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**FUNCTIONS AND THEIR GRAPHS**

2.1 The Cartesian Coordinate System and Straight Lines

Concept Questions page 74

1. a. a < 0 and b > 0. b. a < 0 and b < 0. c. a > 0 and b < 0.

2. The slope of a nonvertical line is m = y2 âˆ’ y1

x2 âˆ’ x1

, where P (x1, y1) and P (x2, y2) are any two distinct points on the

line. The slope of a vertical line is undefined.

Exercises page 74

1. The coordinates of A are (3, 3) and it is located in Quadrant I.

2. The coordinates of B are (âˆ’5, 2) and it is located in Quadrant II.

3. The coordinates of C are (2, âˆ’2) and it is located in Quadrant IV.

4. The coordinates of D are (âˆ’2, 5) and it is located in Quadrant II.

5. The coordinates of E are (âˆ’4, âˆ’6) and it is located in Quadrant III.

6. The coordinates of F are (8, âˆ’2) and it is located in Quadrant IV.

7. A 8. (âˆ’5, 4) 9. E, F, and G 10. E 11. F 12. D

For Exercises 13â€“20, refer to the following figure.

x

1

y

0

13. (_2, 5)

14. (1, 3)

15 (3,_1)

20. (1.2,_3.4)

16. (3,_4)

19. (4.5,_4.5)

18. 3

2 ( ) 5

2 _ ,

17. 7

2 ( ) 8,_

1

21. Referring to the figure shown in the text, we see that m = 2 âˆ’ 0

0 âˆ’ (âˆ’4) = 1

2

.

22. Referring to the figure shown in the text, we see that m = 4 âˆ’ 0

0 âˆ’ 2 = âˆ’2.

49

50 2 FUNCTIONS AND THEIR GRAPHS

23. This is a vertical line, and hence its slope is undefined.

24. This is a horizontal line, and hence its slope is 0.

25. m = y2 âˆ’ y1

x2 âˆ’ x1

= 8 âˆ’ 3

5 âˆ’ 4 = 5. 26. m = y2 âˆ’ y1

x2 âˆ’ x1

= 8 âˆ’ 5

3 âˆ’ 4 = 3

âˆ’1 = âˆ’3.

27. m = y2 âˆ’ y1

x2 âˆ’ x1

= 8 âˆ’ 3

4 âˆ’ (âˆ’2) = 5

6

. 28. m = y2 âˆ’ y1

x2 âˆ’ x1

= âˆ’4 âˆ’ (âˆ’2)

4 âˆ’ (âˆ’2) = âˆ’2

6 = âˆ’1

3

.

29. m = y2 âˆ’ y1

x2 âˆ’ x1

= d âˆ’ b

c âˆ’ a

, provided a /= c.

30. m = y2 âˆ’ y1

x2 âˆ’ x1

= âˆ’b âˆ’ (b âˆ’ 1)

a + 1 âˆ’ (âˆ’a + 1) = âˆ’ âˆ’b âˆ’ b + 1

a + 1 + a âˆ’ 1 = 1 âˆ’ 2b

2a .

31. Because the equation is already in slope-intercept form, we read off the slope m = 4.

a. If x increases by 1 unit, then y increases by 4 units.

b. If x decreases by 2 units, then y decreases by 4 (âˆ’2) = âˆ’8 units.

32. Rewrite the given equation in slope-intercept form: 2x + 3y = 4, 3y = 4 âˆ’ 2x, and so y = âˆ’2

3 x + 4

3 .

a. Because m = âˆ’2

3 , we conclude that the slope is negative.

b. Because the slope is negative, y decreases as x increases.

c. If x decreases by 2 units, then y increases by r

âˆ’2

3

s

(âˆ’2) = 4

3 units.

33. The slope of the line through A and B is âˆ’10 âˆ’ (âˆ’2)

âˆ’3 âˆ’ 1 = âˆ’8

âˆ’4 = 2. The slope of the line through C and D is

1 âˆ’ 5

âˆ’1 âˆ’ 1 = âˆ’4

âˆ’2 = 2. Because the slopes of these two lines are equal, the lines are parallel.

34. The slope of the line through A and B is âˆ’2 âˆ’ 3

2 âˆ’ 2 . Because this slope is undefined, we see that the line is vertical.

