)

m = y2 − y1 — x2 − x1

= −2 − 0 — 0 − 2

= −2 — −2

= 1

b = −2 because the line passes through −2 on the y-axis. An equation of the line is y = x − 2.


Chapter 4

4.1 Monitoring Progress (pp. 177–178)

1. m = 7; b = 2 y = mx + b y = 7x + 2 An equation is y = 7x + 2.

2. m = 1 — 3 ; b = −1

y = mx + b

y = 1 — 3 x + (−1)

y = 1 — 3 x − 1


3. Let (x1, y1) = (0, 1) and (x2, y2) = (4, 3).

m = y2 − y1 — x2 − x1

= 3 − 1 — 4 − 0

= 2 — 4 , or 1 —

2


So, the equation is y = 1 — 2 x + 1.

4. Let (x1, y1) = (0, −1) and (x2, y2) = (5, −3).

m = y2 − y1 — x2 − x1

= −3 − (−1) — 5 − 0

= −3 + 1 — 5 − 0

= − 2 — 5


So, the equation is y =− 2 — 5 x − 1.

5. m = 10 − (−2) — 4 − 0

= 10 + 2 — 4 − 0

= 12 — 4 , or 3


So, an equation is y = 3x − 2.

6. Write g(0) = 9 as (0, 9) and g(8) = 7 as (8, 7). Find the slope of the line through these points.

m = 7 − 9 — 8 − 0

= −2 — 8 , or − 1 —

4


So, a function is g(x) = − 1 — 4 x + 9.

7. Let (x1, y1) = (0, 248) and (x2, y2) = (5, 277).

m = y2 − y1 — x2 − x1

= 277 − 248 — 5 − 0

= 29 — 5 = 5.8

Mega watt hours (millions) =

Initial value +

Rate of change ⋅

Years since 2007

y = 248 + 5.8 ⋅ xy = 248 + 5.8x

The linear model is y = 5.8x + 248.

4.1 Exercises (pp.179–180)

Vocabulary and Core Concept Check

1. A linear function that models a real-life situation is called a linear model.

2. When you are given the slope m and y-intercept b, you can write an equation of the line by substituting them into the slope-intercept form y = mx + b.

Monitoring Progress and Modeling with Mathematics

3. y = mx + b y = 2x + 9 An equation is y = 2x + 9.

4. y = mx + b y = 0x + 5 y = 5 An equation is y = 5.

5. y = mx + b y = −3x + 0 y = −3x An equation is y = −3x.

6. y = mx + b y = −7x + 1 An equation is y = −7x + 1.

7. y = mx + b

y = 2 — 3 x + (−8)

y = 2 — 3 x − 8


8. y = mx + b

y = − 3 — 4 x + (−6)

y = − 3 — 4 x − 6

An equation is y = − 3 — 4 x − 6.

9. Let (x1, y1) = (0, 2) and (x2, y2) = (3, 3).

m = y2 − y1 — x2 − x1

= 3 − 2 — 3 − 0

= 1 — 3




Chapter 4

10. Let (x1, y1) = (0, 3) and (x2, y2) = (4, 2).

m = y2 − y1 — x2 − x1

= 2 − 3 — 4 − 0

= −1 — 4 = − 1 —

4


So, the equation is y = − 1 — 4 x + 3.

11. Let (x1, y1) = (−3, 4) and (x2, y2) = (0, 0).

m = y2 − y1 — x2 − x1

= 0 − 4 — 0 − (−3)

= 0 − 4 — 0 + 3

= −4 — 3

= − 4 — 3


So, the equation is y = − 4 — 3 x + 0, or y = − 4 —

3 x.

12. Let (x1, y1) = (0, −2) and (x2, y2) = (2, 2).

m = y2 − y1 — x2 − x1

= 2 − (−2) — 2 − 0

= 2 + 2 — 2 − 0

= 4 — 2 , or 2


So, the equation is y = 2x −2.

13. m = 10 − 1 — 0 − 3

= 9 — −3

, or −3


So, the equation is y = −3x + 10.

14. m = −5 − 7 — 0 − 2

= −12 — −2

, or 6


So, the equation is y = 6x − 5.

15. m = −4 − (−4) — 0 − 2

= −4 + 4 — 0 − 2

, = 0 — −2

, or 0


So, the equation is y = 0x + (−4), or y = −4.

16. m = −24 − 0 — 0 − (−6)

= −24 − 0 — 0 + 6

= −24 — 6 , or −4


So, the equation is y = −4x −24.

17. m = 1−5 — −1.5 − 0

= −4 — −1.5

= −4 ⋅ 2 — −1.5 ⋅ 2 =

8 — 3



18. m = 2.5−3 — −5 − 0

= −0.5 — −5

= −0.5 ⋅ 2 — −5 ⋅ 2 =

1 — 10


So, the equation is y = 1 — 10

x + 3.

19. Write f (0) = 2 as (0, 2) and f (2) = 4 as (2, 4). Find the slope of the line through these points.

m = 4 − 2 — 2 − 0

= 2 — 2 , or 1


y = mx + b y = 1x + 2 y = x + 2 A function is f (x) = x + 2.

20. Write f (0) = 7 as (0, 7) and f (3) = 1 as (3, 1). Find the slope of the line through these points.

m = 1 − 7 — 3 − 0

= −6 — 3 , or −2


y = mx + b y = −2x + 7 A function is f (x) = −2x + 7.

21. Write f (4) = −3 as (4, −3) and f (0) = −2 as (0, −2). Find the slope of the line through these points.

m = −2 − (−3) — 0 − 4

= −2 + 3 — 0 − 4

= 1 — −4

= − 1 — 4


y = mx + b

y = − 1 — 4 x − 2

A function is f (x) = − 1 — 4 x − 2.

22. Write f (5) = −1 as (5,−1) and f (0) = −5 as (0, −5). Find the slope of the line through these points.

m = −5 − (−1) — 0 − 5

= −5 + 1 — 0 − 5

= −4 — −5

= 4 — 5


y = mx + b

y = 4 — 5 x − 5

A function is f (x) = 4 — 5 x − 5.


Chapter 4

23. Write f (−2) = 6 as (−2, 6) and f (0) = −4 as (0,−4). Find the slope of the line through these points.

m = −4 − 6 — 0 − (−2)

= −4 − 6 — 0 + 2

= −10 — 2 , or −5

Because the line crosses the y-axis at (0,−4), the y-intercept is −4.

y = mx + b y = −5x − 4 A function is f (x) = −5x − 4.

24. Write f (0) = 3 as (0, 3) and f (−6) = 3 as (−6, 3). Find the slope of the line through these points.

m = 3 − 3 — −6 − 0

= 0 — −6

, or 0


y = mx + b y = 0x + 3 y = 3 A function is f (x) = 3.

25. Let (x1, y1) = (1,−1) and (x2, y2) = (0, 1).

m = y2 − y1 — x2 − x1

= 1 − (−1) — 0 − 1

= 1 + 1 — 0 − 1

= 2 — −1

= −2

Because the point (0, 1) is on the y-axis, the y-intercept is 1.

y = mx + b y = −2x + 1 A function is f (x) = −2x + 1.

26. Let (x1, y1) = (−4,−2) and (x2, y2) = (0, 0).

m = y2 − y1 — x2 − x1

= 0 − (−2) — 0 − (−4)

= 0 + 2 — 0 + 4

= 2 — 4 , or 1 —

2

Because the point (0, 0) is on the y-axis, the y-intercept is 0.

y = mx + b

y = 1 — 2 x + 0

y = 1 — 2 x

A function is f (x) = 1 — 2 x.

27. The slope and y-intercept were substituted incorrectly. The slope is 2. So, m = 2. The y-intercept is 7. So, b = 7.

y = mx + b y = 2x + 7 An equation is y = 2x + 7.

28. The coordinates of the points were substituted incorrectly when calculating the slope.

Let (x1, y1) = (0, 4) and (x2, y2) = (5, 1).

Slope = y2 − y1 — x2 − x1

= 1 − 4 — 5 − 0

= −3 — 5 = − 3 —

5

Because the line crosses the y-axis at (0, 4), the y-intercept is 4, which was correct.

y = mx + b

y = − 3 — 5 x + 4

An equation is y = − 3 — 5 x + 4.

29. a. Let (x1, y1) = (0, 3.91) and (x2, y2) = (20, 3.81).

m = y2 − y1 — x2 − x1

= 3.81 − 3.91 — 20 − 0

= −0.1 — 20

= −0.1 ⋅ 10 — 20 ⋅ 10

= − 1 — 200

, or −0.005

World record (in minutes) =

Initial value +

Rate of change ⋅

Years since 1960

y = 3.91 + −0.005 ⋅ xy = 3.91 − 0.005x

A linear model is y = 3.91 − 0.005x. b. y = 3.91 − 0.005x y = 3.91 − 0.005x y = 3.91 − 0.005 (40) y = 3.91 − 0.005 (60) y = 3.91 − 0.2 y = 3.91 − 0.3 y = 3.71 y = 3.61 The model estimates the record time for the men’s mile in

2000 was 3.71 minutes, and the model predicts the record time in 2020 will be 3.61 minutes.

30. a. Words: Total cost

=

Initial fee

+

Rate of Change (cost

per hour)

⋅ Studio

time (in hours)

Variables: Let y be the total cost and x be the time (in hours) spent in the studio.

Linear model: y = 50 + 75 ⋅ x y = 50 + 75x A linear model is y = 50 + 75x. b. y = 50 + 75x y = 50 + 75 (12) y = 50 + 900 y = 950 According to the model, 12 hours of recording time will

cost $950. So, it would be less expensive to purchase a $750 music software program.


Chapter 4

31. Let (x1, y1) = (0,−2) and (x2, y2) = (0, 5).

m = y2 − y1 — x2 − x1

= 5 − (−2) — 0 − 0

= 5 + 2 — 0 − 0

= 7 — 0 ✗

no; Because you cannot divide by 0, the slope is undefi ned. So, you cannot write an equation of the line in slope-intercept form.

