hscc alg1 wsk 04 - schoolwires · 2015. 4. 23. · 4.1 explorations (p.175) 1. a. m = y 2 − y 1...

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

Chapter 4

Chapter 4 Maintaining Mathematical Pro! 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.

2. Solve a simpler problem.

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

3. Solve a simpler problem.

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 " 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 ⋅ x

y = 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


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.


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


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


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 ⋅ x

y = 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 unde" 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. Re# ect (0, 1) and (3, −4) in the x-axis. The re# 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 of" 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 of" 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 of" 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 of" 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 Pro! 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

1012

46

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

234

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

y21

−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

468

10

2

−4

2 41 3−2−1−3−4

(0, )

(3, 0)

212

4.2 Explorations (p.181) 1. a.

x

y

456

23

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

45678

23

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).

4.2 Monitoring Progress (pp. 182–184) 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)

Vocabulary and Core Concept Check 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

Monitoring Progress and Modeling with Mathematics 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


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


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 " 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 " 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

Now plot the y-intercept (0, −5) on the y-axis. 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 con" 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 " rst, and Sibling A runs out of money last.

Maintaining Mathematical Pro! ciency 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 .

4.3 Explorations (p.187) 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 = −3

4x + 32

The " 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 + 3

y = −2.5x + 5

y = −2x + 3

The " 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 " 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 " 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.

4.3 Monitoring Progress (pp. 188–190)

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.


Vocabulary and Core Concept Check 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.

Monitoring Progress and Modeling with Mathematics

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.

Maintaining Mathematical Pro! ciency 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

of" 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 " 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

35404550556065707580

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 " 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.

2. a. Sample answer:

0

2018

22242628

1960 1970 1980 1990 2000 2010 2020Year

Age

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 " 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 " rst marriage increases by 1 year. The y-intercept is 19.75. So, in 1960 the median age of American women at their " 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 " 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 " 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.

0122124126128130132134136138140

0 1 2 3 4 5 6 7 x

y

Years since 2005

PA Statewide Car Crashes

Cras

hes

(tho

usan

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 " 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.

4.4 Monitoring Progress (pp. 196–198) 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

enda

nce

(tho

usan

ds)

The scatter plot shows a positive correlation.

4.

05

10152025

0 1 2 3 4 5 6 7 8 x

y

Age of car (years)

Valu

e (t

hous

ands

of

dol

lars

)

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

Mon

thly

car

pay

men

t (d

olla

rs)

(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 " 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.


Vocabulary and Core Concept Check 1. When data show a positive correlation, the dependent

variable tends to increase as the independent variable increases.

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


Chapter 4

Monitoring Progress and Modeling with Mathematics 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.

01234

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

Num

ber

of b

irth

spe

r 10

00 p

eopl

e

(10, 32.5)

(40, 22.5)

y = − x + 13

2156

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 " 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

ning

s(d

olla

rs)

(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 " 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.

19. a. Sample answer:

450 55 65 75 x

y

55

65

75

45

0

Height (inches)

Arm

spa

n (in

ches

)

(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 " 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 traf" 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 " 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

20406080

100120140160180200220240

15 1893−3−9−15−18

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

Maintaining Mathematical Pro! ciency 25. g(x) = 6x

g(−3) = 6(−3)

= −18

When x = −3, g(x) = −18.

g(x) = 6x

g(0) = 6(0)

= 0

When x = 0, g(x) = 0.

g(x) = 6x

g(4) = 6(4)

= 24

When x = 4, g(x) = 24.


Chapter 4

26. h(x) = −10x

h(−3) = −10(−3)

= 30

When x = −3, h(x) = 30.

h(x) = −10x

h(0) = −10(0)

= 0

When x = 0, h(x) = 0.

h(x) = −10x

h(4) = −10(4)

= −40

When x = 4, h(x) = −40.

27. f (x) = 5x − 8 f (−3) = 5(−3) − 8 = −15 − 8 = −23

When x = −3, f (x) = −23.

f (x) = 5x − 8 f (0) = 5(0) − 8 = 0 − 8 = −8

When x = 0, f (x) = −8.

f (x) = 5x − 8 f (4) = 5(4) − 8 = 20 − 8 = 12

When x = 4, f (x) = 12.

28. v(x) = 14 − 3x

v(−3) = 14 − 3(−3)

= 14 + 9

= 23

When x = −3, v(x) = 23.

v(x) = 14 − 3x

v(0) = 14 − 3(0)

= 14 − 0 = 14

When x = 0, v(x) = 14.

v(x) = 14 − 3x

v(4) = 14 − 3(4)

= 14 − 12

= 2

When x = 4, v(x) = 2.

4.5 Explorations (p. 201) 1. a. The ordered pair (25, 23.3) means that 25 years after

1960, or in 1985, the median age of American women at their " rst marriage was 23.3.

b. Check students’ work.

c. Sample answer: An equation of the line of best " t is y = 0.13x + 19.8. The equation for Exploration 2 in Section 4.4 was y = 1 — 8 x + 19.75. The slope of this equation, 1 — 8 = 0.125, is very close to the actual slope of about 0.126 that was calculated by the linear regression feature of a graphing calculator. The y-intercept, 19.75, of the equation from Section 4.4 is also very close to the calculated y-intercept, 19.845, for the line of best " t. So, the equation from Section 4.4 is a close model for the data, but not exactly the best " tting model.

2. You can use a computer, spreadsheet, or the linear regression feature of a graphing calculator to " nd an exact slope and y-intercept of a line of best " t, or a line that minimizes the distance from each data point to the line.

3.

65707580859095

13 14 15 16 17 18 19 20 21 x

y

Chirps per second

Tem

pera

ture

(°F)

An equation of the line of best " t is y = 3.2746x + 24.984.

y = 3.2746x + 24.984

= 3.2746(19) + 24.984

= 62.2174 + 24.984

= 87.2014

So, when there are 19 chirps per second for striped ground crickets, the outside temperature is about 87° F.


Chapter 4

4.5 Monitoring Progress (pp. 203–205) 1.

x yy-Value

from model Residual

0 850 850 850 − 850 = 0

1 845 840.2 845 − 840.2 = 4.8

2 828 830.4 828 − 830.4 = −2.4

3 798 820.6 798 − 820.6 = −22.6

4 800 810.8 800 − 810.8 = −10.8

5 792 801 792 − 801 = −9

6 785 791.2 785 − 791.2 = −6.2

7 781 781.4 781 − 781.4 = −0.4

8 775 771.6 775 − 771.6 = 3.4

9 760 761.8 760 − 761.8 = −1.8

x

residual10

−20

−10

84

There are more points below the x-axis than there are above the x-axis. So, the equation y = −9.8x + 850 does not model the data well.

2. a. After entering the data from the table into two lists using a graphing calculator, the linear regression feature yields the equation y = −9.6x + 844.5.

12750

870

0

b. The correlation coef" cient is about −0.964. This means that the relationship between the year and the attendance has a strong negative correlation and the equation closely models the data, as shown in the graph.

c. The slope of the line is −9.6. This means the attendance at the amusement park decreases by about 9600 people each year. The y-intercept is 844.5. This means the attendance in 2005 was 844,500 people.

3. y = −9.6x + 844.5

= −9.6(12) + 844.5

= −115.2 + 844.5

= 729.3

So, in 2017, if the trend continues, the attendance at the amusement park will be about 729,300 people.

4. Sample answer: There may be a correlation but no causal relationship. Playing video games does not directly affect your grade point average. However, it is likely that a student, who spends a lot of time playing video games, will probably spend less time studying and therefore have a lower grade point average.


Vocabulary and Core Concept Check 1. A residual is positive when the y-value of the actual data

point is greater than the y-value of the corresponding point on the line of " t that has the same x-value. A residual is negative when the y-value of the actual data point is less than the y-value of the corresponding point on the line of " t that has the same x-value.

2. A scatter plot of the residuals shows how well a model " ts a data set. If the model is a good " t, then the absolute values of the residuals are relatively small, and the residual points will be more or less evenly dispersed about the horizontal axis. If the model is not a good " t, then the residual points will form some type of pattern that suggests the data are not linear.

3. Using a graph or its equation to approximate a value between two known values is called interpolation. Extrapolation also involves using a graph or its equation to approximate a value, but extrapolation is predicting a value outside of the range of known values.

4. The correlation coef" cient r = −0.09 does not belong with the other three, because this value is close to 0 and represents a weak correlation, but the other three are close to either 1 or −1 and represent strong correlations.


Chapter 4

Monitoring Progress and Modeling with Mathematics 5.

x yy-Value

from model Residual

−4 −18 −21 −18 − (−21) = 3

−3 −13 −17 −13 − (−17) = 4

−2 −10 −13 −10 − (−13) = 3

−1 −7 −9 −7 − (−9) = 2

0 −2 −5 −2 − (−5) = 3

1 0 −1 0 − (−1) = 1

2 6 3 6 − 3 = 3

3 10 7 10 − 7 = 3

4 15 11 15 − 11 = 4

x42−2−4

1

2

3

4residual

All of the residual values are above the horizontal axis. So, the equation y = 4x − 5 does not model the data well.

6.

x yy-Value

from model Residual

1 13 10 13 − 10 = 3

2 14 16 14 − 16 = −2

3 23 22 23 − 22 = 1

4 26 28 26 − 28 = −2

5 31 34 31 − 34 = −3

6 42 40 42 − 40 = 2

7 45 46 45 − 46 = −1

8 52 52 52 − 52 = 0

9 62 58 62 − 58 = 4

x

−4

−2

4 6 82

2

4residual

The points are evenly dispersed about the horizontal axis. So, the equation y = 6x + 4 is a good " t.

7.

x yy-Value

from model Residual

−8 9 11.4 9 − 11.4 = −2.4

−6 10 8.8 10 − 8.8 = 1.2

−4 5 6.2 5 − 6.2 = −1.2

−2 8 3.6 8 − 3.6 = 4.4

0 −1 1 −1 − 1 = −2

2 1 −1.6 1 − (−1.6) = 2.6

4 −4 −4.2 −4 − (−4.2) = 0.2

6 −12 −6.8 −12 − (−6.8) = −5.2

8 −7 −9.4 −7 − (−9.4) = 2.4

x

4

−4

84−8

residual

The points are evenly dispersed about the horizontal axis. So, the equation y = −1.3x + 1 is a good " t.

8.

x yy-Value

from model Residual

4 −1 −4 −1 − (−4) = 3

6 −3 −5 −3 − (−5) = 2

8 −6 −6 −6 − (−6) = 0

10 −8 −7 −8 − (−7) = −1

12 −10 −8 −10 − (−8) = −2

14 −10 −9 −10 − (−9) = −1

16 −10 −10 −10 − (−10) = 0

18 −9 −11 −9 − (−11) = 2

20 −9 −12 −9 − (−12) = 3

x

−2

8 12 16 204

2

4residual

The residual points form a ∪-shaped pattern, which suggests the data are not linear. So, the equation y = −0.5x − 2 does not model the data well.


Chapter 4

9.

x yy-Value

from model Residual

1 6 6.1 6 − 6.1 = −0.1

2 5.5 5.4 5.5 − 5.4 = 0.1

3 4.7 4.7 4.7 − 4.7 = 0

4 3.9 4 3.9 − 4 = −0.1

5 3.3 3.3 3.3 − 3.3 = 0

x

−0.1

42

0.1

residual

The points are evenly dispersed about the horizontal axis. So, the equation y = −0.7x + 6.8 is a good " t.

10.

x yy-Value

from model Residual

1 27 28.3 −1.3

2 28 29.6 −1.6

3 36 30.9 5.1

4 28 32.2 −4.2

5 32 33.5 −1.5

6 35 34.8 0.2

x

−4

4 6

4

residual

Most of the residual values are below the horizontal axis. So, the equation y = 1.3x + 27 does not model the data well.