The slope of the line through C and D is 5 âˆ’ 4

âˆ’2 âˆ’ (âˆ’2)

. Because this slope is undefined, we see that this line is also

vertical. Therefore, the lines are parallel.

35. The slope of the line through the point (1, a) and (4, âˆ’2) is m1 = âˆ’2 âˆ’ a

4 âˆ’ 1 and the slope of the line through

(2, 8) and (âˆ’7, a + 4) is m2 = a + 4 âˆ’ 8

âˆ’7 âˆ’ 2 . Because these two lines are parallel, m1 is equal to m2. Therefore,

âˆ’2 âˆ’ a

3 = a âˆ’ 4

âˆ’9 , âˆ’9 (âˆ’2 âˆ’ a) = 3 (a âˆ’ 4), 18 + 9a = 3a âˆ’ 12, and 6a = âˆ’30, so a = âˆ’5.

36. The slope of the line through the point (a, 1) and (5, 8) is m1 = 8 âˆ’ 1

5 âˆ’ a

and the slope of the line through (4, 9) and

(a + 2, 1) is m2 = 1 âˆ’ 9

a + 2 âˆ’ 4

. Because these two lines are parallel, m1 is equal to m2. Therefore, 7

5 âˆ’ a = âˆ’8

a âˆ’ 2

,

7 (a âˆ’ 2) = âˆ’8 (5 âˆ’ a), 7a âˆ’ 14 = âˆ’40 + 8a, and a = 26.

37. Yes. A straight line with slope zero (m = 0) is a horizontal line, whereas a straight line whose slope does not exist

(m cannot be computed) is a vertical line.

2.2 EQUATIONS OF LINES 51

2.2 Equations of Lines

Concept Questions page 82

1. a. y âˆ’ y1 = m (x âˆ’ x1) b. y = mx + b

c. ax + by + c = 0, where a and b are not both zero.

2. a. m1 = m2 b. m2 = âˆ’ 1

m1

Exercises page 82

1. (e) 2. (c) 3. (a) 4. (d) 5. (f) 6. (b)

7. The slope of the line through A and B is

2 âˆ’ 5

4 âˆ’ (âˆ’2) = âˆ’3

6 = âˆ’1

2

. The slope of the line through C and D is

6 âˆ’ (âˆ’2)

3 âˆ’ (âˆ’1) = 8

4 = 2. Because the slopes of these two lines are the negative reciprocals of each other, the lines are

perpendicular.

8. The slope of the line through A and B is âˆ’2 âˆ’ 0

1 âˆ’ 2 = âˆ’2

âˆ’1 = 2. The slope of the line through C and D is

4 âˆ’ 2

âˆ’8 âˆ’ 4 = 2

âˆ’12 = âˆ’1

6

. Because the slopes of these two lines are not the negative reciprocals of each other, the

lines are not perpendicular.

9. An equation of a horizontal line is of the form y = b. In this case b = âˆ’5, so y = âˆ’5 is an equation of the line.

10. An equation of a vertical line is of the form x = a. In this case a = 0, so x = 0 is an equation of the line.

11. We use the point-slope form of an equation of a line with the point (3, âˆ’4) and slope m = 2. Thus

y âˆ’ y1 = m (x âˆ’ x1) becomes y âˆ’ (âˆ’4) = 2 (x âˆ’ 3). Simplifying, we have y + 4 = 2x âˆ’ 6, or y = 2x âˆ’ 10.

12. We use the point-slope form of an equation of a line with the point (2, 4) and slope m = âˆ’1. Thus

y âˆ’ y1 = m (x âˆ’ x1), giving y âˆ’ 4 = âˆ’1 (x âˆ’ 2), y âˆ’ 4 = âˆ’x + 2, and finally y = âˆ’x + 6.

13. Because the slope m = 0, we know that the line is a horizontal line of the form y = b. Because the line passes

through (âˆ’3, 2), we see that b = 2, and an equation of the line is y = 2.