32. Sample answer: You have $20 set aside for your savings, and each week you plan to save another $15 out of your babysitting earnings. Let y be how much you have saved, and x be how long (in weeks) you have been saving.

33. Ax + By = C −Ax + Ax + By = C − Ax By = C − Ax

y = C − Ax — B

y = − A — B

x + C — B

In the equation −6x + 5y = 9, A = −6, B = 5, and C = 9.

So, the rewritten equation is y = 6 — 5 x + 9 —

5 .

So, the slope is m = 6 — 5 , and the y-intercept is b = 9 —

5 .

34. Your friend is correct. If f is a function, then the line is not vertical.

35. Refl ect (0, 1) and (3, −4) in the x-axis. The refl ected points are (x1, y1) = (0, −1) and (x2, y2) = (3, 4).

Slope of line k = y2 − y1 — x2 − x1

= 4 − (−1) — 3 − 0

= 4 + 1 — 3 − 0

= 5 — 3

Because line k passes through the point (0, −1), the y-intercept of line k is −1.

So, an equation that represents line k is y = 5 — 3 x − 1.

36. a. Sample answer: Let (x1, y1) = (0, 8) and (x2, y2) = (9, 10).

m = y2 − y1 — x2 − x1

= 10 − 8 — 9 − 0

= 2 — 9

So, the slope is 2 — 9 , or about 0.22.

Because the line appears to pass through the point (0, 8), the y-intercept is 8.

b. Sample answer: The U.S. box offi ce revenue was about $8 billion in 2000, and it has increased at a rate of about $0.22 billion each year from 2000 to 2012.

c. You can use the slope m = 2 — 9 and y-intercept b = 8 from part (a) to write a linear model that represents this situation. Let y be the approximate U.S. box offi ce revenues (in billions of dollars) and x be the number of years that have passed since the year 2000.

y = mx + b

y = 2 — 9 x + 8

Then, in order to predict the U.S. box offi ce revenue in 2018, substitute 18 for x and solve for y.

y = 2 — 9 x + 8

y = 2 — 9 (18) + 8

y = 4 + 8 y = 12 So, based on this model, the predicted U.S. box offi ce

revenue in 2018 is about $12 billion.

37. Let (x1, y1) = (0, b) and (x2, y2) = (1, b + m).

m = y2 − y1 — x2 − x1

= b + m − b — 1 − 0

= m — 1 = m

Because the line passes through the point (0, b), the y-intercept is b. So, the equation is y = mx + b.

To be sure that the point (−1, b − m) lies on the line, substitute −1 for x and b − m for y.

y = mx + b b − m =

? m(−1) + b

b − m =?

−m + b b − m = b − m ✓ The equation b − m = b − m is always true. So, the point

(−1, b − m) is a solution of the equation y = mx + b and lies on the line.

Maintaining Mathematical Profi ciency

38. 3(x − 15) = x + 11 3(x) − 3(15) = x + 11 3x − 45 = x + 11 − x − x 2x − 45 = 11 + 45 + 45 2x = 56

2x — 2 = 56 —

2

x = 28 The solution is x = 28.

Chapter 4


Chapter 4

39. −4y − 10 = 4(y − 3) −4y − 10 = 4(y) − 4(3) −4y − 10 = 4y − 12 + 4y + 4y − 10 = 8y − 12 + 12 + 12 2 = 8y

2 — 8 = 8y —

8

1 — 4 = y

The solution is y = 1 — 4 .

40. 2(3d + 3) = 7 + 6d 2(3d) + 2(3) = 7 + 6d 6d + 6 = 7 + 6d − 6d − 6d 6 = 7 ✗ The statement 6 = 7 is false. So, there is no solution.

41. −5(4 − 3n) = 10(n − 2) −5(4) − 5(−3n) = 10(n) − 10(2) −20 + 15n = 10n − 20 − 10n − 10n −20 + 5n = −20 + 20 + 20 5n = 0

5n — 5 = 0 —

5

n = 0 The solution is n = 0.

42. −4x + 2y = 16 Let x = 0. Let y = 0. −4x + 2y = 16 −4x + 2y = 16 −4(0) + 2y = 16 −4x + 2(0) = 16 2y = 16 −4x = 16

2y — 2 = 16 —

2 −4x —

−4 = 16 —

−4

y = 8. x = −4 The y-intercept is 8. The x-intercept is −4.

x

y

10

12

4

6

2

−4

21−2−1−3−5−6

(0, 8)

(−4, 0)

43. 3x + 5y = −15 Let x = 0. Let y = 0. 3x + 5y = −15 3x + 5y = −15 3(0) + 5y = −15 3x + 5(0) = −15 5y = −15 3x = −15

5y — 5 = −15 —

5 3x —

3 = −15 —

3

y = −3 x = −5 The y-intercept is −3. The x-intercept is −5.

x

y

2

3

4

1

−2

−4

21−2−1−3−5−6

(0, −3)

(−5, 0)

44. x − 6y = 24 Let x = 0. Let y = 0. x − 6y = 24 x − 6y = 24 0 − 6y = 24 x − 6(0) = 24 −6y = 24 x = 24

−6y — −6

= 24 — −6

The x-intercept is 24.

y = −4 The y-intercept is −4.

x

y2

1

−2−3

−5−6

8 16 24 32

(0, −4)

(24, 0)


Chapter 4

45. −7x − 2y = −21 Let x = 0. Let y = 0. −7x − 2y = −21 −7x − 2y = −21 −7(0) − 2y = −21 −7x − 2(0) = −21 −2y = −21 −7x = −21

−2y — −2

= −21 — −2

−7x — −7

= −21 — −7

y = 21 — 2 x = 3

The y-intercept is 21 — 2 . The x-intercept is 3.

x

y

4

6

8

10

2

−4

2 41 3−2−1−3−4

(0, )

(3, 0)

212


1. a.

x

y

4

5

6

2

3

1

−3−4−5−6

−2

4 5 6 7321−1−3−4−5

Use m = 1 — 2 and the point (3, 2).

y = mx + b

2 = 1 — 2 (3) + b

2 = 3 — 2 + b

− 3 — 2 − 3 —

2

1 — 2 = b ( 2 − 3 — 2 = 4 — 2 − 3 — 2 = 1 — 2 )

The y-intercept is 1 — 2 .

An equation of the line is y = 1 — 2 x + 1 —

2 .

b.

x

y

4

5

6

7

8

2

3

1

−3−4

−2

4 5 6321−1−3−2−4−5−6

Use m = −2 and the point (−4, 6). y = mx + b 6 = −2(−4) + b 6 = 8 + b − 8 − 8 −2 = b The y-intercept is −2. An equation of the line is y = −2x − 2.

2. m = y − y1 — x − x1

m(x − x1) = (y − y1) — (x − x1)

⋅ (x − x1) m(x − x1) = y − y1, or y − y1 = m(x − x1) The point-slope form of a linear equation

is y − y1 = m(x − x1).

3. a. Let (x1, y1) = (4, 175) and m = 25. y − y1 = m(x − x1) A − 175 = 25(t − 4) A = 25t − 100 + 175 A = 25t + 75 The equation is A = 25t + 75. b. Graph y = 25x + 75 using a graphing calculator.

80

0

300

4. When you are given the slope and a point on a line, you can write an equation of the line by substituting the slope for m and the coordinates of the point for x1 and y1 in the point-slope form of an equation y − y1 = m(x − x1).


Chapter 4

5. Sample answer: To write an equation of a line that passes through the point (−1, 2) and has a slope of 3, use the point-slope form of a linear equation, and substitute −1 for x1, 2 for y1, and 3 for m.

y − y1 = m(x − x1) y − 2 = 3[x − (−1)] y − 2 = 3(x + 1) An equation is y − 2 = 3(x + 1).


1. y − y1 = m(x − x1) y − (−1) = −2(x − 3) y + 1 = −2(x − 3) The equation is y + 1 = −2(x − 3).

2. y − y1 = m(x − x1)

y − 0 = − 2 — 3 (x − 4)

y = − 2 — 3 (x − 4)

The equation is y = − 2 — 3 (x − 4).

3. m = 10 − 4 — 3 − 1

= 6 — 2 , or 3

y − y1 = m(x − x1) y − 4 = 3(x − 1) y − 4 = 3(x) − 3(1) y − 4 = 3x − 3 + 4 + 4 y = 3x + 1 The equation is y = 3x + 1.

4. m = −4 − (−1) — 8 − (−4)

= −4 + 1 — 8 + 4

= −3 — 12

, or − 1 — 4

y − y1 = m ( x − x1 )

y − (−4) = − 1 — 4 (x − 8)

y + 4 = − 1 — 4 (x) − 1 —

4 (−8)

y + 4 = − 1 — 4 x + 2

− 4 − 4

y = − 1 — 4 x − 2

The equation is y = − 1 — 4 x − 2.

5. Rewrite g(2) = 3 as (2, 3) and g(6) = 5 as (6, 5).

m = 5 − 3 — 6 − 2

= 2 — 4 , or 1 —

2

y − y1 = m(x − x1)

y − 3 = 1 — 2 (x − 2)

y − 3 = 1 — 2 (x) − 1 —

2 (2)

y − 3 = 1 — 2 x − 1

+ 3 + 3

y = 1 — 2 x + 2

A function is g(x) = 1 — 2 x + 2.

6. 302 − 176 — 6 − 3

= 126 — 3 = 42, 428 − 302 —

9 − 6 = 126 —

3 = 42,

554 − 428 — 12 − 9

= 126 — 3 = 42

C − C1 = m(n − n1) C − 176 = 42(n − 3) C − 176 = 42(n) − 42(3) C − 176 = 42n − 126 + 176 + 176 C = 42n + 50 Because the cost increases at a constant rate, the situation

can be modeled by a linear equation. A linear model is C = 42n + 50.