11. After entering the data from the table into two lists using a graphing calculator, the linear regression feature yields the equation y = 2.1x − 7.5. The correlation coef" cient is about 0.980. This means that the relationship between the independent and dependent variables has a strong positive correlation and the equation closely models the data.

12. After entering the data from the table into two lists using a graphing calculator, the linear regression feature yields the equation y = −1.35x + 7.8. The correlation coef" cient is about −0.886. This means that the relationship between the independent and dependent variables is a reasonably strong negative correlation and the equation models the data well.

13. After entering the data from the table into two lists using a graphing calculator, the linear regression feature yields the equation y = 1.38x + 15.7. The correlation coef" cient is about 0.999. This means that the relationship between the independent and dependent variables has a strong positive correlation and the equation closely models the data.

14. After entering the data from the table into two lists using a graphing calculator, the linear regression feature yields the equation y = −x + 11.25. The correlation coef" cient is about −0.444. This means that the relationship between the independent and dependent variables has a weak negative correlation and the equation does not model the data well.

15. The slope and y-intercept are reversed. Because the equation is in the form y = ax + b, the slope is a = −4.47 and the y-intercept is b = 23.16. An equation of the line of best " t is y = −4.47x + 23.16.

16. Because the correlation coef" cient is a negative number, the correlation is negative. The correlation coef" cient is about −0.999. So, the data have a strong negative correlation.

17. a. After entering the data from the table into two lists using a graphing calculator, the linear regression feature yields the equation y = 381x − 566.

80

2200

0

b. The correlation coef" cient is about 0.989. This means that there is a strong positive correlation between the number of the minutes after an earthquake ended and the number of people who reported the earthquake. So, the equation closely models the data, as shown in the graph.

c. The slope of the line is 381. This means that the number of people who reported an earthquake increased by about 381 each minute after the earthquake ended. The y-intercept is −565.7, which is meaningless in this context. There cannot be a negative number of people.

18. a. After entering the data from the table into two lists using a graphing calculator, the linear regression feature yields the equation y = x + 7.

100

20

0

b. The correlation coef" cient is about 0.619. This means that the relationship between the day and the number of people who volunteer has a weak positive correlation and the equation does not model the data well.


Chapter 4

c. The slope of the line is 1. This means, the number of volunteers increases by 1 each day. The y-intercept is 6.75, which has no meaning in this context because there is no day 0.

19. a. After entering the data from the table into two lists using a graphing calculator, the linear regression feature yields the equation y = −0.18x + 19.7.

b. The correlation coef" cient is about −0.968. This means that the relationship between the mileage of a used automobile and its selling price has a strong negative correlation, and the equation y = −0.18x + 19.7 closely models the data.

c. The slope of the line is −0.18. This means that the price of a car decreases by about $180 for every thousand miles the car has been driven. The y-intercept of the line is 19.7, which has no meaning in this context because a used car cannot have 0 mileage.

d. y = −0.18x + 19.7

15.5 = −0.18x + 19.7

− 19.7 − 19.7

−4.2 = −0.18x

−4.2 — −0.18

= −0.18x — −0.18

23.3 ≈ x So, a car that costs $15,500 will have about 23,300 miles

on it.

e. y = −0.18x + 19.7

= −0.18 (6) + 19.7

= −1.08 + 19.7

= 18.62

So, a car with 6000 miles will cost about $18,620.

20. a. After entering the data from the table into two lists using a graphing calculator, the linear regression feature yields the equation y = 4.9x − 37.7.

b. The correlation coef" cient is about 0.936. This means that the relationship between the lengths and costs of the sailboats has a strong positive correlation, and the equation y = 4.9x − 37.7 closely models the data.

c. The slope of the line is 4.9. This means that the cost of a sailboat increases by about $4900 per foot of a sailboat’s length. The y-intercept of the line is −37.7. However, the y-intercept is meaningless in this context because a sailboat cannot be 0 feet long, and the cost cannot be negative.

d. y = 4.9x − 37.7

= 4.9(20) − 37.7

= 98 − 37.7

= 60.3

So, a sailboat that is 20 feet long will cost about $60,300.

e. y = 4.9x − 37.7

147 = 4.9x − 37.7

+ 37.7 + 37.7

184.7 = 4.9x

37.7 ≈ x

So, a sailboat that costs $147,000 is about 37.7 feet long.

21. There is a negative correlation and a causal relationship because the more time you spend talking on a cell phone, the more you will drain the remaining battery life.

22. There may be a positive correlation but no causal relationship. Growing taller will not cause a toddler to learn new words, but as a toddler grows, he or she will probably be learning more words.

23. There is no correlation between the number of hats you own and the size of your head because buying more hats will not change the size of your head.

24. There may be a positive correlation but no causal relationship. The weight of a dog does not determine the length of its tail.

25. Sample answer: Each week of the summer season, a shop at the beach recorded the numbers x of ice cream cones sold and the numbers y of surf board rentals. This data set has a strong positive correlation but no causal relationship. Selling more ice cream cones will not cause the number of surf board rentals to increase. However, on the days with good weather in the height of the season, both numbers will likely be high, and on days with bad weather, both numbers will likely be low.

26. a. B; The correlation coef" cient is r = 0.98 because the graph shows data with a strong positive correction that is closely modeled by the line.

b. C; The correlation coef" cient is r = −0.97 because the graph shows data with a strong negative correlation that is closely modeled by the line.

c. A; The correlation coef" cient is r = 0 because the graph shows data with no correlation.

d. D; The correlation coef" cient is r = 0.69 because the graph shows data with a weak positive correlation that is roughly modeled by the line.


Chapter 4

27 a. After entering the data from the table into two lists using a graphing calculator, the linear regression feature yields the equation y = −0.08x + 3.8. The correlation coef" cient is about −0.965. This means that the relationship between grade point average and hours spent watching television has a strong negative correlation.

b. The slope of the line is −0.08. This means that a student’s grade point average decreases by about 0.08 for every hour spent watching television each week. The y-intercept of the line is 3.8. This means that a student who watches no television is likely to have a grade point average of about 3.8.

c. y = −0.08x + 3.8

= −0.08 (14) + 3.8

= −1.12 + 3.8

= 2.68

So, a student who watches 14 hours of television each week will have a grade point average of about 2.68.

d. Sample answer: no; Watching a lot of television does not directly cause a student to get lower grades. However, students who spend a lot of time watching television will probably spend less time studying and therefore earn lower grades.

28. y = −0.08x + 3.8

= −0.08(2) + 3.8

= −0.16 + 3.8

= 3.64

yes; The model predicts that a student who watches 2 hours of television will have a grade point average of about 3.64. Because the actual grade point average, 2.4, is signi" cantly less, it will weaken the correlation between the model and the actual data.

29. a. y = 381x − 566 y = 381x − 566

= 381(9) − 566 = 381(15) − 566

= 3429 − 566 = 5715 − 566

= 2863 = 5149

So, using the equation y = 381x − 566 of the model from Exercise 17, the number of people who reported the earthquake 9 minutes after in ended was about 2863 and the number of people who reported the earthquake 15 minutes after it ended was about 5149.

b. The difference between the " rst extrapolation and the actual data value is only 113. The difference between the second extrapolation and the actual data value is 1949. So, the second prediction was less accurate than the " rst. This " ts with what you would expect because the farther removed a value is from the known values, the less con" dence you can have in the accuracy of the prediction. Sometimes trends change over time.

30. Sample answer:

200 40 60 80 x

80

y

40

0

Beach 1

Beach Attendance for One Week

Beac

h 2

As the number of people at Beach 1 increases, the number of people at Beach 2 also increases. A possible correlation coef" cient is about 0.9 because there is a strong positive correlation between the number of people at the two beaches, and the graph shows that the points could " t closely to a line. This does not represent a causal relationship, however. The factors that affect how many people come to the beaches are things such as the temperature outside and whether it is a weekend or holiday.

31. a. After entering the data from the table into two lists using a graphing calculator, the linear regression feature yields the equation y = 513.5x − 298. The correlation coef" cient is about 0.993. This means that the relationship between the year and the numbers of the messages sent has a strong positive correlation and the equation closely models the data.

b. no; Sample answer: The number of text messages sent does not just go up simply because another year has passed. The number of text messages increases each year because cell phone service gets faster, more ef" cient, and less expensive over time as the cell phone companies work to improve their services.

c.

x yy-Value from

model Residual

1 241 215.5 241 − 215.5 = 25.5

2 601 729 601 − 729 = −128

3 1360 1242.5 1360 − 1242.5 = 117.5

4 1806 1756 1806 − 1756 = 50

5 2206 2269.5 2206 − 2269.5 = −63.5

x

−100

42

100

residual

The points are evenly dispersed about the horizontal axis. So, the equation y = 513.5x − 298 is a good " t.


Chapter 4

d. Sample answer: Using the graphing calculator to calculate a correlation coef" cient is faster than calculating and graphing the residuals. Both methods will reveal how closely a model " ts the actual data. The graph of the residuals sometimes reveals why a linear model is not a good " t, if the data " ts a nonlinear model for instance. The correlation coef" cient, however, will simply reveal that the linear model is not a good " t but does not help determine why.

Maintaining Mathematical Pro! ciency 32. nonlinear; As x increases by 1, the y-value alternates between

4 and −4. The rate of change is not constant. So, the function is not linear.

33. linear; As x increases by 2, y decreases by 5. The rate of change is constant. So, the function is linear.

4.6 Explorations (p. 209) 1. a. Number of stars, n 1 2 3 4 5

Number of sides, y 10 20 30 40 50

0102030405060

0 1 2 3 4 5 n

y

The y-values increase by 10. So, each y-value is 10 times the corresponding value of n.

b. n 1 2 3 4 5Number of circles, y 2 3 4 5 6

0123456

0 1 2 3 4 5 n

y

The y-values increase by 1. So, each y-value is 1 more than the corresponding value of n.

c. Number of rows, n 1 2 3 4 5Number of dots, y 2 4 6 8 10

02468

1012

0 1 2 3 4 5 n

y

The y-values increase by 2 each time a row is added. So, each y-value is twice the corresponding value of n.

2. An arithmetic sequence can describe a pattern in which the difference between consecutive terms is the same. For example, if every time you beat a level of a game, you earn 50 points.

3. The number y of atoms increases by 3 for each additional water molecule. Each y-value is 3 times the corresponding x-value. So, 23 molecules of water have 3(23) = 69 atoms.

4.6 Monitoring Progress (pp. 210–213) 1. Position 1 2 3 4 5 6 7

Term −12 0 12 24 36 48 60

+12 +12 +12

The next three terms are 36, 48, and 60.

2. Position 1 2 3 4 5 6 7Term 0.2 0.6 1 1.4 1.8 2.2 2.6

+0.4 +0.4 +0.4

The next three terms are 1.8, 2.2, and 2.6.

3. Position 1 2 3 4 5 6 7

Term 4 3 3 — 4 3 1 —

2 3 1 —

4 3 2 3 —

4 2 1 —

2

+ ( − 1 — 4 ) + ( − 1 —

4 ) + ( − 1 —

4 )

The next three terms are 3, 2 3 — 4 , and 2 1 — 2 .

4. Position, n 1 2 3 4

Term, an 3 6 9 12

02468

10121416

0 1 2 3 4 5 6 7 n

an

(1, 3)

(2, 6)

(3, 9)

(4, 12)

The points lie on a line.