14. We use the point-slope form of an equation of a line with the point (1, 2) and slope m = âˆ’1

2 . Thus

y âˆ’ y1 = m (x âˆ’ x1) gives y âˆ’ 2 = âˆ’1

2 (x âˆ’ 1), 2y âˆ’ 4 = âˆ’x + 1, 2y = âˆ’x + 5, and y = âˆ’1

2 x + 5

2 .

15. We first compute the slope of the line joining the points (2, 4) and (3, 7), obtaining m = 7 âˆ’ 4

3 âˆ’ 2 = 3. Using the

point-slope form of an equation of a line with the point (2, 4) and slope m = 3, we find y âˆ’ 4 = 3 (x âˆ’ 2), or

y = 3x âˆ’ 2.

16. We first compute the slope of the line joining the points (2, 1) and (2, 5), obtaining m = 5 âˆ’ 1

2 âˆ’ 2

. Because this slope

is undefined, we see that the line must be a vertical line of the form x = a. Because it passes through (2, 5), we see

that x = 2 is the equation of the line.

52 2 FUNCTIONS AND THEIR GRAPHS

17. We first compute the slope of the line joining the points (1, 2) and (âˆ’3, âˆ’2), obtaining m = âˆ’2 âˆ’ 2

âˆ’3 âˆ’ 1 = âˆ’4

âˆ’4 = 1.

Using the point-slope form of an equation of a line with the point (1, 2) and slope m = 1, we find y âˆ’ 2 = x âˆ’ 1, or

y = x + 1.

18. We first compute the slope of the line joining the points (âˆ’1, âˆ’2) and (3, âˆ’4), obtaining

m = âˆ’4 âˆ’ (âˆ’2)

3 âˆ’ (âˆ’1) = âˆ’2

4 = âˆ’1

2

. Using the point-slope form of an equation of a line with the point (âˆ’1, âˆ’2) and

slope m = âˆ’1

2 , we find y âˆ’ (âˆ’2) = âˆ’1

2 [x âˆ’ (âˆ’1)], y + 2 = âˆ’1

2 (x + 1), and finally y = âˆ’1

2 x âˆ’ 5

2 .

19. We use the slope-intercept form of an equation of a line: y = mx + b. Because m = 3 and b = 5, the equation is

y = 3x + 5.

20. We use the slope-intercept form of an equation of a line: y = mx + b. Because m = âˆ’2 and b = âˆ’1, the equation

is y = âˆ’2x âˆ’ 1.

21. We use the slope-intercept form of an equation of a line: y = mx + b. Because m = 0 and b = 5, the equation is

y = 5.

22. We use the slope-intercept form of an equation of a line:y = mx + b. Because m = âˆ’1

2 , and b = 3

4 , the equation is

y = âˆ’1

2 x + 3

4 .

23. We first write the given equation in the slope-intercept form: x âˆ’ 2y = 0, so âˆ’2y = âˆ’x, or y = 1

2 x. From this

equation, we see that m = 1

2 and b = 0.

24. We write the equation in slope-intercept form: y âˆ’ 2 = 0, so y = 2. From this equation, we see that m = 0 and

b = 2.

25. We write the equation in slope-intercept form: 2x âˆ’ 3y âˆ’ 9 = 0, âˆ’3y = âˆ’2x + 9, and y = 2

3 x âˆ’ 3. From this

equation, we see that m = 2

3 and b = âˆ’3.

26. We write the equation in slope-intercept form: 3x âˆ’ 4y + 8 = 0, âˆ’4y = âˆ’3x âˆ’ 8, and y = 3

4 x + 2. From this

equation, we see that m = 3

4 and b = 2.

27. We write the equation in slope-intercept form: 2x + 4y = 14, 4y = âˆ’2x + 14, and y = âˆ’2

4 x + 14

4 = âˆ’1

2 x + 7

2 .

From this equation, we see that m = âˆ’1

2 and b = 7

2 .

28. We write the equation in the slope-intercept form: 5x + 8y âˆ’ 24 = 0, 8y = âˆ’5x + 24, and y = âˆ’5

8 x + 3. From

this equation, we conclude that m = âˆ’5

8 and b = 3.

29. We first write the equation 2x âˆ’ 4y âˆ’ 8 = 0 in slope-intercept form: 2x âˆ’ 4y âˆ’ 8 = 0, 4y = 2x âˆ’ 8, y = 1

2 x âˆ’ 2.