4.2 Exercises (pp. 185–186)


1. y − 5 = −2(x + 5) y − y1 = m(x − x1) The slope of the line is m = −2, and the line passes through

the point (x1, y1) = (−5, 5).

2. Let (x, y) be another point on the line, where x ≠ 3. You can write an equation relating x and y using the slope formula with (x1, y1) = (3, −2), (x2, y2) = (x, y), and m = 4.

m = y2 − y1 — x2 − x1

4 = y − (−2) — x − 3

4 = y + 2 — x − 3

4(x − 3) = (y + 2) — (x − 3)

⋅ (x − 3) 4(x − 3) = y + 2 The equation in point-slope form is y + 2 = 4(x − 3).


Chapter 4


3. y − y1 = m(x − x1) y − 1 = 2(x − 2) The equation is y − 1 = 2(x − 2).

4. y − y1 = m(x − x1) y − 5 = −1(x − 3) y − 5 = −(x − 3) The equation is y − 5 = −(x − 3).

5. y − y1 = m(x − x1) y − (−4) = −6(x − 7) y + 4 = −6(x − 7) The equation is y + 4 = −6(x − 7).

6. y − y1 = m(x − x1) y − (−2) = 5[x − (−8)] y + 2 = 5(x + 8) The equation is y + 2 = 5(x + 8).

7. y − y1 = m(x − x1) y − 0 = −3(x − 9) y = −3(x − 9) The equation is y = −3(x − 9).

8. y − y1 = m(x − x1) y − 2 = 4(x − 0) y − 2 = 4x The equation is y − 2 = 4x.

9. y − y1 = m(x − x1)

y − 6 = 3 — 2 [x − (−6)]

y − 6 = 3 — 2 (x + 6)

The equation is y − 6 = 3 — 2 (x + 6).

10. y − y1 = m(x − x1)

y − (−12) = − 2 — 5 (x − 5)

y + 12 = − 2 — 5 (x − 5)

The equation is y + 12 = − 2 — 5 (x − 5).

11. m = 1 − (−3) — 3 − 1

= 1 + 3 — 3 − 1

= 4 — 2 , or 2

y − y1 = m(x − x1) y − 1 = 2(x − 3) y − 1 = 2(x) − 2(3) y − 1 = x − 6 + 1 + 1 y = 2x − 5 The equation is y = 2x − 5.

12. m = −5 − 0 — 1 − (−4)

= −5 − 0 — 1 + 4

= −5 — 5 , or −1

y − y1 = m(x − x1) y − 0 = −1[x − (−4)] y = −1(x + 4) y = −x − 4 The equation is y = −x − 4.

13. m = 4 − 2 — −6 − (−2)

= 4 − 2 — −6 + 2

= 2 — −4

= − 1 — 2

y − y1 = m(x − x1)

y − 2 = − 1 — 2 [x − (−2)]

y − 2 = − 1 — 2 (x + 2)

y − 2 = − 1 — 2 (x) − 1 —

2 (2)

y − 2 = − 1 — 2 x − 1

+ 2 + 2

y = − 1 — 2 x + 1

The equation is y = − 1 — 2 x + 1.

14. m = 4 − 1 — 8 − 4

= 3 — 4

y − y1 = m(x − x1)

y − 1 = 3 — 4 (x − 4)

y − 1 = 3 — 4 (x) − 3 —

4 (4)

y − 1 = 3 — 4 x − 3

+ 1 + 1

y = 3 — 4 x − 2

The equation is y = 3 — 4 x − 2.


Chapter 4

15. m = 12 − 2 — 2 − 7

= 10 — − 5

, or −2

y − y1 = m(x − x1) y − 2 = −2(x − 7) y − 2 = −2(x) − 2(−7) y − 2 = −2x + 14 + 2 + 2 y = −2x + 16 The equation is y = −2x + 16.

16. m = 1 − (−2) — 12 − 6

= 1 + 2 — 12 − 6

= 3 — 6 = 1 —

2

y − y1 = m(x − x1)

y − (−2) = 1 — 2 (x − 6)

y + 2 = 1 — 2 (x) − 1 —

2 (6)

y + 2 = 1 — 2 x − 3

− 2 − 2

y = 1 — 2 x − 5

The equation is y = 1 — 2 x − 5.

17. m = −7 − (−1) — 3 − 6

= −7 + 1 — 3 − 6

= −6 — −3

, or 2

y − y1 = m(x − x1) y − (−1) = 2(x − 6) y + 1 = 2(x − 6) y + 1 = 2(x) − 2(6) y + 1 = 2x − 12 − 1 − 1 y = 2x − 13 The equation is y = 2x − 13.

18. m = −5 − 5 — −4 − (−2)

= −5 − 5 — −4 + 2

= −10 — −2

, or 5

y − y1 = m(x − x1) y − 5 = 5[x − (−2)] y − 5 = 5(x + 2) y − 5 = 5(x) + 5(2) y − 5 = 5x + 10 + 5 + 5 y = 5x + 15 The equation is y = 5x + 15.

19. m = −9 − (−9) — −3 − 1

= −9 + 9 — −3 − 1

= 0 — −4

, or 0

y − y1 = m(x − x1) y − (−9) = 0(x − 1) y + 9 = 0 − 9 − 9 y = −9 The equation is y = −9.

20. m = 13 − 19 — 5 − (−5)

= 13 − 19 — 5 + 5

= −6 — 10

, or − 3 — 5

y − y1 = m(x − x1)

y − 13 = − 3 — 5 (x − 5)

y − 13 = − 3 — 5 (x) − 3 —

5 (−5)

y − 13 = − 3 — 5 x + 3

+ 13 + 13

y = − 3 — 5 x + 16

The equation is y = − 3 — 5 x + 16.

21. Rewrite f (2) = −2 as (2, −2) and f (1) = 1 as (1, 1).

m = 1 − (−2) — 1 − 2

= 1 + 2 — 1 − 2

= 3 — −1

, or −3

y − y1 = m(x − x1) y − 1 = −3(x − 1) y − 1 = −3(x) − 3(−1) y − 1 = −3x + 3 + 1 + 1 y = −3x + 4 A function is f (x) = −3x + 4.

22. Rewrite f (5) = 7 as (5, 7) and f (−2) = 0 as (−2, 0).

m = 0 − 7 — −2 − 5

= −7 — −7

, or 1

y − y1 = m(x − x1) y − 7 = 1(x − 5) y − 7 = 1(x) − 1(5) y − 7 = x − 5 + 7 + 7 y = x + 2 A function is f (x) = x + 2.


Chapter 4

23. Rewrite f (−4) = 2 as (−4, 2) and f (6) = −3 as (6, −3).

m = −3 − 2 — 6 − (−4)

= −3 − 2 — 6 + 4

= −5 — 10

, or − 1 — 2

y − y1 = m(x − x1)

y − (−3) = − 1 — 2 (x − 6)

y + 3 = − 1 — 2 (x) − 1 —

2 (−6)

y + 3 = − 1 — 2 x + 3

− 3 − 3

y = − 1 — 2 x

A function is f (x) = − 1 — 2 x.

24. Rewrite f (−10) = 4 as (−10, 4) and f (−2) = 4 as (−2, 4).

m = 4 − 4 —— −2 − (−10)

= 4 − 4 — −2 + 10

= 0 — 8 , or 0

y − y1 = m(x − x1) y − 4 = 0[x − (−2)] y − 4 = 0 + 4 + 4 y = 4 A function is f (x) = 4.

25. Rewrite f (−3) = 1 as (−3, 1) and f (13) = 5 as (13, 5).

m = 5 − 1 — 13 − (−3)

= 5 − 1 — 13 + 3

= 4 — 16

, or 1 — 4

y − y1 = m(x − x1)

y − 5 = 1 — 4 (x − 13)

y − 5 = 1 — 4 (x) − 1 —

4 (13)

y − 5 = 1 — 4 x − 13 —

4

+ 5 + 5

y = 1 — 4 x + 7 —

4 ( − 13 — 4 + 5 = − 13 — 4 + 20 — 4 = 7 — 4 )

A function is f (x) = 1 — 4 x + 7 —

4 .

26. Rewrite f (−9) = 10 as (−9, 10) and f (−1) = −2 as (−1, −2).

m = −2 − 10 — −1 − (−9)

= −2 − 10 — −1 + 9

= −12 — 8 , or − 3 —

2

y − y1 = m(x − x1)

y − (−2) = − 3 — 2 [x − (−1)]

y + 2 = − 3 — 2 (x + 1)

y + 2 = − 3 — 2 (x) − 3 —

2 (1)

y + 2 = − 3 — 2 (x) − 3 —

2

− 2 − 2

y = − 3 — 2 x − 7 —

2 ( − 3 — 2 − 2 = − 3 — 2 − 4 — 2 = − 7 — 2 )

A function is f (x) = − 3 — 2 x − 7 —

2 .

27. 5 − (−1) — 4 − 2

= 5 + 1 — 4 − 2

= 6 — 2 = 3, 15 − 5 —

6 − 4 = 10 —

2 = 5,

29 − 15 — 8 − 6

= 14 — 2 = 7

47 − 29 — 10 − 8

= 18 — 2 = 9

Because the y-values are not changing at a constant rate, the data cannot be modeled by a linear equation.

28. 10 − 16 — −1 − (−3)

= 10 − 16 — −1 + 3

= −6 — 2 = −3,

4 − 10 — 1 − (−1)

= 4 − 10 — 1 + 1

= −6 — 2 = −3,

−2 − 4 — 3 − 1

= −6 — 2 = −3, −8 − (−2) —

5 − 3 = −8 + 2 —

5 − 3 = −6 —

2 = −3

y − y1 = m(x − x1) y − 4 = −3(x − 1) y − 4 = −3(x) − 3(−1) y − 4 = −3x + 3 + 4 + 4 y = −3x + 7 Because the y-values decrease at a constant rate, the data can

be modeled by a linear equation. A linear model is y = −3x + 7.