5. Position, n 1 2 3 4

Term, an 4 2 0 −2

n

an

12345

−3−2−1

4 5 6 7321

(1, 4)

(2, 2)

(4, −2)

(3, 0)


⤻ ⤻

⤻⤻

⤻

⤻

⤻⤻⤻


Chapter 4

6. Position, n 1 2 3 4Term, an 1 0.8 0.6 0.4

00.20.40.60.81.01.21.41.6

0 1 2 3 4 5 6 7 n

an

(1, 1)(2, 0.8)

(3, 0.6)(4, 0.4)


7. Position, n 1 2 3 4Term, an 2 4 7 11

Consecutive terms do not have a common difference. So, the sequence of terms 2, 4, 7, 11, . . . is not arithmetic.

8. an = a1 + (n − 1)d

an = 4 + (n − 1)(1)

an = 4 + n(1) − 1(1)

an = 4 + n − 1 an = n + 3

An equation for the nth term is an = n + 3.

an = n + 3

a25 = 25 + 3

= 28

The 25th term of the arithmetic sequence is 28.

9. an = a1 + (n − 1)d

an = 8 + (n − 1)(8)

an = 8 + n(8) − 1(8)

an = 8 + 8n − 8 an = 8n

An equation for the nth term is an = 8n.

an = 8n

a25 = 8(25)

= 200


10. an = a1 + (n − 1)d

an = 1 + (n − 1)(−1)

an = 1 + n(−1) − 1(−1)

an = 1 − n + 1

an = −n + 2

An equation for the nth term is an = −n + 2.

an = −n + 2

a25 = −25 + 2

= −23

The 25th term of the arithmetic sequence is −23.

11. a. f (n) = a1 + (n − 1)d

f (n) = 7 + (n − 1)(2)

f (n) = 7 + n(2) − 1(2)

f (n) = 7 + 2n − 2 f (n) = 2n + 5

The function f (n) = 2n + 5 represents the arithmetic sequence.

b.

02468

10121416

0 1 2 3 4 5 6 7 8 n

an

Games

Tota

l cos

t (d

olla

rs)

c. f (n) = 2n + 5

29 = 2n + 5

− 5 − 5

24 = 2n

24 — 2 = 2n —

2

12 = n

You can play 12 games.

4.6 Exercises (pp. 214−216)

Vocabulary and Core Concept Check 1. The graph of an arithmetic sequence is a graph of a linear

function whose domain is the set of positive integers.

2. The one that is different is “" nd the difference between the terms a2 and a4.” This answer is 16 − 10 = 6. The answer of all the others is 13 − 10 = 3.

Monitoring Progress and Modeling with Mathematics 3. Position 1 2 3 4

Term 2 15 28 41

+13 +13 +13


4. Position 1 2 3 4Term 18 12 6 0

+(−6) +(−6) +(−6)


5. 18 − 13 = 5, 23 − 18 = 5, 28 − 23 = 5

The common difference is 5.

6. 150 − 175 = −25, 125 − 150 = −25, 100 − 125 = −25

The common difference is −25.

⤻+2

⤻+3

⤻+4

⤻⤻ ⤻

⤻ ⤻⤻


Chapter 4

7. −12 − (−16) = −12 + 16 = 4, −8 − (−12) = −8 + 12 = 4,

−4 − (−8) = −4 + 8 = 4


8. 3 2 — 3 − 4 = 3 2 —

3 − 3 3 —

3 = − 1 —

3 , 3 1 —

3 − 3 2 —

3 = − 1 —

3 ,

3 − 3 1 — 3 = 2 3 —

3 − 2 4 —

3 = − 1 —

3

The common difference is − 1 — 3 .

9. 5 − 6.5 = −1.5, 3.5 − 5 = −1.5, 2 − 3.5 = −1.5

The common difference is −1.5.

10. −7 − (−16) = −7 + 16 = 9, 2 − (−7) = 2 + 7 = 9, 11 − 2 = 9


11. Position 1 2 3 4 5 6 7Term 19 22 25 28 31 34 37

+3 +3 +3


12. Position 1 2 3 4 5 6 7Term 1 12 23 34 45 56 67

+11 +11 +11


13. Position 1 2 3 4 5 6 7Term 16 21 26 31 36 41 46

+5 +5 +5


14. Position 1 2 3 4 5 6 7Term 60 30 0 −30 −60 −90 −120

+(−30) +(−30) +(−30)

The next three terms are −60, −90, and −120.

15. Position 1 2 3 4 5 6 7Term 1.3 1 0.7 0.4 0.1 −0.2 −0.5

+(−0.3) +(−0.3) +(−0.3)

The next three terms are 0.1, −0.2, and −0.5.

16. Position 1 2 3 4 5 6 7

Term 5 — 6 2 —

3 1 —

2 1 —

3 1 —

6 0 − 1 —

6

+ ( − 1 — 6 ) + ( − 1 —

6 ) + ( − 1 —

6 )

The next three terms are 1 — 6 , 0, and − 1 — 6 .


048

121620242832

0 1 2 3 4 5 6 7 n

an

(1, 4)

(2, 12)

(3, 20)

(4, 28)

18. Position, n 1 2 3 4Term, an −15 0 15 30

n

an

1020304050

−10−20−30

4 5 6 732

(3, 15)

(4, 30)

(1, −15)

(2, 0)

19. Position, n 1 2 3 4Term, an −1 −3 −5 −7

0

−8−7−6−5−4−3−2−1

0 1 2 3 4 5 6 7 n

an

(1, −1)

(2, −3)

(3, −5)

(4, −7)


01020304050607080

0 1 2 3 4 5 6 7 n

an

(1, 2)

(2, 19)

(3, 36)

(4, 53)

⤻⤻⤻

⤻⤻⤻

⤻⤻⤻

⤻⤻ ⤻

⤻ ⤻ ⤻

⤻ ⤻⤻


Chapter 4

21. Position, n 1 2 3 4

Term, an 0 4 1 — 2 9 13 1 —

2

02468

10121416

0 1 2 3 4 5 6 7 n

an

(1, 0)

(3, 9)

(4, 13 )12

(2, 4 )12

22. Position, n 1 2 3 4Term, an 6 5.25 4.5 3.75

012345678

0 1 2 3 4 5 6 7 8 n

an

(1, 6)(2, 5.25)

(4, 3.75)

(3, 4.5)

23. The points do not lie on a line. So, the sequence 1, 4, 1, 4, . . . is not arithmetic.


+7 +7 +7

Consecutive terms have a common difference of 7. So, the graph represents the arithmetic sequence 5, 12, 19, 26, . . . .


+(−15) +(−15) +(−15)

Consecutive terms have a common difference of −15. So, the graph represents the arithmetic sequence 70, 55, 40, 25, . . . .


+8 +6 +4

Consecutive terms do not have a common difference, and the points do not lie on a line. So, the sequence 2, 10, 16, 20, . . . is not arithmetic.

27. Position 1 2 3 4Term 13 26 39 52

+13 +13 +13

The sequence is arithmetic with a common difference of 13.

28. Position 1 2 3 4Term 5 9 14 20

+4 +5 +6

Consecutive terms do not have a common difference. So, the sequence is not arithmetic.

29. Position 1 2 3 4Term 48 24 12 6

+(−24) +(−12) +(−6)

Consecutive terms do not have a common difference. So, the sequence is not arithmetic.

30. Position 1 2 3 4Term 87 81 75 69

+(−6) +(−6) +(−6)

The sequence is arithmetic with a common difference of −6.

31. Position 1 2 3 4Term 4 6 8 10

+2 +2 +2

Consecutive terms have a common difference of 2. So, the sequence 4, 6, 8, 10, . . . is arithmetic.

32. Position 1 2 3 4Term 2 5 8 11

+3 +3 +3

Consecutive terms have a common difference of 3.So, the sequence 2, 5, 8, 11, . . . is arithmetic.

33. d = −4 − (−5) = −4 + 5 = 1

an = a1 + (n − 1)d

an = −5 + (n − 1)(1)

an = −5 + n(1) − 1(1)

an = −5 + n − 1

an = n − 6

An equation for the nth term of the arithmetic sequence is an = n − 6.

an = n − 6 a10 = 10 − 6 = 4


⤻ ⤻⤻

⤻ ⤻⤻

⤻ ⤻⤻

⤻ ⤻⤻

⤻ ⤻⤻

⤻ ⤻⤻

⤻ ⤻⤻

⤻ ⤻⤻

⤻ ⤻⤻


Chapter 4

34. d = −9 − (−6) = −9 + 6 = −3

an = a1 + (n − 1)d

an = −6 + (n − 1)(−3)

an = −6 + n(−3) − 1(−3)

an = −6 − 3n + 3

an = −3n − 3 An equation for the nth term of the arithmetic sequence

is an = −3n − 3.

an = −3n − 3 a10 = −3(10) − 3 = −30 − 3 = −33


35. d = 1 − 1 — 2

= 1 — 2

an = a1 + (n − 1)d

an = 1 — 2 + (n − 1) ( 1 —

2 )

an = 1 — 2 + n ( 1 —

2 ) − 1 ( 1 —

2 )

an = 1 — 2 + 1 —

2 n − 1 —

2

an = 1 — 2 n

An equation for the nth term of the arithmetic sequence is an = 1 — 2 n.

an = 1 — 2 n

a10 = 1 — 2 (10)

= 5


36. d = 110 − 100 = 10

an = a1 + (n − 1)d

an = 100 + (n − 1)(10)

an = 100 + n(10) − 1(10)

an = 100 + 10n − 10

an = 10n + 90

An equation for the nth term of the arithmetic sequence is an = 10n + 90.

an = 10n + 90

a10 = 10(10) + 90

= 100 + 90

= 190


37. d = 0 − 10 = −10

an = a1 + (n − 1)d

an = 10 + (n − 1)(−10)

an = 10 + n(−10) − 1(−10)

an = 10 − 10n + 10

an = −10n + 20

An equation for the nth term of the arithmetic sequence is an = −10n + 20.

an = −10n + 20

a10 = −10(10) + 20

= −100 + 20

= −80


38. d = 4 — 7 − 3 —

7 = 1 —

7

an = a1 + (n − 1)d

an = 3 — 7 + (n − 1) ( 1 —

7 )

an = 3 — 7 + n ( 1 —

7 ) − 1 ( 1 —

7 )

an = 3 — 7 + 1 —

7 n − 1 —

7

an = 1 — 7 n + 2 —

7

An equation for the nth term of the arithmetic sequence is an = 1 — 7 n + 2 — 7 .

an = 1 — 7 n + 2 —

7

a10 = 1 — 7 (10) + 2 —

7

= 10 — 7 + 2 —

7

= 12 — 7 = 1 5 —

7

The 10th term of the arithmetic sequence is 1 5 — 7 .

39. Because consecutive terms are decreasing by 1, the common difference should be negative.

2, 1, 0, −1, . . .

+(−1) +(−1) +(−1)

The common difference is −1.

⤻ ⤻ ⤻


Chapter 4

40. The equation an = a1 + (n − 1)d should be used to write an equation for the nth term of the arithmetic sequence.

14, 22, 30, 38, . . .

d = 22 − 14 = 8

an = a1 + (n − 1)d

an = 14 + (n − 1)(8)

an = 14 + n(8) − 1(8)

an = 14 + 8n − 8

an = 8n + 6

An equation for the nth term of the arithmetic sequence is an = 8n + 6.

41. a1 = 3

d = 1.5(3) = 4.5

Position, n 1 2 3 4Term, an 3 7.5 12 16.5

The next three terms are 7.5, 12, and 16.5.

0369

1215182124

0 1 2 3 4 5 6 7 8 n

an

(1, 3)

(2, 7.5)

(3, 12)

(4, 16.5)

42. Position, n 1 2 3 4 5Term, an 10 12 14 16 18

The " rst " ve terms are 10, 12, 14, 16, and 18.