Now the required line is parallel to this line, and hence has the same slope. Using the point-slope form of an

equation of a line with m = 1

2 and the point (âˆ’2, 2), we have y âˆ’ 2 = 1

2 [x âˆ’ (âˆ’2)] or y = 1

2 x + 3.

30. The slope of the line passing through (âˆ’2, âˆ’3) and (2, 5) is m = 5 âˆ’ (âˆ’3)

2 âˆ’ (âˆ’2) = 8

4 = 2. Thus, the required equation

is y âˆ’ 3 = 2 [x âˆ’ (âˆ’1)], y = 2x + 2 + 3, or y = 2x + 5.

2.2 EQUATIONS OF LINES 53

31. We first write the equation 3x + 4y âˆ’ 22 = 0 in slope-intercept form: 3x + 4y âˆ’ 22 = 0, so 4y = âˆ’3x + 22

and y = âˆ’3

4 x + 11

2 Now the required line is perpendicular to this line, and hence has slope 4

3 (the negative

reciprocal of âˆ’3

4 ). Using the point-slope form of an equation of a line with m = 4

3 and the point (2, 4), we have

y âˆ’ 4 = 4

3 (x âˆ’ 2), or y = 4

3 x + 4

3 .

32. The slope of the line passing through (âˆ’2, âˆ’1) and (4, 3) is given by m = 3 âˆ’ (âˆ’1)

4 âˆ’ (âˆ’2) = 3 + 1

4 + 2 = 4

6 = 2

3

, so

the slope of the required line is m = âˆ’3

2 and its equation is y âˆ’ (âˆ’2) = âˆ’3

2 (x âˆ’ 1), y = âˆ’3

2 x + 3

2 âˆ’ 2, or

y = âˆ’3

2 x âˆ’ 1

2 .

33. A line parallel to the x-axis has slope 0 and is of the form y = b. Because the line is 6 units below the axis, it passes

through (0, âˆ’6) and its equation is y = âˆ’6.

34. Because the required line is parallel to the line joining (2, 4) and (4, 7), it has slope m = 7 âˆ’ 4

4 âˆ’ 2 = 3

2

. We also know

that the required line passes through the origin (0, 0). Using the point-slope form of an equation of a line, we find

y âˆ’ 0 = 3

2 (x âˆ’ 0), or y = 3

2 x.

35. We use the point-slope form of an equation of a line to obtain y âˆ’ b = 0 (x âˆ’ a), or y = b.

36. Because the line is parallel to the x-axis, its slope is 0 and its equation has the form y = b. We know that the line

passes through (âˆ’3, 4), so the required equation is y = 4.

37. Because the required line is parallel to the line joining (âˆ’3, 2) and (6, 8), it has slope m = 8 âˆ’ 2

6 âˆ’ (âˆ’3) = 6

9 = 2

3

. We

also know that the required line passes through (âˆ’5, âˆ’4). Using the point-slope form of an equation of a line, we

find y âˆ’ (âˆ’4) = 2

3 [x âˆ’ (âˆ’5)], y = 2

3 x + 10

3 âˆ’ 4, and finally y = 2

3 x âˆ’ 2

3 .

38. Because the slope of the line is undefined, it has the form x = a. Furthermore, since the line passes through (a, b),

the required equation is x = a.

39. Because the point (âˆ’3, 5) lies on the line kx + 3y + 9 = 0, it satisfies the equation. Substituting x = âˆ’3 and y = 5

into the equation gives âˆ’3k + 15 + 9 = 0, or k = 8.

40. Because the point (2, âˆ’3) lies on the line âˆ’2x + ky + 10 = 0, it satisfies the equation. Substituting x = 2 and

y = âˆ’3 into the equation gives âˆ’2 (2) + (âˆ’3) k + 10 = 0, âˆ’4 âˆ’ 3k + 10 = 0, âˆ’3k = âˆ’6, and finally k = 2.