Chapter 4

29. 1.4 − 1.2 — 1 − 0

= 0.2 — 1 = 0.2, 1.6 − 1.4 —

2 − 1 = 0.2 —

1 = 0.2,

2 − 1.6 — 4 − 2

= 0.4 — 2 = 0.2

y − y1 = m(x − x1) y − 2 = 0.2(x − 4) y − 2 = 0.2(x) − 0.2(4) y − 2 = 0.2x − 0.8 + 2 + 2 y = 0.2x + 1.2 Because the y-values increase at a constant rate, the data

can be modeled by a linear equation. A linear model is y = 0.2x + 1.2.

30. 15 − 18 — 2 − 1

= −3 — 1 = −3, 12 − 15 —

4 − 2 = 3 —

2 , 9 − 12 —

8 − 4 = 3 —

4

The y-values are decreasing at a constant rate, but the x-values are not increasing at a constant rate. The rate of change is not constant. So, the data cannot be modeled by a linear equation.

31. In point-slope form, the slope is multiplied by the quantity x − x1. So, the equation is y − y1 = m(x − x1).

m = 10 − 4 — 3 − 5

= 6 — −2

= −3

y − y1 = m(x − x1) y − 4 = −3(x − 5) y − 4 = −3(x) − 3(−5) y − 4 = −3x + 15 + 4 + 4 y = −3x + 19 A linear function is f (x) = −3x + 19.

32. The values that are substituted for x1 and y1 should be from the same point.

m = 3 − 2 — 4 − 1

= 1 — 3

Let (x1, y1) = (1, 2).

y − y1 = m(x − x1)

y − 2 = 1 — 3 (x − 1)

An equation is y − 2 = 1 — 3 (x − 1).

33. a.

Words: Total cost (in dollars) =

Initial cost

(for fi rst 1000)

+

Change per

additional 1000

⋅Number of additional stickers ordered

Variables: Let y be the total cost (in dollars) and x be the number (in thousands) of stickers ordered.

Equation: y = 225 + 80 ⋅ (x − 1) y = 225 + 80x − 80 y = 80x + 145 A linear equation that models this situation is

y = 80x + 145. b. y = 80x + 145 = 80(9) + 145 = 720 + 145 = 865 The total cost of 9000 stickers is $865.

34. a. 450 − 246 — 4 − 2

= 204 — 2 = 102, 654 − 450 —

6 − 4 = 204 —

2 = 102,

858 − 654 — 8 − 6

= 204 — 2 = 102

Because the rate of change is constant, the situation can be modeled by a linear equation.

b. Let C be the total cost of renting the beach house and d be the number of days of the rental.

C − C1 = m(d − d1) C − 246 = 102(d − 2) C − 246 = 102(d) − 102(2) C − 246 = 102d − 204

+ 246 + 246 C = 102d + 42 Because the situation can be modeled by the linear

equation C = 102d + 42, you know that the daily fee is the rate of change, or m = $102, and the processing fee is the initial charge, or b = $42.

c. 102d + 42 ≤ 1200 − 42 − 42 102d ≤ 1158

102d — 102

≤ 1158 — 102

d ≤ about 11.4 Because you cannot rent the beach house for part of a day,

the maximum number of days you can rent the beach house is 11 days.


Chapter 4

35. Sample answer: Because the equation is in point-slope form y − y1 = m (x − x1), we can see immediately that the line has a slope of m = 3 — 2 and passes through the point (x1, y1) = (4, 1). So, one way to graph the equation is to plot the point (4, 1) and then use the slope to plot a second point. Draw a line through these two points.

A second way to graph this equation is to fi rst rewrite the equation in slope-intercept form.

y − 1 = 3 — 2 (x − 4)

y − 1 = 3 — 2 (x) − 3 —

2 (4)

y − 1 = 3 — 2 x − 6

+ 1 + 1

y = 3 — 2 x − 5

The y-intercept is −5, so plot the point (0, −5). Then use the slope to plot a second point. Draw a line through these two points.

36. Sample answer:

Let m = 2 — 5 and (x1, y1) = (12, −5).

y − y1 = m (x − x1)

y − (−5) = 2 — 5 (x − 12)

y + 5 = 2 — 5 (x) − 2 —

5 (12)

y + 5 = 2 — 5 x − 24 —

5

− 5 − 5

y = 2 — 5 x − 49 —

5 ( − 24 — 5 − 5 = − 24 — 5 − 25 — 5 = − 49 — 5 )

Two ways to represent this function are y + 5 = 2 —

5 (x − 12) and f (x) = 2 —

5 x − 49 —

5 .

37. Sample answer: In order to write an equation of a the line that passes through two points that are not on the y-axis, use the point-slope form to write the equation because you can quickly use the coordinates of the two points to calculate the slope and use one of the points to write the equation in point-slope form. In order to use slope-intercept form, you would have to calculate both the slope and the y-intercept.

38. a. Sample answer: The y-intercept of the graph of the linear function appears to be negative, because if you connect the points with a straightedge, the line crosses the y-axis below the origin.

b. Sample answer: The coordinates of the two points appear to be about ( 4, 5 — 3 ) and (8, 4).

You can use these coordinates to calculate the slope of the line. Then, you can either use slope-intercept form to calculate the value of b, or you can write an equation of the line in point-slope form and rewrite the equation in slope-intercept form to identify the value of b.

m = 4 − 5 __

3 —

8 − 4 =

7 — 3 —

4 = 7 —

12

y = mx + b or y − y1 = m(x − x1)

4 = 7 — 12

(8) + b y − 4 = 7 — 12

(x − 8)

4 = 14 — 3 + b y − 4 = 7 —

12 (x) − 7 —

12 (8)

− 14 ___ 3 _____ − 2 __ 3

= − 14 ___ 3 _____ b y − 4 =

7 — 12

x − 14 — 3

( 4 − 14 — 3 = 12 — 3 − 14 — 3 = − 2 — 3 ) + 4 + 4 y = 7 —

12 x − 2 —

3

So, b = − 2 — 3 , which confi rms that the y-intercept of the

graph is negative.

39. Sample answer: The graph of y − 1 = 2(x + 3) is the graph of y = 2x translated 1 unit up and 3 units left. So, the graph of y − k = m(x − h) is a translation h units to the right andk units upward of the graph of y = mx.


Chapter 4

40. a. Because the graph of Sibling A’s spending crosses the y-axis at (0, 8), Sibling A’s initial value is $80.

If you insert a row at the beginning of Sibling B’s table by subtracting 1 in the x column and adding $25 in the y column, you get (0, 125). So, Sibling B’s initial value is $125.

Because Sibling C’s equation, y = −22.5x + 90, is in slope-intercept form y = mx + b, you know that b = 90, or Sibling C’s initial value is $90.

So, Sibling B had the most money initially, and therefore received the most money for the holiday. Sibling A had the least amount of money initially and therefore received the least.

b. Calculate the slope of the line that represents Sibling A’s spending.

m = 20 − 50 — 4 − 2 =

−30 —

2 , or −15

The slope is m = −15. So, Sibling A spends $15 per week. Calculate the rate of change in the table that represents

Sibling B’s spending.

m = 75 − 100 — 2 − 1 =

−25 —

1 , or −25

The rate of change is m = −25. So, Sibling B spends $25 per week.

Because Sibling C’s equation, y = −22.5x + 90, is in slope-intercept form y = mx + b, you know that m = −22.5. So, Sibling C spends $22.50 per week.

So, Sibling B spends money at the fastest rate, and Sibling A spends money at the slowest rate.

c. Sibling A: Because the line crosses the x-axis between 5 and 6, Sibling A runs out of money after more than 5 weeks.

Sibling B: If you add one more row at the end of the table by adding 1 in the x column and subtracting $25 in the y column, you get (5, 0). So, Sibling B runs out of money after exactly 5 weeks.

Sibling C: Let y = 0. y = −22.5x + 90 0 = −22.5x + 90 − 90 − 90 −90 = −22.5x

−90 — −22.5

= −22.5x — −22.5

4 = x So, Sibling C runs out of money after exactly 4 weeks. So, Sibling C runs out of money fi rst, and Sibling A runs

out of money last.


41. The reciprocal of 5 is 1 — 5 .

42. The reciprocal of −8 is − 1 — 8 .

43. The reciprocal of − 2 — 7 is − 7 —

2 .

44. The reciprocal of 3 — 2 is 2 —

3 .


1. a. 3x + 4y = 6 3x − 3x + 4y = 6 − 3x 4y = 6 − 3x

4y — 4 = 6 − 3x —

4

y = 3 — 2 − 3 —

4 x, or y = − 3 —

4 x + 3 —

2

3x + 4y = 12 3x − 3x + 4y = 12 − 3x 4y = 12 − 3x

4y — 4 = 12 − 3x —

4

y = 3 − 3 — 4 x, or y = − 3 —

4 x + 3

4x + 3y = 12 4x − 4x + 3y = 12 − 4x 3y = 12 − 4x

3y — 3 = 12 − 4x —

3

y = 4 − 4 — 3 x, or y = − 4 —

3 x + 4

9

−6

6

−9

y = −43x + 4

y = −34x + 3y = −34x + 32

The fi rst two lines, y = − 3 — 4 x + 3 — 2 and y = −

3 — 4 x + 3, are

parallel because they are always the same distance apart. Note that they have the same slope.