0369

1215182124

0 1 2 3 4 5 6 7 8 n

an

(1, 10)(2, 12)

(3, 14)(4, 16)

(5, 18)

43. a. The next three " gures are:

b. The 20th " gure in the sequence is a regular polygon with 22 sides.

44. a. The next three " gures are:

b. The 20th " gure in the sequence is a circle with 40 equal sections.

45. a. d = 10 − 5 = 5

f (n) = a1 + (n − 1)d

f (n) = 5 + (n − 1)(5)

f (n) = 5 + n(5) − 1(5)

f (n) = 5 + 5n − 5 f (n) = 5n

A function that represents the arithmetic sequence is f (n) = 5n.

b.

0369

1215182124

0 1 2 3 4 5 6 7 8 n

an

Minutes after Midnight January 1st

Tota

l bab

ies

born

(1, 5)

(2, 10)

(3, 15)

(4, 20)

c. f (n) = 5n

100 = 5n

100 — 5 = 5n —

5

20 = n

It takes about 20 minutes for 100 babies to be born.


Chapter 4

46. a. d = 40 − 48 = −8

an = a1 + (n − 1)d

an = 56 + (n − 1)(−8)

an = 56 + n (−8) − 1(−8)

an = 56 − 8n + 8

an = −8n + 64

A function that represents the arithmetic sequence is an = −8n + 64.

b. an = −8n + 64

16 = −8n + 64

− 64 − 64

−48 = −8n

−48 — −8 = −8n —

−8

6 = n

The movie earns $16 million in week 6.

c. 56 + 48 + 40 + 32 + 24 + 16 + 8 = 224

The movie earns $224 million overall.

47. Position, n 1 2 3Term, an 1 5 9

+4 +4

f (n) = a1 + (n − 1)d

f (n) = 1 + (n − 1)(4)

f (n) = 1 + n(4) − 1(4)

f (n) = 1 + 4n − 4 f (n) = 4n − 3 f (30) = 4(30) − 3 = 120 − 3 = 117

Because the areas of consecutive " gures have a common difference of 4, the areas form an arithmetic sequence 1, 5, 9, . . . . A function that represents the arithmetic sequence is f (n) = 4n − 3. The 30th " gure has an area of 117 square inches.

48. Position, n 1 2 3Term, an 4 9 25

+5 +16

Because the sequence of areas 4, 9, 25, . . . does not have a common difference, the sequence is not arithmetic.

49. The domain and range of an arithmetic sequence are both discrete because they consist of individual numbers, not a range of numbers.

50. no; If the " rst term of the sequence is negative and the output values constantly increase, or if the " rst term of the sequence is positive and the output values constantly decrease, the range contains positive and negative numbers.

51. Sample answer:

Position 1 2 3 4Term 6 3 0 −3

+(−3) +(−3) +(−3)

an = a1 + (n − 1)d An arithmetic sequence with a common difference of −3 is 6, 3, 0, −3, . . . . An equation for the nth term is an = −3n + 9.

an = 6 + (n − 1)(−3)

an = 6 + n(−3) − 1(−3)

an = 6 − 3n + 3

an = −3n + 9

Position 1 2 3 4Term −5 −8 −11 −14

+(−3) +(−3) +(−3)

an = a1 + (n − 1)d Another arithmetic sequence with a common difference of −3 is −5, −8, −11, −14, . . . . An equation for the nth term is an = −3n − 2.

an = −5 + (n − 1)(−3)

an = −5 + n(−3) − 1(−3)

an = −5 −3n + 3

an = −3n − 2

52. Sample answer: An algebra class had 22 students at the beginning of the term, and then after the " rst week, one new student joined the class each week for 5 more weeks. The sequence of how many students were in the class at the end of each of the " rst 6 weeks, 22, 23, 24, 25, 26, 27, is arithmetic with a common difference of 1.

53. 20 + 19 + 18 + 17 + 16 + 15 + 14 = 119

The pile has 119 logs in total.

54. a. Size 1 2 3 4Cost 20,000 32,500 45,000 57,500

+12,500 +12,500 +12,500

yes; Consecutive terms have a common difference of $12,500. So, the sequence 20,000, 32,500, 45,000, 57,500, . . . is arithmetic.

b. Sample answer: You could write an equation for the nth term in the sequence and then " nd a6.

an = a1 + (n − 1)d an = 12,500n − 7500

an = 20,000 + (n − 1)(12,500) a6 = 12,500(6) − 7500

an = 20,000 + n(12,500) − 1(12,500) = 75,000 − 7500

an = 20,000 + 12,500n − 12,500 = 67,500

an = 12,500n − 7500 A six-page An equation of the nth term advertisement would of the arithmetic sequence is cost $67,500.an = 12,500n − 7500.

⤻ ⤻

⤻ ⤻

⤻ ⤻⤻

⤻ ⤻⤻

⤻ ⤻⤻


Chapter 4

55. d = 41 − 23 — 4 −1

= 18 — 3 = 6

Position 1 2 3 4Term 23 29 35 41

+6 +6 +6

f (n) = a1 + (n − 1)d

f (n) = 23 + (n − 1)(6)

f (n) = 23 + n (6) − 1(6)

f (n) = 23 + 6n − 6

f (n) = 6n + 17

A function that represents the arithmetic sequence is f (n) = 6n + 17.

56.

Stop number

1 2 3 4 5 6 7 8 9

Time 6:00 6:12 6:24 6:36 6:48 7:00 7:12 7:24 7:36

+12 +12 +12 +12 +12 +12 +12 +12

7:36 − 7:29 = :07

You must wait 7 minutes for the train.

57. a. 3x + 6 − (x + 6)

= 3x + 6 − x − 6

= 2x

5x + 6 − (3x + 6)

= 5x + 6 − 3x − 6

= 2x

7x + 6 − (5x + 6)

= 7x + 6 − 5x − 6

= 2x

yes; Because consecutive terms have a common difference of 2x, the sequence is arithmetic.

b. 3x + 1 − (x + 1)

= 3x + 1 − x − 1

= 2x

9x + 1 − (3x + 1)

= 9x + 1 − 3x − 1

= 6x

27x + 1 − (9x + 1)

= 27x + 1 − 9x − 1

= 18x

no; Because consecutive terms do not have a common difference, the sequence is not arithmetic.

Maintaining Mathematical Pro! ciency 58. x + 8 ≥ −9

− 8 − 8

x ≥ −17

The solution is x ≥ −17.

−30 −25 −20 −15 −10 −5 0 5 10

59. 15 < b − 4

+ 4 + 4

19 < b

The solution is b > 19.

−10 −5 0 5 10 15 20 25 30

60. t − 21 < − 12

+ 21 + 21

t < 9

The solution is t < 9.

420 6 8 10 12 14 16

61. 7 + y ≤ 3

− 7 − 7

y ≤ −4

The solution is y ≤ −4.

−8 −7 −6 −5 −4 −3 −2 −1 0

62. x −2 −1 0 1 2h(x) 6 3 0 3 6

x

y

45678

3

4321−2−1−3−4

h(x) = 3*x*

The function h is of the form y = a f (x), where a = 3. So, the graph of h is a vertical stretch of the graph by a factor of 3. The domain is all real numbers. The range is y ≥ 0.

⤻ ⤻⤻

⤻ ⤻ ⤻⤻⤻⤻ ⤻⤻


Chapter 4

63. x 3 4 5 6 7v(x) 2 1 0 1 1

x

y

45678

23

1

4 5 6 7 8321

v(x) = *x − 5*

The function v is of the form y = f (x − h), where h = 5. So, the graph of v is a horizontal translation 5 units right of the graph of f.

The domain is all real numbers. The range is y ≥ 0.

64. x −2 −1 0 1 2

g(x) 3 2 1 2 3

x

y

45678

3

4321−2−1−3−4

g(x) = *x* + 1

The function g is of the form y = f (x) + k, where k = 1. So, the graph of g is a vertical translation 1 unit up of the graph of f.

The domain is all real numbers. The range is y ≥ 1.

65. x −2 −1 0 1 2

r(x) −4 −2 0 −2 −4

x

y

−3−4−5−6−7−8

4321−2−1−3−4

r(x) = −2*x*

The function r is of the form y = −a f (x), where a = 2. So, the graph of r is a vertical stretch of the graph of f by a factor of 2 and a re# ection in the x-axis.

The domain is all real numbers. The range is y ≤ 0.

4.7 Explorations (p. 217) 1. a. yes; No vertical line can be drawn through more than one

point on the graph. So, each x-value is paired with exactly one y-value, and the graph represents y as a function of x.

b. If x = 0, then y = 0 because there is a closed dot at (0, 0).

c. Use (−2, 2) and (−1, 1).

m = 1 − 2 — −1 − (−2)

= 1 − 2 — −1 + 2

= −1 — 1 , or −1

y − y1 = m(x − x1)

y − 1 = −1(x − (−1))

y − 1 = −1(x + 1)

y − 1 = −1(x) − 1(1)

y − 1 = −x − 1 + 1 + 1

y = −x

So, f (x) = −x, if x ≤ 0.

d. Use (1, 2) and (2, 2).

m = 2 − 2 — 2 − 1

= 0 — 1 , or 0

y − y1 = m(x − x1)

y − 2 = 0(x − 1)

y − 2 = 0

+ 2 + 2

y = 2

So, f (x) = 2, if x > 0.

e. f (x) = { −x, if x ≤ 0 2, if x > 0

2. a. yes; No vertical line can be drawn through more than one point on the graph. So, each x-value is paired with exactly one y-value, and the graph represents y as a function of x.

b. f (x) = { −2, if −6 ≤ x < −3 0, if −3 ≤ x < 0

2, if 0 ≤ x < 3 4, if 3 ≤ x < 6

3. Sample answer: In order to describe a function that is represented by more than one equation, you can write an equation for each portion and give the range of domain values for which the equation describes the graph. Combine all of the equations and corresponding ranges of domain values into one function with a bracket { symbol.


Chapter 4

4. For x ≤ 0, use (−2, 2) and (−1, 1).

m = 1 − 2 — −1 − (−2)

= 1 − 2 — −1 + 2

= −1 — 1 , or −1

y − y1 = m(x − x1)

y − 2 = −1[x − (−2)]

y − 2 = −1(x + 2)

y − 2 = −1(x) − 1(2)

y − 2 = −x − 2

+ 2 + 2

y = −x

For x > 0, use (1, 1) and (2, 2).

m = 2 − 1 — 2 − 1

= 1 — 1 , or 1

y − y1 = m(x − x1)

y − 1 = 1(x − 1)

y − 1 = 1(x) − 1(1)

y − 1 = x − 1

+ 1 + 1

y = x

So, the graph can be represented by the function

f (x) = { −x, if x ≤ 0 x, if x > 0 .

4.7 Monitoring Progress (pp. 218–221) 1. f (x) = 3

f (−8) = 3

The value of f is 3 when x = −8.

2. f (x) = x + 2 f (−2) = −2 + 2 f (−2) = 0 The value of f is 0 when x = −2.

3. f (x) = x + 2 f (0) = 0 + 2 f (0) = 2 The value of f is 2 when x = 0.



6. f (x) = 4x f (10) = 4(10) f (10) = 40 The value of f is 40 when x = 10.

7. Closed Open

x −3 −1 0 0 1 2

y −2 0 1 0 −1 −2

x

y21

−3−4−5−6

−2

4321−3−4

The domain is all real numbers. The range is y ≤ 1.

8. Open Closed

x −2 −1 0 0 0.5 1

y −4 −3 −2 0 2 4

x

y8

46

2

−6−8

−4

4321−2−1−3−4

The domain is all real numbers. The range is y < −2 or y ≥ 0.