54 2 FUNCTIONS AND THEIR GRAPHS

41. 3x âˆ’ 2y + 6 = 0. Setting y = 0, we have 3x + 6 = 0

or x = âˆ’2, so the x-intercept is âˆ’2. Setting x = 0, we

have âˆ’2y + 6 = 0 or y = 3, so the y-intercept is 3.

x

y

2

4

0

42. 2x âˆ’ 5y + 10 = 0. Setting y = 0, we have 2x + 10 = 0

or x = âˆ’5, so the x-intercept is âˆ’5. Setting x = 0, we

have âˆ’5y + 10 = 0 or y = 2, so the y-intercept is 2.

x

y

2

4

_4 _2 0

2

_2

_6

43. x + 2y âˆ’ 4 = 0. Setting y = 0, we have x âˆ’ 4 = 0 or

x = 4, so the x-intercept is 4. Setting x = 0, we have

2y âˆ’ 4 = 0 or y = 2, so the y-intercept is 2.

x

y

4

0

_2

2

_2 2 4 6

44. 2x + 3y âˆ’ 15 = 0. Setting y = 0, we have

2x âˆ’ 15 = 0, so the x-intercept is 15

2 . Setting x = 0,

we have 3y âˆ’ 15 = 0, so the y-intercept is 5.

x

y

4

0

_2

2

_4 4 8

6

12

45. y + 5 = 0. Setting y = 0, we have 0 + 5 = 0, which

has no solution, so there is no x-intercept. Setting

x = 0, we have y + 5 = 0 or y = âˆ’5, so the

y-intercept is âˆ’5.

x

y

0

_2

_2 2 4

_4

_4

_6

2

46. âˆ’2x âˆ’ 8y + 24 = 0. Setting y = 0, we have

âˆ’2x + 24 = 0 or x = 12, so the x-intercept is 12.

Setting x = 0, we have âˆ’8y + 24 = 0 or y = 3, so the

y-intercept is 3.

x

y

4

0

_2

2

_4 4 8 12 16

2.2 EQUATIONS OF LINES 55

47. Because the line passes through the points (a, 0) and (0, b), its slope is m = b âˆ’ 0

0 âˆ’ a = âˆ’b

a

. Then, using the

point-slope form of an equation of a line with the point (a, 0), we have y âˆ’ 0 = âˆ’b

a

(x âˆ’ a) or y = âˆ’b

a

x + b,

which may be written in the form b

a

x + y = b. Multiplying this last equation by

1

b

, we have x

a

+ y

b = 1.

48. Using the equation x

a

+ y

b = 1 with a = 3 and b = 4, we have x

3 + y

4 = 1. Then 4x + 3y = 12, so 3y = 12 âˆ’ 4x

and thus y = âˆ’4

3 x + 4.

49. Using the equation x

a

+ y

b = 1 with a = âˆ’2 and b = âˆ’4, we have âˆ’x

2 âˆ’ y

4 = 1. Then âˆ’4x âˆ’ 2y = 8,

2y = âˆ’8 âˆ’ 4x, and finally y = âˆ’2x âˆ’ 4.

50. Using the equation x

a

+ y

b = 1 with a = âˆ’1

2 and b = 3

4 , we have x

âˆ’1/2 + y

3/4 = 1, 3

4 x âˆ’ 1

2 y =

r

âˆ’1

2

s r3

4

s

,

âˆ’1

2 y = âˆ’3

4 x âˆ’ 3

8 , and finally y = 2

r

3

4 x + 3

8

s

= 3

2 x + 3

4 .

51. Using the equation x

a

+ y

b = 1 with a = 4 and b = âˆ’1

2 , we have x

4 + y

âˆ’1/2 = 1, âˆ’1

4 x + 2y = âˆ’1, 2y = 1

4 x âˆ’ 1,

and so y = 1

8 x âˆ’ 1

2 .

52. The slope of the line passing through A and B is m = âˆ’2 âˆ’ 7

2 âˆ’ (âˆ’1) = âˆ’9

3 = âˆ’3, and the slope of the line passing

through B and C is m = âˆ’9 âˆ’ (âˆ’2)

5 âˆ’ 2 = âˆ’7

3

. Because the slopes are not equal, the points do not lie on the same line.

53. The slope of the line passing through A and B is m = 7 âˆ’ 1

1 âˆ’ (âˆ’2) = 6

3 = 2, and the slope of the line passing through

B and C is m = 13 âˆ’ 7

4 âˆ’ 1 = 6

3 = 2. Because the slopes are equal, the points lie on the same line.