Chapter 4

b. 5x + 2y = 6 5x − 5x + 2y = 6 − 5x 2y = 6 − 5x

2y — 2 = 6 − 5x —

2

y = 3 − 5 — 2 x, or y = − 5 —

2 x + 3

2x + y = 3 2x − 2x + y = 3 − 2x y = 3 − 2x, or y = −2x + 3 2.5x + y = 5 2.5x − 2.5x + y = 5 − 2.5x y = 5 − 2.5x, or y = −2.5x + 5

9

−6

6

−9

y = −52x + 3y = −2.5x + 5

y = −2x + 3

The fi rst line y = − 5 — 2 x + 3 and the third line y = −2.5x + 5 are parallel because they are always the same distance apart. Note that they have the same slopes because m = − 5 — 2 = −2.5.

2. a. 3x + 4y = 6 3x − 3x + 4y = 6 − 3x 4y = 6 − 3x

4y — 4 = 6 − 3x —

4

y = 3 — 2 − 3 —

4 x, or y = − 3 —

4 x + 3 —

2

3x − 4y = 12 3x − 3x − 4y = 12 − 3x −4y = 12 − 3x

−4y — −4

= 12 − 3x — −4

y = −3 + 3 — 4 x, or y = 3 —

4 x − 3

4x − 3y = 12 4x − 4x − 3y = 12 − 4x −3y = 12 − 4x

−3y — −3

= 12 − 4x — −3

y = −4 + 4 — 3 x, or y = 4 —

3 x − 4

9

−6

6

−9

y = −34x + 32

y = 43x − 4

y = 34 x − 3

The fi rst line y = − 3 — 4 x + 3 — 2 and the third line y =

4 — 3 x − 4

are perpendicular because they intersect at a right angle. Note that their slopes, m = − 3 — 4 and m =

4 — 3 are negative

reciprocals.

b. 2x + 5y = 10 2x − 2x + 5y = 10 − 2x 5y = 10 − 2x

5y — 5 = 10 − 2x —

5

y = 2 − 2 — 5 x, or y = − 2 —

5 x + 2

−2x + y = 3 −2x + 2x + y = 3 + 2x y = 3 + 2x, or y = 2x + 3

2.5x − y = 5 2.5x − 2.5x − y = 5 − 2.5x −y = 5 − 2.5x

−y — −1

= 5 − 2.5x — −1

y = −5 + 2.5x, or y = 2.5x − 5

9

−6

6

−9

y = 2.5x − 5y = 2x + 3

y = −25x + 2

The fi rst line y = − 2 — 5 x + 2 and the third line y = 2.5x − 5 are perpendicular because they intersect at a right angle. Note that the slope of the third line is 2.5 = 5 — 2 . So, the slopes of the perpendicular lines are negative reciprocals.

3. Parallel lines are in the same plane and are always the same distance apart. Perpendicular lines are in the same plane and intersect at right angles.

4. The parallel lines in Exploration 1 have the same slope. This will always be true because lines that are increasing or decreasing at the same rate will never intersect.

5. The perpendicular lines in Exploration 2 have slopes that are reciprocals of each other and have opposite signs. When two lines have slopes that are negative reciprocals, the lines will always be perpendicular.


1. Line a: m = −1 − 3 — −6 − (−5)

= −1 − 3 — −6 + 5

= −4 — −1

= 4

Line b: m = −7 − (−2) — 2 − 3

= −7 + 2 — 2 − 3

= −5 — −1

= 5

Lines a and b do not have the same slope. So, they are not parallel.


Chapter 4

2. y = mx + b

2 = 1 — 4 (−4) + b

2 = −1 + b + 1 + 1 3 = b Using m = 1 — 4 and b = 3, an equation of the parallel line is

y = 1 — 4 x + 3.

3. Line a: 2x + 6y = −3 2x − 2x + 6y = −3 − 2x 6y = −3 − 2x

6y — 6 = −3 − 2x —

6

y = − 1 — 2 − 1 —

3 x, or y = − 1 —

3 x − 1 —

2

Line b: y = 3x − 8 Line c: −6y + 18x = 9 −6y + 18x − 18x = 9 − 18x −6y = 9 − 18x

−6y — −6

= 9 − 18x — −6

y = − 3 — 2 + 3x, or y = 3x − 3 —

2

Lines b and c have slopes of 3. So, they are parallel. Line a has a slope of − 1 — 3 , which is the negative reciprocal of 3. So, it is perpendicular to lines b and c.

4. y − y1 = m(x − x1)

y − 5 = 1 — 3 [x − (−3)]

y − 5 = 1 — 3 (x + 3)

y − 5 = 1 — 3 (x) + 1 —

3 (3)

y − 5 = 1 — 3 x + 1

+ 5 + 5

y = 1 — 3 x + 6

An equation of the perpendicular line is y = 1 — 3 x + 6.

5. The slope of the shoreline is m = 1 − 3 — 4 − 1 = − 2 — 3 .

y − y1 = m(x − x1)

y − 3 = − 2 — 3 (x − 9)

y − 3 = − 2 — 3 x + 6

+ 3 + 3

y = − 2 — 3 x + 9

An equation that represents the path of the boat is y = − 2 — 3 x + 9.



1. Nonvertical parallel lines have the same slope.

2. The slope of the other line is 7 — 5 . The slopes, − 5 — 7 and

7 — 5 ,

are negative reciprocals of each other, so the lines are perpendicular.


3. Line a: m = 3 − 1 — 0 − (−3)

= 3 − 1 — 0 + 3

= 2 — 3

Line b: m = 2 − 0 — 3 − 0

= 2 — 3

Line c: m = 0 − (−3) — 3 − (−2)

= 0 + 3 — 3 + 2

= 3 — 5

Lines a and b have the same slope. So, they are parallel.

4. Line a: m = 0 − 5 — 2 − 0

= − 5 — 2

Line b: m = 0 − 4 — 5 − 3

= −4 — 2 , or −2

Line c: m = 4 − 6 — 5 − 4

= −2 — 1 , or −2

Because they have the same slope, lines b and c are parallel.

5. Line a: m = 0 − (−2) — 1 − (−1)

= 0 + 2 — 1 + 1

= 2 — 2 , or 1

Line b: m = −2 − 2 — 2 − 4

= −4 — −2

, or 2

Line c: m = 1 − 2 — −1 − 0

= −1 — −1

, or 1

Because they have the same slope, lines a and c are parallel.

6. Line a: m = 9 − 3 — 1 − (−1)

= 9 − 3 — 1 + 1

= 6 — 2 , or 3

Line b: m = 14 − 12 — −1 − (−2)

= 14 − 12 — −1 + 2

= 2 — 1 , or 2

Line c: m = 10 − 8 — 6 − 3

= 2 — 3

Because none of the lines have the same slope, none are parallel.


Chapter 4

7. Line a: 4y + x = 8 4y + x − x = 8 − x 4y = 8 − x

4y — 4 = 8 − x —

4

y = 2 − 1 — 4 x, or y = − 1 —

4 x + 2

Line a has a slope of − 1 — 4 .

Line b: 2y + x = 8 2y + x − x = 8 − x 2y = 8 − x

2y — 2 = 8 − x —

2

y = 4 − 1 — 2 x, or y = − 1 —

2 x + 4

Line b has a slope of − 1 — 2 .

Line c: 2y = −3x + 6

2y — 2 = −3x + 6 —

2

y = − 3 — 2 x + 3

Line c has a slope of − 3 — 2 .

Because none of the lines have the same slope, none are parallel.

8. Line a: 3y − x = 6 3y − x + x = 6 + x 3y = 6 + x

3y — 3 = 6 + x —

3

y = 2 + 1 — 3 x, or y = 1 —

3 x + 2

The slope of line a is 1 — 3 .

Line b: 3y = x + 18

3y — 3 = x + 18 —

3

y = 1 — 3 x + 6

The slope of line b is 1 — 3 .

Line c: 3y − 2x = 9 3y − 2x + 2x = 9 + 2x 3y = 9 + 2x

3y — 3 = 9 + 2x —

3

y = 3 + 2 — 3 x, or y = 2 —

3 x + 3

The slope of line c is 2 — 3 .

Because lines a and b have the same slope, they are parallel.

9. y = mx + b 3 = 2(−1) + b 3 = −2 + b + 2 + 2 5 = b Using m = 2 and b = 5, an equation of the parallel line is

y = 2x + 5.

10. y − y1 = m(x − x1) y − 2 = −5(x − 1) y − 2 = −5(x) − 5(−1) y − 2 = −5x + 5 + 2 + 2 y = −5x + 7 An equation of the parallel line is y = −5x + 7.

11. 3y − x = −12 3y − x + x = −12 + x 3y = −12 + x

3y — 3 = −12 + x —

3

y = −4 + 1 — 3 x, or y = 1 —

3 x − 4

y = mx + b

2 = 1 — 3 (18) + b

2 = 6 + b − 6 − 6 −4 = b Using m = 1 — 3 and b = −4, an equation of the parallel line

is y = 1 — 3 x − 4.

12. 2y = 3x + 10

2y — 2 = 3x + 10 —

2

y = 3 — 2 x + 5

y − y1 = m(x − x1)

y − (−5) = 3 — 2 (x − 2)

y + 5 = 3 — 2 (x) − 3 —

2 (2)

y + 5 = 3 — 2 x − 3

− 5 − 5

y = 3 — 2 x − 8

An equation of the parallel line is y = 3 — 2 x − 8.


Chapter 4

13. Line a: m = −4 − (−1) — −5 − (−3)

= −4 + 1 — −5 + 3

= −3 — −2

, or 3 — 2

Line b: m = −6 − (−4) — −2 − (−6)

= −6 + 4 — −2 + 6

= −2 — 4 , or − 1 —

2

Line c: m = −1 − (−6) — 0 − (−3)

= −1 + 6 — 0 + 3

= 5 — 3

Because none of the slopes are the same or negative reciprocals of each other, none of the lines are parallel or perpendicular.