9. For x ≤ 0, use (−2, 1) and (−1, 0).

m = 0 − 1 — −1 − (−2)

= 0 − 1 — −1 + 2

= −1 — 1 , or −1

y − y1 = m(x − x1)

y − 0 = −1[x − (−1)]

y = −1(x + 1)

y = −1(x) − 1(1)

y = −x − 1

For x > 0, use (1, 3) and (2, 4).

m = 4 − 3 — 2 − 1

= 1 — 1 , or 1

y − y1 = m(x − x1)

y − 3 = 1(x − 1)

y −3 = 1(x) − 1(1)

y − 3 = x − 1

+ 3 + 3

y = x + 2

So, a piecewise function for the graph is

f (x) = { −x − 1, if x ≤ 0 x + 2, if x > 0 .


Chapter 4

10. For x ≤ −2, use (−4, 2) and (−3, 1).

m = 1 − 2 — −3 − (−4)

= 1 − 2 — −3 + 4

= −1 — 1 , or −1

y − y1 = m(x − x1)

y − 2 = −1[x − (−4)]

y − 2 = −1(x + 4)

y − 2 = −1(x) − 1(4)

y − 2 = −x − 4

+ 2 + 2

y = −x − 2

For −2 < x < 1, y = 2.

For x ≥ 1, use (2, 1) and (3, 3).

m = 3 − 1 — 3 − 2

= 2 — 1 , or 2

y − y1 = m(x − x1)

y − 1 = 2(x − 2)

y − 1 = 2(x) − 2(2)

y − 1 = 2x − 4

+ 1 + 1

y = 2x − 3


f (x) =

{ −x − 2, if x ≤ −2 2, if −2 < x < 1. 2x − 3, if x ≥ 1

11. Number of days

Total cost (dollars)

0 < x ≤ 1 100

1 < x ≤ 2 150

2 < x ≤ 3 200

3 < x ≤ 4 250

f (x) = { 100, if 0 < x ≤ 1 150, if 1 < x ≤ 2

200, if 2 < x ≤ 3 250, if 3 < x ≤ 4

Wood Chipper Rental

0

100

200

300

0 1 2 3 4 x

y

Number of days

Tota

l cos

t (d

olla

rs)

12 a. g(x) = a ∣ x − h ∣ + k

g(x) = a ∣ x − 5 ∣ + 4

0 = a ∣ 3 − 5 ∣ + 4

0 = a ∣ −2 ∣ + 4

0 = a(2) + 4

0 = 2a + 4

− 4 − 4

−4 = 2a

−4 — 2 = 2a —

2

−2 = a

So, the function g(x) = −2 ∣ x − 5 ∣ + 4 represents the path of the reference beam.

b. g(x) = −2 ∣ x − 5 ∣ + 4

g(x) = { −2[−(x − 5)] + 4, if x − 5 < 0 −2(x − 5) + 4, if x − 5 ≥ 0

g(x) = −2[−(x − 5)] + 4, if x − 5 < 0

+ 5 + 5

g(x) = −2(−x + 5) + 4,

g(x) = −2(−x) − 2(5) + 4,

g(x) = 2x − 10 + 4,

g(x) = 2x − 6, if x < 5

g(x) = −2(x − 5) + 4, if x − 5 ≥ 0

+ 5 + 5

g(x) = −2(x) − 2(−5) + 4,

g(x) = −2x + 10 + 4,

g(x) = −2x + 14, if x ≥ 5

So, a piecewise function for g(x) = −2 ∣ x − 5 ∣ + 4 is

g(x) = { 2x − 6, if x < 5 −2x + 14, if x ≥ 5.


Vocabulary and Core Concept Check 1. Sample answer: A piecewise function is a function de" ned

by two or more equations. Each “piece” of the function applies to a different part of its domain. A step function is a type of a piecewise function, speci" cally one that is de" ned by a constant value over each part of its domain.


Chapter 4

2. Sample answer:

x

y

4

2

4f(x) = *x − 2*

The graph of an absolute function, such as the graph of f (x) = ∣ x − 2 ∣ shown, is symmetric about the line x = h, or x = 2 in this example. So, the graph can be “split” at x = 2, and a piecewise function can have one equation de" ne the graph for x < 2 and another equation de" ne the graph for x ≥ 2.

Monitoring Progress and Modeling with Mathematics 3. f (x) = 5x − 1

f (−3) = 5(−3) − 1

f (−3) = −15 − 1

f (−3) = −16

The value of f is −16 when x = −3.

4. f (x) = x + 3 f (−2) = −2 + 3 f (−2) = 1

The value of f is 1 when x = −2.

5. f (x) = x + 3 f (0) = 0 + 3 f (0) = 3

The value of f is 3 when x = 0.

6. f (x) = x + 3 f (5) = 5 + 3 f (5) = 8


7. g(x) = −x + 4 g(−4) = −(−4) + 4 g(−4) = 4 + 4 g(−4) = 8

The value of g is 8 when x = −4.

8. g(x) = −x + 4 g(−1) = −(−1) + 4 g(−1) = 1 + 4 g(−1) = 5

The value of g is 5 when x = −1.

9. g(x) = 3

g(0) = 3

The value of g is 3 when x = 0.

10. g(x) = 3

g(1) = 3


11. g(x) = 2x − 5

g(2) = 2(2) − 5

g(2) = 4 − 5

g(2) = −1

The value of g is −1 when x = 2.

12. g(x) = 2x − 5

g(5) = 2(5) − 5

g(5) = 10 − 5

g(5) = 5


13. Because 2 < 4 ≤ 5, use g(x) = 65x − 20.

g(x) = 65x − 20

g(4) = 65(4) − 20

g(4) = 260 − 20

g(4) = 240

You travel 240 miles in 4 hours on a trip.

14. Because 25 ≤ 26 < 50, use c(x) = 15.80x + 20.

c(x) = 15.80x + 20

c(26) = 15.80(26) + 20

c(26) = 410.8 + 20

c(26) = 430.8

The total cost of ordering 26 shirts is $430.80.

15. Open Closed

x −4 0 2 2 4 6

y 4 0 −2 −4 −2 0

x

y21

−3−4−5−6

−2

4 5 6321−2−1

The domain is all real numbers.

The range is y ≥ −4.


Chapter 4

16. Closed Open

x −5 −4 −3 −3 −2 0

y −10 −8 −6 6 4 0

x

y16

8

−16

−8

4−4−8−12


The range is y < 6.

17. x −3 −2 −1 0 1

y 7 4 1 2 3

x

y

45678

3

1

4 5 6321−2−1


The range is y ≥ 1.

18. x 0 2 4 5 6

y 8 10 12 16 20

x

y

16

24

32

8 124−4


The range is all real numbers.

19. Open Closed Closed Open

x −4 −3 −3 0 1 3 3 4 5

y 1 1 −4 −1 0 2 −2 −4 −6

x

y4

−8

−12

−4

84−4−8


The range is y ≤ 2.

20. Closed Open Open Closed

x −3 −2 −1 −1 0 1 2 2 3 4

y −5 −3 −1 3 2 1 0 −3 −3 −3

x

y4

21

−3−4

−2

4321−2−3−4


The range is y ≤ −1 or 0 < y < 3.

21. Because 5 ≥ 5, when x = 5, f (x) = x + 8 should be used.

f (x) = x + 8 f (5) = 5 + 8 f (5) = 13


22. Because x ≤ −2 and x > −2, there should be a closed dot at (−2, 4) and an open dot at (−2, 1).

x

y

456

23

−2

21−2−1−3−4−5


Chapter 4

23. For x < 0, use (−2, 0) and (−1, 1).

m = 1 − 0 — −1 − (−2)

= 1 − 0 — −1 + 2

= 1 — 1 , or 1

y − 0 = 1[x − (−2)]

y = 1(x + 2)

y = 1(x) + 1(2)

y = x + 2

For x ≥ 0, y = 2.


f (x) = { x + 2, if x < 0 2, if x ≥ 0 .

24. For x ≤ 0, y = −3.

For x > 0, use (1, 0) and (2, −3).

m = −3 − 0 — 2 − 1

= −3 — 1 , or −3

y − y1 = m(x − x1)

y − 0 = −3(x − 1)

y = −3(x) − 3(−1)

y = −3x + 3


f (x) = { −3, if x ≤ 0 −3x + 3, if x > 0 .

25. For x < 4, use (2, −2) and (3, −3).

m = −3 − (−2) — 3 − 2

= −3 + 2 — 3 − 2

= −1 — 1 , or −1

y − y1 = m(x − x1)

y − (−2) = −1(x − 2)

y + 2 = −1(x) − 1(−2)

y + 2 = −x + 2

− 2 − 2

y = −x

For x ≥ 4, use (5, −4) and (6, −5).

m = −5 − (−4) — 6 − 5

= −5 + 4 — 6 − 5

= −1 — 1 , or −1

y − y1 = m(x − x1)

y − (−4) = −1(x − 5)

y + 4 = −1(x − 5)

y + 4 = −1(x) −1(−5)

y + 4 = −x + 5

− 4 − 4

y = −x + 1


f (x) = { −x, if x < 4 −x + 1, if x ≥ 4 .

26. For x ≤ −2, use (−4, −6) and (−3, −4).

m = −4 − (−6) — −3 − (−4)

= −4 + 6 — −3 + 4

= 2 — 1 , or 2

y − y1 = m(x − x1)

y − (−4) = 2[x − (−3)]

y + 4 = 2(x + 3)

y + 4 = 2(x) + 2(3)

y + 4 = 2x + 6

− 4 − 4

y = 2x + 2

For x > 2, use (0, −1) and (2, 0).

m = 0 − (−1) — 2 − 0

= 0 + 1 — 2 − 0

= 1 — 2

y − y1 = m(x − x1)

y − 0 = 1 — 2 (x − 2)

y = 1 — 2 (x) − 1 —

2 (2)

y = 1 — 2 x − 1


f (x) = { 2x + 2, if x ≤ −2 1 — 2 x − 1, if x > 2

.

27. For x ≤ −2, y = 1.

For −2 < x ≤ 0, use (−1, −2) and (0, 0).

m = 0 − (−2) — 0 − (−1)

= 0 + 2 — 0 + 1

= 2 — 1 , or 2

y − y1 = m(x − x1)

y − 0 = 2(x − 0)

y = 2(x)

y = 2x

For x > 0, use (2, 1) and (4, 0).

m = 0 − 1 — 4 − 2

= −1 — 2

y − y1 = m(x − x1)

y − 0 = − 1 — 2 (x − 4)

y = − 1 — 2 (x) − 1 —

2 (−4)

y = − 1 — 2 x + 2


f (x) = { 1, if x ≤ −2 2x, if −2 < x ≤ 0

− 1 — 2 x + 2, if x > 0

.


Chapter 4

28. For x ≤ −1, use (−3, 1) and (−2, 2).

m = 2 − 1 — −2 − (−3)

= 2 − 1 — −2 + 3

= 1 — 1 , or 1

y − y1 = m(x − x1)

y − 2 = 1[x − (−2)]

y − 2 = 1(x + 2)

y − 2 = 1(x) + 1(2)

y − 2 = x + 2

+ 2 + 2

y = x + 4

For −1 < x < 3, use the open dots at (−1, 0) and (3, −1).

m = −1 − 0 — 3 − (−1)

= −1 − 0 — 3 + 1

= −1 — 4

y − y1 = m(x − x1)

y − 0 = − 1 — 4 [x − (−1)]

y = − 1 — 4 (x + 1)

y = − 1 — 4 (x) − 1 —

4 (1)

y = − 1 — 4 x − 1 —

4

For x ≥ 3, y = −3.


f (x) =

{ x + 4, if x ≤ −1 − 1 —

4 x − 1 —

4 , if −1 < x < 3

−3, if x ≥ 3 .