54. a.

_20

0

20

40

_20 20 C

60

F b. The slope is 9

5 . It represents the change in â—¦F per unit change in â—¦C.

c. The F-intercept of the line is 32. It corresponds to 0 in â—¦C, so it is

the freezing point in â—¦F.

55. a.

20

40

60

80

0 5 10 15 t (years)

100

y (% of total capacity) b. The slope is 1.9467 and the y-intercept is 70.082.

c. The output is increasing at the rate of 1.9467% per year. The

output at the beginning of 1990 was 70.082%.

d. We solve the equation 1.9467t + 70.082 = 100, obtaining

1.9467t = 29.918 and t â‰ˆ 15.37. We conclude that the plants were

generating at maximum capacity during April 2005.

56. a. y = 0.0765x b. $0.0765 c. 0.0765 (65,000) = 4972.50, or $4972.50.

56 2 FUNCTIONS AND THEIR GRAPHS

57. a. y = 0.55x b. Solving the equation 1100 = 0.55x for x, we have x = 1100

0.55 = 2000.

58. a. Substituting L = 80 into the given equation, we have

W = 3.51 (80) âˆ’ 192 = 280.8 âˆ’ 192 = 88.8, or 88.8 British

tons.

b.

0

40

80

W (tons)

20 40 60 80 L (feet)

59. Using the points (0, 0.68) and (10, 0.80), we see that the slope of the required line is

m = 0.80 âˆ’ 0.68

10 âˆ’ 0 = 0.12

10 = 0.012. Next, using the point-slope form of the equation of a line, we have

y âˆ’ 0.68 = 0.012 (t âˆ’ 0) or y = 0.012t + 0.68. Therefore, when t = 14, we have y = 0.012 (14) + 0.68 = 0.848,

or 84.8%. That is, in 2004 womenâ€™s wages were 84.8% of menâ€™s wages.

60. a, b.

0

2

4

6

8

1234 x (years)

y (millions) c. Using P1 (0, 3.9) and P2 (4, 7.8), we find

m = 7.8 âˆ’ 3.9

4 âˆ’ 0 = 3.9

4 = 0.975. Thus, an equation is

y âˆ’ 3.9 = 0.975 (x âˆ’ 0) or y = 0.975x + 3.9.

d. If x = 3, then y = 0.975 (3) + 3.9 = 6.825. Thus, the

number of systems installed in 2005 (when x = 3) is

6,825,000 which is close to the projection of 6.8 million.

61. a, b.

0

5

10

15

20

1 23456 x (years)

y ($b) c. m = 18.8 âˆ’ 7.9

6 âˆ’ 0

â‰ˆ 1.82, so y âˆ’ 7.9 = 1.82 (x âˆ’ 0), or

y = 1.82x + 7.9.

d. y = 1.82 (5) + 7.9 â‰ˆ 17 or $17 billion. This agrees with

the actual data for that year.

62. a, b.

100

110

120

130

140

150

160

0 58 60 62 64 66 68 70 72 74 x (lb)

y (in.) c. Using the points (60, 108) and (72, 152), we see that the

slope of the required line is m = 152 âˆ’ 108

72 âˆ’ 60 = 44

12 = 11

3 .

Therefore, an equation is y âˆ’ 108 = 11

3 (x âˆ’ 60),

y = 11

3 x âˆ’ 11

3 (60) + 108 = 11

3 x âˆ’ 220 + 108, or

y = 11

3 x âˆ’ 112.

d. Using the equation from part c, we find

y = 11

3 (65) âˆ’ 112 = 1261

3 , or 1261

3 pounds.

2.2 EQUATIONS OF LINES 57

63. a, b.

190

200

210

220

230

240

250

0 20 40 60 80 100 x

y ($) c. Using the points (0, 200) and (100, 250), we see that the

slope of the required line is m = 250 âˆ’ 200

100 = 1

2

.

Therefore, an equation is y âˆ’ 200 = 1

2 x or y = 1

2 x + 200.

d. The approximate cost for producing 54 units of the

commodity is 1

2 (54) + 200, or $227.