14. Line a: m = 0 − 1 — 2 − (−1)

= 0 − 1 — 2 + 1

= −1 — 3

Line b: m = 4 − 5 — 3 − 0

= −1 — 3

Line c: m = 5 − 0 — 2 − 0

= 5 — 2

Lines a and b have the same slope. So, they are parallel. None of the slopes are negative reciprocals of each other. So, none of the lines are perpendicular.

15. Line a: m = 3 − 1 — 0 − (−2)

= 3 − 1 — 0 + 2

= 2 — 2 , or 1

Line b: m = 4 − 1 — 6 − 4

= 3 — 2

Line c: m = 1 − 3 — 4 − 1

= −2 — 3

None of the lines have the same slope. So, none are parallel. Line b’s slope of 3 — 2 and line c’s slope of −

2 — 3 are negative

reciprocals. So, lines b and c are perpendicular.

16. Line a: m = 13 − 10 — 4 − 2

= 3 — 2

Line b: m = 12 − 9 — 6 − 4

= 3 — 2

Line c: m = 9 − 10 — 4 − 2

= −1 — 2

Lines a and b have slopes of 3 — 2 . So, they are parallel. None of the lines have slopes that are negative reciprocals of each other. So, none are perpendicular.

17. Line a: 4x − 3y = 2 4x − 4x − 3y = 2 − 4x −3y = 2 − 4x

−3y — −3

= 2 − 4x — −3

y = − 2 — 3 + 4 —

3 x, or y = 4 —

3 x − 2 —

3

Line b: y = 4 — 3 x + 2

Line c: 4y + 3x = 4 4y + 3x − 3x = 4 − 3x 4y = 4 − 3x

4y — 4 = 4 − 3x —

4

y = 1 − 3 — 4 x, or y = − 3 —

4 x + 1

Lines a and b have slopes of 4 — 3 . So, they are parallel. Line c has a slope of − 3 — 4 , the negative reciprocal of

4 — 3 . So, line c is

perpendicular to lines a and b.

1 8. Line a: y = 6x − 2 Line b: 6y = −x

6y — 6 = −x —

6

y = − 1 — 6 x

Line c: y + 6x = 1 y + 6x − 6x = 1 − 6x y = 1 − 6x, or y = −6x + 1 Line a’s slope of 6 and line b’s slope of − 1 — 6 are

negative reciprocals of each other. So, lines a and b are perpendicular. None of the lines have the same slope. So, none are parallel.

19. y − y1 = m(x − x1) y − 10 = −2(x − 7) y − 10 = −2(x) − 2(−7) y − 10 = −2x + 14 + 10 + 10 y = −2x + 24 An equation of the perpendicular line is y = −2x + 24. 20. y = mx + b

−1 = − 3 — 4 (−4) + b

−1 = 3 + b − 3 − 3 −4 = b

Using m = − 3 — 4 and b = −4, an equation of the perpendicular

line is y = − 3 — 4 x − 4.


Chapter 4

21. 2y = 8x − 6

2y — 2 = 8x − 6 —

2

y = 4x − 3 y − y1 = m(x − x1)

y − 3 = − 1 — 4 [x − (−3)]

y − 3 = − 1 — 4 (x + 3)

y − 3 = − 1 — 4 (x) − 1 —

4 (3)

y − 3 = − 1 — 4 x − 3 —

4

+ 3 + 3

y = − 1 — 4 x + 9 —

4 ( − 3 — 4 + 3 = − 3 — 4 + 12 — 4 = 9 — 4 )

An equation of the perpendicular line is y = − 1 — 4 x + 9 —

4 .

22. 2y + 4x = 12 2y + 4x − 4x = 12 − 4x 2y = 12 − 4x

2y — 2 = 12 − 4x —

2

y = 6 − 2x, or y = −2x + 6 y = mx + b

1 = 1 — 2 (8) + b

1 = 4 + b − 4 − 4 −3 = b Using m = 1 — 2 and b = −3, an equation of the perpendicular

line is y = 1 — 2 x − 3.

23. m = 2 − 6 — 2 − 1

= −4 — 1 , or −4

a. y − y1 = m(x − x1) y − 3 = −4(x − 4) y − 3 = −4(x) − 4(−4) y − 3 = −4x + 16 + 3 + 3 y = −4x + 19 An equation of the parallel line is y = −4x + 19.

b. y = mx + b

3 = 1 — 4 (4) + b

3 = 1 + b − 1 − 1 2 = b Using m = 1 — 4 and b = 2, an equation of the perpendicular

line is y = 1 — 4 x + 2.

24. m = −1 − (−4) — 2 − 1

= −1 + 4 — 2 − 1

= 3 — 1 , or 3

a. y = mx + b −2 = 3(3) + b −2 = 9 + b − 9 − 9 −11 = b Using m = 3 and b = −11, an equation of the parallel

line is y = 3x − 11. b. y − y1 = m(x − x1)

y − (−2) = − 1 — 3 (x − 3)

y + 2 = − 1 — 3 (x) − 1 —

3 (−3)

y + 2 = − 1 — 3 x + 1

− 2 − 2

y = − 1 — 3 x − 1

An equation of the perpendicular line is y = − 1 — 3 x − 1.

25. Parallel lines have the same slope, not negative reciprocal slopes.

y − y1 = m(x − x1)

y − 3 = 1 — 4 (x − 1)

y − 3 = 1 — 4 (x) − 1 —

4 (1)

y − 3 = 1 — 4 x − 1 —

4

+ 3 + 3

y = 1 — 4 x + 11 —

4 ( − 1 — 4 + 3 = − 1 — 4 + 12 — 4 = 11 — 4 )

An equation of the parallel line is y = 1 — 4 x + 11 —

4 .


Chapter 4

26. The reciprocal of the slope was used, but the slopes of perpendicular lines are negative reciprocals of each other.

y − y1 = m(x − x1) y − (−5) = −3(x − 4) y + 5 = −3(x) − 3(−4) y + 5 = −3x + 12 − 5 − 5 y = −3x + 7 An equation of the perpendicular line is y = −3x + 7.

27. m = 3 − (−1) — 0 − (−3)

= 3 + 1 — 0 + 3

= 4 — 3

y − y1 = m(x − x1)

y − 0 = − 3 — 4 (x − 2)

y = − 3 — 4 (x) − 3 —

4 (−2)

y = − 3 — 4 x + 3 —

2

An equation that represents the new pipe is y = − 3 — 4 x + 3 —

2 .

28. m = 4 − 0 — 11 − 8

= 4 — 3

y − y1 = m(x − x1)

y − 5 = 4 — 3 (x − 4)

y − 5 = 4 — 3 (x) − 4 —

3 (4)

y − 5 = 4 — 3 x − 16 —

3

+ 5 + 5

y = 4 — 3 x − 1 —

3 ( − 16 — 3 + 5 = − 16 — 3 + 15 — 3 = − 1 — 3 )

An equation that represents the new bike path is y = 4 — 3 x − 1 — 3 .

29.

x

y

4

6

8

10

2

04 6 820

A

B

C

D

slope of — AB = 4 − 2 — 6 − 2

= 2 — 4 , or 1 —

2

slope of — BC = 10 − 4 — 8 − 6

= 6 — 2 , or 3

slope of — CD = 10 − 8 — 8 − 4

= 2 — 4 , or 1 —

2

slope of — AD = 8 − 2 — 4 − 2

= 6 — 2 , or 3

a. yes; Opposite sides, — AB and — CD , have the same slope. So, the sides are parallel. Similarly, opposite sides, — BC and — AD , are parallel because they have the same slope. So, quadrilateral ABCD is a parallelogram.

b. no; The slopes of adjacent sides — AB and — BC , 1 — 2 and 3, are not negative reciprocals. So, the sides are not perpendicular. None of the other pairs of adjacent sides are perpendicular either. So, quadrilateral ABCD is not a rectangle.

30. 6y = −2x + 4 2y = ax + 5

6y — 6 = −2x + 4 —

6 2y —

2 = ax + 5 —

2

y = − 1 — 3 x + 2 —

3 y = a —

2 x + 5 —

2

− 1 — 3 = a —

2 3 = a —

2

2 ⋅ ( − 1 — 3 ) = 2 ⋅ a — 2 2 ⋅ 3 = 2 ⋅ a — 2 − 2 —

3 = a 6 = a

If a = − 2 — 3 , then the lines will be parallel, because their slopes will both be − 1 — 3 . If a = 6, then the lines will have slopes of − 1 — 3 and 3, which are negative reciprocals of each other. So, the lines are perpendicular.

31. m = 8 − 0 — 0 − (−4)

= 8 − 0 — 0 + 4

= 8 — 4 , or 2

m = 8 − 16 — 0 − (−4)

= 8 − 16 — 0 + 4

= −8 — 4 , or −2

no; The slopes of the two segments of the puck’s path are 2 and −2, which are opposites but not reciprocals of each other. So, they are not perpendicular.


Chapter 4

32. a. yes; The graph of the total amount paid by Student B crosses the y-axis at a point that is greater than the y-intercept of Student A’s graph. So, Student B had a greater initial cost, or registration fee.

b. no; The lines are parallel. They have the same slope and are increasing at the same rate. So, they pay the same monthly fee.

33. never; The slopes of perpendicular lines must be negative reciprocals of each other. So, if one is positive, the other must be negative.

34. always; All vertical lines are parallel to each other, and because the y-axis is a vertical line, it is parallel to all vertical lines.

35. sometimes; If two lines have slopes that are negative reciprocals of each other, then they are perpendicular. If they also have the same y-intercept, then they intersect on the y-axis to form right angles.

36. Sample answer:

x

y

4

6

8

2

04 6 820

Math

Club

The equations of the lines that form the long sides of the “equal sign” are y = 3, y = 4.5, y = 4.75, and y = 6. They all have a slope of 0. So, they are all parallel. The lines that form the long sides of the “multiplication sign” are y = x + 1, y = x − 1, y = −x + 8, and y = −x + 10. The slopes of these lines are either 1 or −1, which are negative reciprocals of each other. These four lines form four pairs of perpendicular lines, such as y = x + 1 and y = −x + 8, and two pairs of parallel lines, such as y = x + 1 and y = x − 1.