29. A step function for the graph is

f (x) = { −5, if −5 ≤ x < −3 −3, if −3 ≤ x < −1

−1, if −1 ≤ x < 1 .

30. A step function for the graph is f (x) = { 4, if 0 < x ≤ 1 3, if 1 < x ≤ 2

2, if 2 < x ≤ 3 1, if 3 < x ≤ 4 .

31.

x

y

45678

23

1

4 5 6 7 8321

The domain is 0 ≤ x < 8.

The range is 3, 4, 5, and 6.

32. x

y

−8

−12

−16

−4

4 5 6 7 8321

The domain is 1 < x ≤ 5.

The range is −10, −8, −6, and −4.

33.

x

y

8

12

16

4

8 12 164

The domain is 1 < x ≤ 12.

The range is 1, 5, 6, and 9.

34.

x

y4

2

−4

−2

−2−4−6−8

The domain is −6 ≤ x < 0.

The range is −2, −1, 0, and 1.


Chapter 4

35. Number of team

members

Total cost (dollors)

f (x) = { 180, if 0 < x ≤ 5 210, if 5 < x ≤ 6

240, if 6 < x ≤ 7 270, if 7 < x ≤ 8 300, if 8 < x ≤ 9

0 < x ≤ 5 180

5 < x ≤ 6 210

6 < x ≤ 7 240

7 < x ≤ 8 270

8 < x ≤ 9 300

0

50

100

150

200

250

300

0 2 4 6 8 10 12 p

y

Number of people on team

Tota

l cos

t (d

olla

rs)

36. Number of hours

Total cost (dollors)

0 < x ≤ 1 4

f (x) = { 4, if 0 < x ≤ 1 8, if 1 < x ≤ 2

12, if 2 < x ≤ 3 15, if 3 < x ≤ 24 1 < x ≤ 2 8

2 < x ≤ 3 12

3 < x ≤ 24 15

0

4

8

12

16

0 4 8 12 16 20 24 x

y

Time parked (hours)

Tota

l cos

t (d

olla

rs)

37. A piecewise function for y = ∣ x ∣ + 1 is

g(x) = { −x + 1, if x < 0 x + 1, if x ≥ 0 .

38. A piecewise function for y = ∣ x ∣ − 3 is

g(x) = { −x − 3, if x < 0 x − 3, if x ≥ 0 .

39. g(x) = { −(x − 2), if x − 2 < 0 x − 2, if x − 2 ≥ 0

g(x) = −(x − 2), if x − 2 < 0 g(x) = x − 2, if x − 2 ≥ 0 + 2 + 2 + 2 + 2

g(x) = −x + 2, if x < 2 g(x) = x − 2, if x ≥ 2 So, a piecewise function for y = ∣ x − 2 ∣ is

g(x) = { −x + 2, if x < 2 x − 2, if x ≥ 2 .

40. g(x) = { −(x + 5), if x + 5 < 0 x + 5, if x + 5 ≥ 0

g(x) = −(x + 5), if x + 5 < 0 g(x) = x + 5, if x + 5 ≥ 0 − 5 − 5 − 5 − 5

g(x) = −x − 5, if x < −5 g(x) = x + 5, if x ≥ −5

So, a piecewise function for y = ∣ x + 5 ∣ is

g(x) = { −x − 5, if x < −5 x + 5, if x ≥ −5 .

41. g(x) = { 2[−(x + 3)], if x + 3 < 0 2(x + 3), if x + 3 ≥ 0

g(x) = 2[−(x + 3)], if x + 3 < 0

g(x) = 2(−x − 3), − 3 − 3

g(x) = 2(−x) − 2(3),

g(x) = −2x − 6, if x < −3

g(x) = 2(x + 3), if x + 3 ≥ 0

g(x) = 2(x) + 2(3), − 3 − 3

g(x) = 2x + 6, if x ≥ −3

So, a piecewise function for y = 2 ∣ x + 3 ∣ is

g(x) = { −2x − 6, if x < −3 2x + 6, if x ≥ −3 .

42. g(x) = { 4[−(x − 1)], if x − 1 < 0 4(x − 1), if x − 1 ≥ 0

g(x) = 4[−(x − 1)], if x − 1 < 0

g(x) = 4(−x + 1), + 1 + 1

g(x) = 4(−x) + 4(1),

g(x) = −4x + 4, if x < 1

g(x) = 4(x − 1), if x − 1 ≥ 0

g(x) = 4(x) − 4(1), + 1 + 1

g(x) = 4x − 4, if x ≥ 1

So, a piecewise function for y = 4 ∣ x − 1 ∣ is

g(x) = { −4x + 4, if x < 1 4x − 4, if x ≥ 1 .


Chapter 4

43. g(x) = { −5[−(x − 8)], if x − 8 < 0 −5(x − 8), if x − 8 ≥ 0

g(x) = −5[−(x − 8)], if x − 8 < 0

g(x) = −5(−x + 8), + 8 + 8

g(x) = −5(−x) − 5(8),

g(x) = 5x − 40, if x < 8

g(x) = −5(x − 8), if x − 8 ≥ 0

g(x) = −5(x) − 5(−8), + 8 + 8

g(x) = −5x + 40, if x ≥ 8

So, a piecewise function for y = −5 ∣ x − 8 ∣ is

g(x) = { 5x − 40, if x < 8 −5x + 40, if x ≥ 8 .

44. g(x) = { −3[−(x + 6)], if x + 6 < 0 −3(x + 6), if x + 6 ≥ 0

g(x) = −3[−(x + 6)], if x + 6 < 0

g(x) = −3(−x − 6), − 6 − 6 g(x) = −3(−x) −3(−6),

g(x) = 3x + 18, if x < −6

g(x) = −3(x + 6), if x + 6 ≥ 0

g(x) = −3(x) − 3(6), − 6 − 6 g(x) = −3x − 18, if x ≥ −6

So, a piecewise function for y = −3 ∣ x + 6 ∣ is

g(x) = { 3x + 18, if x < −6 −3x − 18, if x ≥ −6 .

45. g(x) = { −[−(x − 3)] + 2, if x − 3 < 0 −(x − 3) + 2 if x − 3 ≥ 0

g(x) = −[−(x − 3)] + 2, if x − 3 < 0

g(x) = −(−x + 3) + 2, + 3 + 3

g(x) = x − 3 + 2,

g(x) = x − 1, if x < 3

g(x) = −(x − 3) + 2, if x − 3 ≥ 0

g(x) = −x + 3 + 2 + 3 + 3

g(x) = −x + 5 if x ≥ 3

So, a piecewise function for y = − ∣ x − 3 ∣ + 2 is

g(x) = { x − 1, if x < 3 −x + 5, if x ≥ 3 .

46. g(x) = { 7[−(x + 1)] − 5, if x + 1 < 0 7(x + 1) − 5, if x + 1 ≥ 0

g(x) = 7[−(x + 1)] − 5, if x + 1 < 0

g(x) = 7(−x − 1) − 5, − 1 − 1

g(x) = 7(−x) − 7(1) − 5,

g(x) = −7x − 7 − 5,

g(x) = −7x − 12, if x < −1

g(x) = 7(x + 1) − 5, if x + 1 ≥ 0

g(x) = 7(x) + 7(1) − 5, − 1 − 1

g(x) = 7x + 7 − 5,

g(x) = 7x + 2, if x ≥ −1

So, a piecewise function for y = 7 ∣ x + 1 ∣ − 5 is

g(x) = { −7x − 12, if x < −1 7x + 2, if x ≥ −1 .

47. a. g(x) = a ∣ x − h ∣ + k

g(x) = a ∣ x − 3 ∣ + 0

2 = a ∣ 4 − 3 ∣ + 0

2 = a ∣ 1 ∣ 2 = a(1)

2 = a

So, the function g(x) = 2 ∣ x − 3 ∣ represents the path of the sunlight that re# ects off the water.

b. g(x) = { 2[−(x − 3)], if x − 3 < 0 2(x − 3), if x − 3 ≥ 0

g(x) = 2[−(x − 3)], if x − 3 < 0

g(x) = 2(−x + 3), + 3 + 3

g(x) = 2(−x) + 2(3),

g(x) = −2x + 6, if x < 3

g(x) = 2(x − 3), if x − 3 ≥ 0

g(x) = 2(x) − 2(3), + 3 + 3

g(x) = 2x − 6, if x ≥ 3 So, a piecewise function for g(x) = 2 ∣ x − 3 ∣ is

g(x) = { −2x + 6, if x < 3 2x − 6, if x ≥ 3 .


Chapter 4

48. a. g(x) = a ∣ x − h ∣ + k

g(x) = a ∣ x − 6 ∣ + 4

2 = a ∣ 9 − 6 ∣ + 4

2 = a ∣ 3 ∣ + 4

2 = a(3) + 4

2 = 3a + 4

− 4 − 4

−2 = 3a

−2 — 3

= 3a — 3

− 2 — 3 = a

So, the function g(x) = − 2 — 3 ∣ x − 6 ∣ + 4 represents the path

of the golf ball.

b. g(x) = { − 2 — 3 [−(x − 6)] + 4, if x − 6 < 0

− 2 —

3 (x − 6) + 4, if x − 6 ≥ 0

g(x) = − 2 — 3 (−x + 6) + 4, if x − 6 < 0

g(x) = − 2 — 3 (−x) − 2 —

3 (6) + 4, + 6 + 6

g(x) = 2 — 3 x − 4 + 4,

g(x) = 2 — 3 x, if x < 6

g(x) = − 2 — 3 (x − 6) + 4, if x − 6 ≥ 0

g(x) = − 2 — 3 (x) − 2 —

3 (−6) + 4, + 6 + 6

g(x) = − 2 — 3 x + 4 + 4,

g(x) = − 2 — 3 x + 8, if x ≥ 6

So, a piecewise function for g(x) = − 2 — 3 ∣ x − 6 ∣ + 4 is

g(x) = { 2 — 3 x, if x < 6

− 2 —

3 x + 8, if x ≥ 6

.

49. a. For x ≤ 3, use (0, 1) and (2, 2).

m = 2 − 1 — 2 − 0

= 1 — 2

y − y1 = m(x − x1)

y − 1 = 1 — 2 (x − 0)

y − 1 = 1 — 2 x

+ 1 + 1

y = 1 — 2 x + 1

So, f (x) = 1 — 2 x + 1 represents the graph for x ≤ 3.

f (x) = 1 — 2 x + 1

f (−10) = 1 — 2 (−10) + 1

= −5 + 1

= −4

So, f (−10) = −4.

b. For x > 3, use (4, 2) and (5, 3).

m = 3 − 2 — 5 − 4

= 1 — 1 , or 1

y − y1 = m(x − x1)

y − 2 = 1(x − 4)

y − 2 = 1(x) − 1(4)

y − 2 = x − 4

+ 2 + 2

y = x − 2

So, f (x) = x − 2 represents the graph for x > 3.

f (x) = x − 2

f (8) = 8 − 2

= 6

So, f (8) = 6.

50. a. When the inequality symbols are changed, the open dot at (2, 4) becomes a closed dot, and the closed dot at (2, −3) becomes an open dot. The domain is still all real numbers. The range of the original equation was y < 4, but when the open dot at (2, 4) becomes a closed dot, the range becomes y ≤ 4.

b. Because both parts of the graph meet at the point (1, 2), changing the inequality symbols will not change the graph. So, the domain and range will also stay the same.


Chapter 4

51. x −x + 2 y

−4 −(−4) + 2 6

−3 −(−3) + 2 5

−2 −(−2) + 2 4

x |x| y

Open dot: −2 ∣ −2 ∣ 2

−1 ∣ −1 ∣ 1

0 ∣ 0 ∣ 0

1 ∣ 1 ∣ 1

2 ∣ 2 ∣ 2

x

y

45678

23

1

4321−2−1−3−4


The range is y ≥ 0.

52. a. It costs more to make 100 copies than 101 copies. The closed dot at (100, 10) represents a cost of $10 for 100 copies. The point that represents the cost of 101 copies is on the next portion of the graph, and it has a y-value that is less $10.

b. Yes: The portion of the graph that is farthest to the right has a point on it with a y-value of $40, and the corresponding x-value is greater than 500. So, with $40, you can buy more than 500 copies.

53. Domain Range

−4 ≤ x < −3 −4

−3 ≤ x < −2 −3

−2 ≤ x < −1 −2

−1 ≤ x < 0 −1

0 ≤ x < 1 0

1 ≤ x < 2 1

2 ≤ x < 3 2

3 ≤ x < 4 3

x

y4

23

1

−3−4

4321−2−3−4

The greatest integer function is both a piecewise function and a step function. It is a piecewise function because each “piece” of the function applies to a different part of the domain. More speci" cally, it is a step function, because it is de" ned by a constant value over each part of its domain, and therefore the graph consists of a series of horizontal line segments.

54. x 2x − 2 y 1 2(1) − 2 0 2 2(2) − 2 2 3 2(3) − 2 4

x y 3 −3 4 −3 5 −3

Because there are closed dots at both (3, 4) and (3, −3), the same x-value, 3, is paired with two different y-values, −3 and 4. So, it is not a function. If you replace ≤ with < or ≥ with >, then one will be a closed dot and the other will be an open dot so that no vertical line can be drawn through more than one point on the graph, which means that y will represent a function.


Chapter 4

55. a. x y y1 1(1) 12 1(2) 2

3 2 + 2(3 − 2) 4

4 2 + 2(4 − 2) 6

8 2 + 2(8 − 2) 14

9 14 + 1(9 − 8) 15

f (x) = { 1(x), if 0 ≤ x ≤ 2 2 + 2(x − 2), if 2 < x ≤ 8

14 + 1(x − 8), if 8 < x ≤ 9

So, a piecewise function that represents the depth of snow is

f (x) = { x, if 0 ≤ x ≤ 2 2x − 2, if 2 < x ≤ 8

x + 6, if 8 < x ≤ 9 .

02468

10121416

0 1 2 3 4 5 6 7 8 9 x

y

Time (hours)

Dep

th o

f sn

ow (i

nche

s)

b. no; The point (9, 15) represents the total accumulation of 15 inches in 9 hours. There were 2 inches of accumulation in the " rst 2 hours, an additional 12 inches in the next 6 hours, and 1 more inch in the last hour for a total of 2 + 12 + 1 = 15 inches.

Maintaining Mathematical Pro! ciency 56. A number r is greater than −12 and no more than 13.

r > −12 and r ≤ 13

An inequality is −12 < r ≤ 13.

−16 −12 −8 −4 0 4 8 12 16

57. A number t is less than equal to 4 or no less than 18.

t ≤ 4 or t ≥ 18

An inequality is t ≤ 4 or t ≥ 18.

−8 −4 0 4 8 12 16 20 24

58.

x

y

3

−3−4

−2

4321−2−3−4

h(x) = 4x + 3

f(x) = x

The transformations are a vertical stretch by a factor of 4 and then a vertical translation 3 units up.

59.

x

y21

−3−4−5−6

−2

−2−3−4−5−6−8

h(x) = −x − 8

f(x) = x

Sample answer: The translations are a re# ection in the y-axis followed by a vertical translation 8 units down.

60.

x

y

456

23

1

−2

4 5 6 7 8321

h(x) = − x + 5

f(x) = x

12

Sample answer: The translations are a vertical shrink by a factor of 1 — 2 , followed by a re# ection in the y-axis and then a vertical translation 5 units up.

4.4–4.7 What Did You Learn? (p. 225) 1. Sample answer: You could use the Internet or an almanac to

search for statistics or other data about your favorite sport or pastime.

2. Sample answer: You " rst subtract the values of consecutive terms to " nd the common difference. Then you repeatedly add this amount to the last term in order to get the next. If you write an equation for an in the form an = a1 + (n − 1)d, then you can use this equation to calculate the value of a term, such as the 50th term, a50, without knowing the values of all the terms before it.

3. Use the de" nitions of a piecewise function and a step function. A piecewise function is broken up into two or more “pieces” that are each de" ned by a separate equation over the corresponding part of the domain. A step function is a type of piecewise function, whose “pieces” are horizontal segments, each of which are de" ned by a different constant value.

{ {{ {


Chapter 4

Chapter 4 Review (pp. 226–228) 1. Let (x1, y1) = (0, 1) and (x2, y2) = (4, −1).

m = y2 − y1 — x2 − x1

= −1 − 1 — 4 − 0

= −2 — 4 , or − 1 —

2


So, the equation is y = − 1 — 2 x + 1.

2. y − y1 = m (x − x1)

y − 7 = −1(x − 4)

The equation is y − 7 = −(x − 4).

3. Write f (10) = 5 as (10, 5) and f (2) = −3 as (2, −3). Find the slope of the line through these points.

m = −3 − 5 — 2 − 10

= −8 — −8

, or 1

y − y1 = m(x − x1)

y − 5 = 1(x − 10)

y − 5 = 1(x) − 1(10)

y − 5 = x − 10 + 5 + 5

y = x − 5

A function is f (x) = x − 5.

4. Write f (3) = −4 as (3, −4) and f (5) = −4 as (5, −4). Find the slope of the line through these points.

m = −4 − (−4) — 5 − 3

= −4 + 4 — 5 − 3

= 0 — 2 , or 0

y − y1 = m(x − x1)

y − (−4) = 0(x − 3)

y + 4 = 0 − 4 − 4 y = −4

A function is f (x) = −4.


m = 3 − 8 — 9 − 6

= −5 — 3

y − y1 = m(x − x1)

y − 8 = − 5 — 3 (x − 6)

y − 8 = − 5 — 3 (x) − 5 —

3 (−6)

y − 8 = − 5 — 3 x + 10

+ 8 + 8

y = − 5 — 3 x + 18

A function is f (x) = − 5 — 3 x + 18.

6. Line a: m = 3 − 4 — 4 − 0

= −1 — 4

Line b: m = 0 − 1 — 4 − 0

= −1 — 4

Line c: m = 4 − 0 — 4 − 2

= 4 — 2 , or 2

Lines a and b have slopes of − 1 — 4 . So, they are parallel. None of the lines have negative reciprocal slopes, so none are perpendicular.

7. Line a: 2x − 7y = 14

2x − 2x − 7y = 14 − 2x

−7y = 14 − 2x

−7y — −7

= 14 − 2x — −7

y = −2 + 2 — 7 x, or y = 2 —

7 x − 2

Line b: y = 7 — 2 x − 8

Line c: 2x + 7y = −21

2x − 2x + 7y = −21 − 2x

7y = −21 − 2x

7y — 7 = −21 − 2x —

7

y = −3 − 2 — 7 x, or y = − 2 —

7 x − 3

The slope of line b is 7 — 2 . The slope of line c is − 2 — 7 , which is the negative reciprocal of 7 — 2 . So, lines b and c are perpendicular. The slope of line a is 2 — 7 , which is not the same as either of the other two slopes. So, none of the lines are parallel.

8. y − y1 = m(x − x1)

y − 5 = −4(x − 1)

y − 5 = −4(x) − 4(−1)

y − 5 = −4x + 4 + 5 + 5 y = −4x + 9 An equation of the parallel line is y = −4x + 9.

9. 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 perpendicular line is y = 1 — 2 x − 4.

10. The roasting time for a 12-pound turkey is 4 hours.


Chapter 4

11. Sample answer:

Use (8, 3.0) and (20, 5.5).

The slope of the line is m = 5.5 − 3.0 — 20 − 8

= 2.5 — 12

= 5 — 24

.

y − y1 = m(x − x1)

y − 3 = 5 — 24

(x − 8)

y − 3 = 5 — 24

(x) − 5 — 24

(8)

y − 3 = 5 — 24

x − 5 — 3

+ 3 + 3

y = 5 — 24

x + 4 — 3

An equation of a line that models the data is y = 5 — 24 x + 4 — 3 . The slope of this line is 5 — 24 ,which means that the roasting time increases by about 5 — 24 hour for each pound the weight of the turkey increases. The y-intercept is 4 — 3 , which has no meaning in this context because the weight of a turkey cannot be 0 pounds.

12.

x yy-Value

from model Residual

64 9 8.5 9 − 8.5 = 0.5

62 7 7.5 7 − 7.5 = −0.5

70 12 11.5 12 − 11.5 = 0.5

63 8 8 8 − 8 = 0

72 13 12.5 13 − 12.5 = 0.5

68 9.5 10.5 9.5 − 10.5 = −1

66 9 9.5 9 − 9.5 = −0.5

74 13.5 13.5 13.5 − 13.5 = 0

68 10 10.5 10 − 10.5 = −0.5

59 6.5 6 6.5 − 6 = 0.5

x

1

−1

62 64 66 68 70 72 7460

residual

The points are evenly dispersed about the horizontal axis. So, the equation y = 0.50x − 23.5 is a good " t.

13. a. y = 0.50x − 23.5

9 = 0.50x − 23.5 + 23.5 + 23.5

32.5 = 0.50x

32.5 — 0.50

= 0.50x — 0.50

65 = x

A student whose shoe size is 9 is about 65 inches tall.

b. y = 0.50x − 23.5

y = 0.50(60) − 23.5

y = 30 − 23.5

y = 6.5

A student, who is 60 inches tall, has a shoe size of about 6.5.

14. no; Height does not determine shoe size.

15. d = 10 − 11 = −1

an = a1 + (n − 1)d

an = 11 + (n − 1)(−1)

an = 11 + n(−1) − 1(−1)

an = 11 − n + 1 an = −n + 12

An equation for the nth term of the arithmetic sequence is an = −n + 12.

an = −n + 12

a30 = −(30) + 12

a30 = −30 + 12

a30 = −18


16. d = 12 − 6 = 6

an = a1 + (n − 1)d

an = 6 + (n − 1)(6)

an = 6 + n(6) − 1(6)

an = 6 + 6n − 6

an = 6n

An equation for the nth term of the arithmetic sequence is an = 6n.

an = 6n

a30 = 6(30)

a30 = 180


17. d = −6 − (−9) = −6 + 9 = 3

an = a1 + (n − 1)d

an = −9 + (n − 1)(3)

an = −9 + n(3) − 1(3)

an = −9 + 3n − 3

an = 3n − 12

An equation for the nth term of the arithmetic sequence is an = 3n − 12.

an = 3n − 12

a30 = 3(30) − 12

a30 = 90 − 12

a30 = 78



Chapter 4

18. a. Because 0 ≤ 0, use y = 3 — 2 x + 3.

y = 3 — 2 x + 3

y = 3 — 2 (0) + 3

= 0 + 3 = 3

The value of y is 3 when x = 0.

b. Because 5 > 0, use y = −2x.

y = −2x

y = −2(5)

= −10

The value of y is −10 when x = 5.

19. x x + 6 y

−2 −2 + 6 4 open dot:−1 −1 + 6 5

0 0 + 6 6

x −3x y 0 −3(0) 0 1 −3(1) −3 2 −3(2) −6

x

y

456

23

1

−2

21−2−1−3−4−5


The range is y ≤ 6.