37. yes; Each x-value is paired with exactly one y-value. So, the relation is a function.

38. no; The same x-value, 1, is paired with two different y-values, 4 and 6, and the same x-value, −1, is paired with multiple y-values, 2, 5, and 6. So, the relation is not a function.

4.1–4.3 What Did You Learn? (p. 193)

1. Sample answer: The graph shows the increase in U.S. box offi ce revenue over time.

2. Sample answer: The slopes were the same, so the lines were parallel. The constants in the point-slope form indicated the graph was translated vertically and horizontally.

3. Sample answer: The diagram was used to identify two points on each segment of the puck’s path. Then fi nd the slope of each segment and compare the slopes to determine whether they are negative reciprocals. Because they are not, you know that your friend is incorrect about them being perpendicular.

4.1–4.3 Quiz (p.194)

1. m = 3 − (−2) — 1 − 0

= 3 + 2 — 1 − 0

= 5 — 1 , or 5

Because the line crosses the y-axis at (0, −2), the y-intercept is −2. So, the equation is y = 5x − 2.

2. m = 4 − 5 — 3 − 0

= −1 — 3

Because the line crosses the y-axis at (0, 5), the y-intercept is

5. So, the equation is y = − 1 — 3 x + 5.

3. m = 0 − 4 — 0 − (−2)

= 0 − 4 — 0 + 2

= −4 — 2 , or −2

Because the line crosses the y-axis at (0, 0), the y-intercept is 0. So, the equation is y = −2x + 0, or y = −2x.

4. m = −1 − 5 — 1 − (−2)

= −1 − 5 — 1 + 2

= −6 — 3 , or −2

y − y1 = m(x − x1) or y − y1 = m(x − x1) y − (−1) = −2(x − 1) y − 5 = −2[x − (−2)] y + 1 = −2(x − 1) y − 5 = −2(x + 2)

So, an equation is y + 1 = −2(x − 1) or y − 5 = −2(x + 2).

5. m = −1 − (−2) — 2 − (−3)

= −1 + 2 — 2 + 3

= 1 — 5

y − y1 = m(x − x1) or y − y1 = m(x − x1)

y − (−2) = 1 — 5 (x − (−3)) y − (−1) = 1 —

5 (x − 2)

y + 2 = 1 — 5 (x + 3) y + 1 = 1 —

5 (x − 2)

So, an equation is y + 2 = 1 — 5 (x + 3) or y + 1 = 1 —

5 (x − 2).

6. m = 4 − 0 — 4 − 1

= 4 — 3

y − y1 = m(x − x1) or y − y1 = m(x − x1)

y − 0 = 4 — 3 (x − 1) y − 4 = 4 —

3 (x − 4)

y = 4 — 3 (x − 1)

So, an equation is y = 4 — 3 (x − 1) or y − 4 = 4 —

3 (x − 4).


Chapter 4

7. Rewrite f (0) = 2 as (0, 2) and f (5) = −3 as (5, −3).

m = −3 − 2 — 5 − 0

= −5 — 5 , or −1

y − y1 = m(x − x1) y − 2 = −1(x − 0) y − 2 = −1(x) y − 2 = −x y − 2 + 2 = −x + 2 y = −x + 2 So, a linear function is f (x) = −x + 2.

8. Rewrite f (−1) = −6 as (−1, −6) and f (4) = −6 as (4, −6).

m = −6 − (−6) — 4 − (−1)

= −6 + 6 — 4 + 1

= 0 — 5 , or 0

y − y1 = m(x − x1) y − (−6) = 0(x − 4) y + 6 = 0 − 6 − 6 y = −6 So, a linear function is f (x) = −6.

9. Rewrite f (−3) = −2 as (−3, −2) and f (−2) = 3 as (−2, 3).

m = 3 − (−2) — −2 − (−3)

= 3 + 2 — −2 + 3

= 5 — 1 , or 5

y = mx + b −2 = 5(−3) + b −2 = −15 + b + 15 + 15 13 = b Using m = 5 and b = 13, a linear function is f (x) = 5x + 13.

10. Line a: m = 1 − 2 — 2 − (−2)

= 1 − 2 — 2 + 2

= −1 — 4

Line b: m = 0 − (−8) — 3 − 1

= 0 + 8 — 3 − 1

= 8 — 2 , or 4

Line c: m = −2 − (−3) — 0 − (−4)

= −2 + 3 — 0 + 4

= 1 — 4

None of the lines have the same slope. So, none of the lines are parallel. The slopes of lines a and b, − 1 — 4 and 4 respectively, are negative reciprocals of each other. So, lines a and b are perpendicular.

11. Line a: 2x + 6y = −12 2x − 2x + 6y = −12 − 2x 6y = −12 − 2x

6y — 6 = −12 − 2x —

6

y = −2 − 1 — 3 x, or y = − 1 —

3 x − 2

Line b: y = 3 — 2 x − 5

Line c: 3x − 2y = −4 3x − 3x − 2y = −4 − 3x −2y = −4 − 3x

−2y — −2

= −4 − 3x — −2

y = 2 + 3 — 2 x, or y = 3 —

2 x + 2

Lines b and c each have a slope of 3 — 2 . So, they are parallel. None of the lines have slopes that are negative reciprocals. So, none are perpendicular.

12. Let (x1, y1) = (1, −1) and (x2, y2) = (2, 2).

m = y2 − y1 — x2 − x1

= 2 − (−1) — 2 − 1

= 2 + 1 — 2 − 1

= 3 — 1 , or 3

a. y − y1 = m(x − x1) y − 2 = 3(x − 6) y − 2 = 3(x) − 3(6) y − 2 = 3x − 18 + 2 + 2 y = 3x − 16 So, an equation of the parallel line is y = 3x − 16. b. y − y1 = m(x − x1)

y − 2 = − 1 — 3 (x − 6)

y − 2 = − 1 — 3 (x) − 1 —

3 (−6)

y − 2 = − 1 — 3 x + 2

+ 2 + 2 y = − 1 —

3 x + 4

So, an equation of the perpendicular line is y = − 1 — 3 x + 4.


Chapter 4

13. Let (x1, y1) = (0, 2) and (x2, y2) = (2, 1).

m = y2 − y1 — x2 − x1

= 1 − 2 — 2 − 0

= −1 — 2

a. y = mx + b −3 = − 1 —

2 (−2) + b

−3 = 1 + b − 1 − 1 −4 = b Using m = − 1 — 2 and b = −4, an equation of the parallel line

is y = − 1 — 2 x − 4.

b. y = mx + b −3 = 2(−2) + b −3 = −4 + b + 4 + 4 1 = b Using m = 2 and b = 1, an equation of the perpendicular

line is y = 2x + 1.

14. Let (x1, y1) = (−3, 3) and (x2, y2) = (−2, −1).

m = y2 − y1 — x2 − x1

= −1 − 3 — −2 − (−3)

= −1 − 3 — −2 + 3

= − 4 — 1 , or − 4

a. y − y1 = m(x − x1) y − 0 = − 4[x − (−4)] y = − 4(x + 4) y = − 4(x) − 4(4) y = − 4x − 16 So, an equation of the parallel line is y = − 4x − 16. b. y − y1 = m(x − x1)

y − 0 = 1 — 4 [x − (−4)]

y = 1 — 4 (x + 4)

y = 1 — 4 (x) + 1 —

4 (4)

y = 1 — 4 x + 1

So, an equation of the perpendicular line is y = 1 — 4 x + 1.

15. a. Words: Total cost =

Initial fee +

Charge per month ⋅

Number of months

Variable: Let C be the total cost of setting up and maintaining a website and m be the number of months it is maintained.

Linear model: C = 48 + 44 ⋅ m A linear model that represents this situation is C = 48 + 44m. b. C = 48 + 44m = 48 + 44(6) = 48 + 264 = 312

The total cost of setting up a website and maintaining it for 6 months is $312.

c. C = 48 + 44m C = 62m 620 = 48 + 44m 620 = 62m

− 48 − 48 620 — 62

= 62m — 62

572 = 44m 10 = m

572 — 44

= 44m — 44

13 = m At the original company, you can set up and maintain a

website for 13 months, but at the second company, you can only set up and maintain a website for 10 months. So, with $620, you can maintain the website for a longer time at the original company.

16. 150 − 155 — 10 − 8

= −5 — 2 , 145 − 150 —

12 − 10 = −5 —

2 ,

140 − 145 — 14 − 12

= −5 — 2 , 130 − 140 —

16 − 14 = −5 —

2

w − w1 = m(t − t1)

w − 150 = − 5 — 2 (t − 10)

w − 150 = − 5 — 2 (t) − 5 —

2 (−10)

w − 150 = − 5 — 2 t + 25

+ 150 + 150

w = − 5 — 2 t + 175

Because the amount of water decreases at a constant rate, the situation can be modeled by a linear equation. A linear model is w = − 5 — 2 t + 175.


Chapter 4

4.4 Explorations (p. 195)

1. a. Sample answer:

Ages of Married Couples

30

0

35

40

45

50

55

60

65

70

75

80

300 35 40 45 50 55 60 65 70 75 80 x

y

Husband’s age

Wif

e’s

age

y = x − 2

Notice that as the husband’s age increases, the wife’s age also increases at an approximately constant rate. So, the trend can be modeled by a line with a positive slope. Draw a line that appears to fi t the data closely. Draw the line through the middle of the points so that there are approximately as many points above the line as below it. Then write an equation using two points on the line such as (40, 38) and (75, 73).