20. x 4x + 2 y x 2x − 6 y

−6 4(−6) + 2 −22 −4 2(−4) − 6 −14

−5 4(−5) + 2 −18 −3 2(−3) − 6 −12

−4 4(−4) + 2 −14 −2 2(−2) − 6 −10

x

y8

−16

84−4−8

−24

The domain is all red numbers. The range is all real numbers.

21. A piecewise function for y = ∣ x ∣ + 15 is

g(x) = { −x + 15, if x < 0 x + 15, if x ≥ 0

.

22. g(x) = { 4[−(x + 5)], if x + 5 < 0 4(x + 5), if x + 5 ≥ 0

g(x) = 4[−(x + 5)], if x + 5 < 0

g(x) = 4(−x − 5) − 5 − 5

g(x) = 4(−x) − 4(5)

g(x) = −4x − 20, if x < −5

g(x) = 4(x + 5), if x + 5 ≥ 0

g(x) = 4(x) + 4(5) − 5 − 5

g(x) = 4x + 20, if x ≥ −5

A piecewise function for y = 4 ∣ x + 5 ∣ is

g(x) = { −4x − 20, if x < −5 4x + 20, if x ≥ −5 .

23. g(x) = { 2[−(x + 2)] − 3, if x + 2 < 0 2(x + 2) − 3, if x + 2 ≥ 0

g(x) = 2[−(x + 2)] − 3, if x + 2 < 0

g(x) = 2(−x − 2) − 3 − 2 − 2

g(x) = 2(−x) − 2(2) − 3

g(x) = −2x − 4 − 3

g(x) = −2x − 7, if x < −2

g(x) = 2(x + 2) − 3, if x + 2 ≥ 0

g(x) = 2(x) + 2(2) − 3 − 2 − 2

g(x) = 2x + 4 − 3

g(x) = 2x + 1, if x ≥ −2

So, a piecewise function for y = 2 ∣ x + 2 ∣ − 3 is

g(x) = { −2x − 7, if x < −2 2x + 1, if x ≥ −2 .

24. Number of days Total cost

f (x) = { 65, if 0 < x ≤ 1 100, if 1 < x ≤ 2

135, if 2 < x ≤ 3

0 < x ≤ 1 65

1 < x ≤ 2 100

2 < x ≤ 3 135

Popcorn Machine Rental

020406080

100120140160

0 1 2 3 4 x

y

Days

Tota

l cos

t (d

olla

rs)


Chapter 4

Chapter 4 Test (p. 229)

1. x 2x + 4 y x 1 — 3 x − 1 y

−3 2(−3) + 4 −2 open dot: −1 1 — 3 (−1) − 1 −1 1 —

3

−2 2(−2) + 4 0 0 1 — 3 (0) − 1 −1

−1 2(−1) + 4 2 3 1 — 3 (3) − 1 0

x

y4

−8

−12

−4

84−4−8

The domain is all real numbers. The range is all real numbers.

2.

x

y21

−2

2 4 6 8 10 12

The domain is 0 ≤ x ≤ 12 and the range is −2, −1, 0, 1.

3. y = mx + b

y = 2 — 5 x + (−7)


4. m = −3 − 6 — 3 − 0

= −9 — 3 , or −3

y − y1 = m(x − x1)

y − 6 = −3(x − 0)

y − 6 = −3(x)

y − 6 = −3x

+ 6 + 6

y = −3x + 6 The equation is y = −3x + 6.

5. y − y1 = m(x − x1)

y − (−8) = 3(x − (−2))

y + 8 = 3(x + 2)

y + 8 = 3(x) + 3(2)

y + 8 = 3x + 6 − 8 − 8 y = 3x − 2

The equation of the parallel line is y = 3x − 2.

6. y = mx + b 1 = −4(1) + b 1 = −4 + b + 4 + 4 5 = b

Using m = −4 and b = 5, the equation of the perpendicular line is y = −4x + 5.

7. y − y1 = m(x − x1)

y − 2 = 10(x − 6)

The equation is y − 2 = 10(x − 6).

8. m = −1 − 2 — 6 − (−3)

= −1 − 2 — 6 + 3

= −3 — 9 , or − 1 —

3

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

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

3 (x − 6)

y − 2 = − 1 — 3 (x + 3) y + 1 = − 1 —

3 (x − 6)

The equation is y − 2 = − 1 — 3 (x + 3) or y + 1 = − 1 — 3 (x − 6).

9. a1 = 42 and d = 3

an = a1 + (n − 1)d

an = 42 + (n − 1)(3)

an = 42 + n(3) − 1(3)

an = 42 + 3n − 3

an = 3n + 39

a. an = 3n + 39

a25 = 3(25) + 39

a25 = 114

So, row 25 has 114 seats.

b. an = 3n + 39

90 = 3n + 39

− 39 − 39

51 = 3n

51 — 3 = 3n —

3

17 = n

So, row 17 has 90 seats.


Chapter 4

10. a. Neighborhood Festival Attendance

0

400

800

1200

1600

0 1000 2000 3000 4000 x

y

Advertising (dollars)

Year

ly a

tten

danc

e

The scatter plot shows a positive correlation.

b. Sample answer: A line through (1000, 500) and (2000, 700) models the data.

The slope of the line is m = 700 − 500 —— 2000 − 1000

= 200 — 1000

= 1 — 5 .

y − y1 = m(x − x1)

y − 500 = 1 — 5 (x − 1000)

y − 500 = 1 — 5 (x) − 1 —

5 (1000)

y − 500 = 1 — 5 x − 200

+ 500 + 500

y = 1 — 5 x + 300

An equation of the line of " t is y = 1 — 5 x + 300.

c. Sample answer: The slope of the line is 1 — 5 . This means that the yearly attendance at the festival increases by about 1 person for every $5 spent on advertising. The y-intercept is 300, meaning that if no money is spent on advertising, about 300 people would attend the festival.

11. a. After entering the data from the table into two lists using a graphing calculator, the linear regression feature yields the equation y = 0.19x + 309.

b. The correlation coef" cient is about 0.943. This means that there is a strong positive correlation between the amount spent on advertising and the yearly attendance of the festival. So, the equation y = 0.19x + 309 closely models the data.

c. Sample answer: Because the data show a strong positive correlation and the line of best " t closely models the data, you would expect the scatter plot of the residuals to show a relatively even distribution of residuals on both sides of the x-axis.

d. There may be a casual relationship in the data but the correlation may be caused by other factors, such as the quality of the attractions each year.

e. y = 0.19x + 309

2000 = 0.19x + 309

− 309 − 309

1691 = 0.19x

8900 = x

In order to get 2000 people to attend the festival, $8900 should be spent on advertising.

12. Line 1: y − c = ax

y − c + c = ax + c y = ax + c Line 2: ay = −x − b

ay — a = −x − b —

a

y = −1 — a x − b —

a

Line 3: ax + y = d

ax − ax + y = d − ax

y = d − ax, or y = −ax + d

The slope of line 1 is a. The slope of line 2 is − 1 — a , which is the negative reciprocal of a. So, lines 1 and 2 are perpendicular. The slope of line 3 is −a, which is the opposite of a and the reciprocal of − 1 — a , but neither the same nor the negative reciprocal of either. So, line 3 is neither parallel nor perpendicular to the other two lines.

13. Sample answer:

f (x) =

{ −1, if x ≤ −3 x, if −3 < x ≤ 1 a, if x > 1

Chapter 4 Standards Assessment (pp. 230–231) 1. C; d = 21 − 24 = −3

an = a1 + (n − 1)d

an = 24 + (n − 1)(−3)

an = 24 + n(−3) − 1(−3)

an = 24 − 3n + 3 an = 27 − 3n


Chapter 4

2. 3x + 6 ≤ 8 + 2x 5x − 5 ≥ 7x − 9

− 2x − 2x −5x − 5x

x + 6 ≤ 8 −5 ≥ 2x − 9

− 6 − 6 + 9 + 9 x ≤ 2 4 ≥ 2x

4 — 2 ≥ 2x —

2

2 ≥ x

12 − 3x ≤ 18

− 12 − 12

−3x ≤ 6

−3x — −3

≥ 6 — 3

x ≥ −2

−2 − 3 — 2 x ≥ −3 − x

+ 3 — 2 x + 3 —

2 x

−2 ≥ −3 + 1 — 2 x ( −1 + 3 —

2 = − 2 —

2 + 3 —

2 = 1 —

2 )

+ 3 + 3

1 ≥ 1 — 2 x

2 ⋅ 1 ≥ 2 ⋅ 1 — 2 x

2 ≥ x

Three of the inequalities are equivalent because they have the same solution, x ≤ 2. They are 3x + 6 ≤ 8 + 2x, 5x − 5 ≥ 7x − 9, and −2 − 3 — 2 x ≥ −3 − x.

3. Situation Correlation Causation

Yes No Yes No

a. X X

b. X X

c. X X

d. X X

a. Increasing the price of a pair of pants will cause fewer people to buy them.

b. Increasing the number of cell phones in a city will not cause the number of taxis in the city to increase.

c. A persons IQ, or intelligence level, is not related to the person’s athletic ability.

d. Spending more time studying affects the score earned.

4. g(x) = f (x + 2) is of the form y = f (x − h), where h = −2. So, the graph of g is a horizontal translation 2 units left of the graph of f, which is represented by the green line.

h(x) = f (3x) is of the form y = f (ax), where a = 3. So, the graph of h is a horizontal shrink of the graph of f by a factor of 1 ÷ 3 = 1 — 3 , which is represented by the red line.

k(x) = f (x) + 4 is of the form y = f (x) + k, where k = 4. So, the graph of h is a vertical translation 4 units up of the graph of f, which is represented by the purple line.

q(x) = −f (x) is of the form y = −f (x). So, the graph of q is a re# ection in the x-axis of the graph of f, which is represented by the blue line.

5. m = −3 − 1 — −2 − 6

= −4 — −8

, or 1 — 2

y = mx + b

1 = 1 — 2 (6) + b

1 = 3 + b

− 3 − 3 −2 = b

Using m = 1 — 2 and b = −2, the equation of the line is y = 1 — 2 x − 2.

6. A piecewise function that represents the graph is

y = { 2x + 3, if x < 0 2x − 3, if x ≥ 0 .

The left portion of the graph has a y-intercept of positive 3, and an open dot where x = 0. The right portion of the graph has a y-intercept of negative 3 and a closed dot where x = 0.

7. Sample Answer:

My function: x −3 −2 −1 0 1 y −4 2 3 4 0

Friend’s relation: x −3 −2 −3 −2 −3

y 0 1 2 3 4

8. Let x be how much your total bill is before the discount.

0.80x = x − 5

−x −x

−0.2x = −5 (0.80 − 1.00 = −0.2)

−0.2x — −0.2

= −5 — −0.2

x = 25

Both coupons will save you the same amount of money on a total bill of $25.


Chapter 4

9. a.

x yy-Value

from model Residual

54 40 112 40 − 112 = −72

60 120 130 120 − 130 = −10

68 180 154 180 − 154 = 26

72 260 166 260 − 166 = 94

78 280 184 280 − 184 = 96

84 260 202 260 − 202 = 58

92 220 226 220 − 226 = −6

98 180 244 180 − 244 = −64

The points that appear on a scatter plot of the residuals are (92, −6), (78, 96), (60, −10), (84, 58), (98, −64), (72, 94), (54, −72), and (68, 26).

b. no; The residual points form a ∩-shaped pattern, which suggests the data are not linear. So, the equation y = 3x − 50 does not model the data well.

hscc alg1 wsk 04 - schoolwires · 2015. 4. 23. · 4.1 explorations (p.175) 1. a. m = y 2 − y 1...

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