The slope of the line is m = 73 − 38 — 75 − 40

= 35 — 35

, or 1.

y − y1 = m(x − x1) y − 38 = 1(x − 40) y − 38 = x − 40 + 38 + 38 y = x − 2 An equation of a line that approximates the data is

y = x − 2. b. Sample answer: Based on the equation, married couples are

usually close to the same age, but women are, on average, a couple of years younger than their husband. The slope of the line is 1 because their ages increase at the same rate.


0

20

18

22

24

26

28

1960 1970 1980 1990 2000 2010 2020

Year

Ag

e

Ages of American Women at First Marriage

y = x + 19.7518

Notice that the ages are increasing at an approximately constant rate over time. So, the trend can be modeled by a line with a positive slope. Draw a line that appears to fi t the data closely. Draw the line through the middle of the points so that there are approximately as many points above the line as below it. Then write an equation using two points on the line such as (10, 21) and (50, 26).


= 5 — 40

, or 1 — 8 .

y − y1 = m(x − x1)

y − 21 = 1 — 8 (x − 10)

y − 21 = 1 — 8 (x) − 1 —

8 (10)

y − 21 = 1 — 8 x − 5 —

4

+ 21 + 21

y = 1 — 8 x + 19.75

An equation of a line that approximates the data is y = 1 — 8 x + 19.75.

b. Sample answer: The slope is 1 — 8 . So, every 8 years, the median age of American women at their fi rst marriage increases by 1 year. The y-intercept is 19.75. So, in 1960 the median age of American women at their fi rst marriage was just under 20 years old.

c. y = 1 — 8 x + 19.75

y = 1 — 8 (60) + 19.75

y = 7.5 + 19.75 y = 27.25 So, if the trend continues, the median age of American

women at their fi rst marriage in the year 2020 will be just over 27 years old.

3. Sample answer: Scatter plots can show trends in data so that you can identify any correlations between data sets. If the data can be modeled by a line, then you can draw a line that appears to fi t the data closely. You can use two points on the line to write an equation in slope-intercept form. So, you will know the approximate rate of change, or slope, of the data and an initial value if applicable.


Chapter 4

4. Sample answer: The data in the graph is from the Pennsylvania Department of Transportation.

0

122

124

126

128

130

132

134

136

138

140

0 1 2 3 4 5 6 7 x

y

Years since 2005

PA Statewide Car Crashes

Cra

shes

(th

ou

san

ds)

The number of crashes is decreasing at a fairly constant rate over time. So, the data can be modeled by a line with a negative slope. Estimate a line through the middle of the plotted data, then use two points on that line to fi nd the equation. Use the points (6, 124.9) and (7, 122.8).

The slope of the line is m = 122.8 − 124.9 —— 7 − 6

= −2.1 — 1 ,

or −2.1.

y − y1 = m(x − x1) y − 124.9 = −2.1 (x − 6) y − 124.9 = −2.1x − 2.1(−6) y − 124.9 = −2.1x + 12.6 + 124.9 + 124.9 y = −2.1x + 137.5

An equation of a line that approximates the data is y = −2.1x + 137.5.


1. The smoothie has 260 calories.

2. The smoothie has 52 grams of sugar.

3.

0

1

2

3

4

5

6

0 66 68 70 72 74 76 78 80 82 84 86 88 90 92 94 x

y

Temperature (°F)

Att

end

ance

(th

ou

san

ds)

The scatter plot shows a positive correlation.

4.

0

5

10

15

20

25

0 1 2 3 4 5 6 7 8 x

y

Age of car (years)

Val

ue

(th

ou

san

ds

of

do

llars

)

The scatter plot shows a negative correlation.

5. Sample answer:

14000 1600 1800 2000 2200

320

360

400

440

280

0

Monthly income (dollars)

Monthly Income vs. Monthly Car Payment

Mo

nth

ly c

ar p

aym

ent

(do

llars

)

(1700, 340)

(1900, 380)

y = x15

x

y

Use (1700, 340) and (1900, 380).

The slope of the line is m = 380 − 340 —— 1900 − 1700

= 40 — 200

= 1 — 5

.

y − y1 = m(x − x1)

y − 340 = 1 — 5 (x − 1700)

y − 340 = 1 — 5 (x) − 1 —

5 (1700)

y − 340 = 1 — 5 x − 340

+ 340 + 340

y = 1 — 5 x

An equation of the line of fi t is y = 1 — 5 x. The slope of the line is 1 — 5 .

This means that a person’s monthly car payment increases by about $100 for every $500 increase in their monthly income, or about $0.20 per dollar. The y-intercept is 0, meaning that a person with no monthly income, has no car payment.



1. When data show a positive correlation, the dependent variable tends to increase as the independent variable increases.

2. A line of fi t is a line drawn on a scatter plot that is close to most of the points of a data set.


Chapter 4


3. (16, 6) 4. (3, 14) 5. (7, 12) 6. (8, 17)

7. a. The laptop with a hard drive capacity of 8 gigabytes costs $1100.

b. The $1200 laptop has a hard drive capacity of 12 gigabytes.

c. The price tends to increase as the hard drive capacity increases.

8. a. The pitcher with an earned run average of 4.2 has a winning percentage of 0.600.

b. The pitcher with a winning percentage of 0.33 has an earned run average of 5.0.

c. The winning percentage tends to decrease as the earned run average increases.

9. The y-values tend to increase as the x-values increase. So, the scatter plot shows a positive correlation.

10. The points show no pattern. So, the scatter plot shows no correlation.

11. The points show no pattern. So, the scatter plot shows no correlation.

12. The y-values tend to decrease as the x-values increase. So, the scatter plot shows a negative correlation.

13.

0

1

2

3

4

0 1 2 3 4 x

y

The points show no pattern. So, the scatter plot shows no correlation.

14.

0

20

40

60

80

0 1 2 3 4 5 6 7 8 9 10 x

y

As the x-values increase, the y-values decrease. So, the scatter plot shows a negative correlation.

15. Sample answer:

200 40 x

y

30

40

20

0

Years since 1960

World Birth Rates

Nu

mb

er o

f b

irth

sp

er 1

000

peo

ple

(10, 32.5)

(40, 22.5)

y = − x + 132156

a. Use (10, 32.5) and (40, 22.5).

The slope of the line is m = 22.5 − 32.5 — 40 − 10

= −10 — 30

, or − 1 — 3 .

y − y1 = m(x − x1)

y − 32.5 = − 1 — 3 (x − 10)

y − 32.5 = − 1 — 3 (x) − 1 —

3 (−10)

y − 32.5 = − 1 — 3 x + 10 —

3

+ 32.5 + 32.5

y = − 1 — 3 x + 215 —

6

An equation of the line of fi t is y = − 1 — 3 x + 215 —

6 .

b. The slope of the line is − 1 — 3 . This means that every 3 years the world birth rate decreases by 1 birth per 1000 people. The y-intercept is 215 — 6 , or about 35.8, which is close to the actual initial birth rate of 35.4 in 1960.

16. Sample answer:

20 4 6

80

120

40

0

Hours worked

Working in Food Service

Tota

l ear

nin

gs

(do

llars

)

(1, 18)

(0, 0)y = 18x

a. Use (0, 0) and (1, 18).


= 18 — 1

, or 18.

y − y1 = m(x − x1) y − 0 = 18(x − 0) y = 18x An equation of the line of fi t is y = 18x. b. The slope of the line is 18. This means that the server

earns about $18 per hour. The y-intercept is 0, which means that if the server does not work any hours, then the server will not make any money.


Chapter 4

17. Sample answer:

High temperature (degrees Fahrenheit) 32 43 50 65 73 80

Attendance at exercise facility 100 95 83 72 63 55

I randomly chose 6 days out of the year. Then I compared the high temperature in my town with the number of people who went to a local indoor exercise facility to work out. The data would probably show a negative correlation as with this sample data set, because as the weather gets warmer, more people exercise outside and fewer people go to the gym.

18. no; Both the independent variable x and the dependent variable y are decreasing. So, as you go backward through the chart, both are increasing. Therefore, the data shows a positive correlation.


450 55 65 75 x

y

55

65

75

45

0

Height (inches)

Arm

sp

an (

inch

es)

(65, 65)

(50, 50)

Height vs. Arm Span

y = x

Use (50, 50) and (65, 65).


= 15 — 15

, or 1.

y − y1 = m(x − x1) y − 50 = 1(x − 50) y − 50 = 1(x) − 1(50) y − 50 = x − 50 + 50 + 50 y = x An equation of the line of fi t is y = x. b. The slope of the line is 1. This means that a person’s arm

span increases by about 1 inch for every 1 inch increase in height. The y-intercept is 0. The y-intercept has no meaning in this context, because the height cannot be 0 inches.

20. Sample answer: A car is going around the curve of an entrance ramp at 20 miles per hour. Then, as the ramp starts to straighten out, the car begins to accelerate. For every 1 second of acceleration, the car’s speed increases by about 5 miles per hour until the car merges with traffi c on the highway. The data could look something like this:

Elapsed time (hours), x 0 1 2 3 4 5 6 7 8 9

Speed (miles per hour), y 20 26 29 35 41 43 49 56 62 66

21. Sample answer: A scatter plot is the best way to display a relationship between two sets of numerical data, such as age and time.

22. Sample answer: Choose points that are on or close to the line, such as (0, 10), (17, 135), (13, 107) and (21, 163).

23. no; A line of fi t is used to model a trend in the data and should be close to most of the data points. If there is no trend, it is not possible to draw a line that is close to most of the points.

24.

x

y

20

40

60

80

100

120

140

160

180

200

220

240

15 1893−3−9−15−18

The data follows a curve that decreases and then increases (a parabola). The data cannot fi t a line because the data points do not have a linear trend.


Chapter 4 · 2016. 10. 10. · Chapter 4 3. Solve a simpler problem. Suppose you drive 250 miles in...

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Transcript of Chapter 4 · 2016. 10. 10. · Chapter 4 3. Solve a simpler problem. Suppose you drive 250 miles in...