64. a. Using the points (0, 5) and (4, 25), the slope of the required

line is m = 25 âˆ’ 5

4 âˆ’ 0 = 20

4 = 5. Next, using the point-slope

form of an equation of a line, we have y âˆ’ 5 = 5 (t âˆ’ 0) or

y = 5t + 5.

c. When t = 2, y = 5 (2) + 5 = 15. We conclude that 15 percent

of homes had digital TV services at the beginning of 2001.

b.

0

5

10

15

20

25

1 2345 t

y

65. a, b.

0

2

4

6

8

1 2345

10

x (years)

y ($m) c. The slope of L is m = 9.0 âˆ’ 5.8

5 âˆ’ 1 = 3.2

4 = 0.8. Using the

point-slope form of an equation of a line, we have

y âˆ’ 5.8 = 0.8 (x âˆ’ 1) = 0.8x âˆ’ 0.8, or y = 0.8x + 5.

d. Using the equation from part c with x = 9, we have

y = 0.8 (9) + 5 = 12.2, or $12.2 million.

66. True. The slope of the line is given by âˆ’2

4 = âˆ’1

2 .

67. False. Substituting x = âˆ’1 and y = 1 into the equation gives 3 (âˆ’1) + 7 (1) = 4, and this is not equal to the

right-hand side of the equation. Therefore, the equation is not satisfied and so the given point does not lie on the line.

68. True. If (1, k) lies on the line, then x = 1, y = k must satisfy the equation. Thus 3 + 4k = 12, or k = 9

4 .

Conversely, if k = 9

4 , then the point (1, k) =

r

1, 9

4

s

satisfies the equation. Thus, 3 (1) + 4

r

9

4

s

= 12, and so the

point lies on the line.

69. True. The slope of the line Ax + By + C = 0 is âˆ’ A

B . (Write it in slope-intercept form.) Similarly, the slope of the

line ax + by + c = 0 is âˆ’a

b

. They are parallel if and only if âˆ’ A

B = âˆ’a

b

, that is, if Ab = a B, or Ab âˆ’ aB = 0.

70. False. Let the slope of L1 be m1 > 0. Then the slope of L2 is m2 = âˆ’ 1

m1

< 0.

71. True. The slope of the line ax + by + c1 = 0 is m1 = âˆ’a

b

. The slope of the line bx âˆ’ ay + c2 = 0 is m2 = b

a

.

Because m1m2 = âˆ’1, the straight lines are indeed perpendicular.

58 2 FUNCTIONS AND THEIR GRAPHS

72. True. Set y = 0 and we have Ax + C = 0 or x = âˆ’C/A, and this is where the line intersects the x-axis.

73. Writing each equation in the slope-intercept form, we have y = âˆ’a1

b1

x âˆ’ c1

b1

(b1 /= 0) and y = âˆ’a2

b2

x âˆ’ c2

b2

(b2 /= 0). Because two lines are parallel if and only if their slopes are equal, we see that the lines are parallel if and

only if âˆ’a1

b1

= âˆ’a2

b2

, or a1b2 âˆ’ b1a2 = 0.

74. The slope of L1 is m1 = b âˆ’ 0

1 âˆ’ 0 = b. The slope of L2 is m2 = c âˆ’ 0

1 âˆ’ 0 = c. Applying the Pythagorean theorem to

O AC and OCB gives (O A)

2 = 12 + b2 and (O B)

2 = 12 + c2. Adding these equations and applying the

Pythagorean theorem to OBA gives (AB)

2 = (O A)

2 + (O B)

2 = 12 + b2 + 12 + c2 = 2 + b2 + c2. Also,

(AB)

2 = (b âˆ’ c)

2, so (b âˆ’ c)

2 = 2 + b2 + c2, b2 âˆ’ 2bc + c2 = 2 + b2 + c2, and âˆ’2bc = 2, 1 = âˆ’bc. Finally,

m1m2 = b Â· c = bc = âˆ’1, as was to be shown.

Technology Exercises page 89

Graphing Utility

1.

-10 -5 0 5 10

-10

-5

0

5

10 2.

-10 -5 0 5 10

-10

-5

0

5

10

3.

-10 -5 0 5 10

-10

-5

0

5

10 4.

-10 -5 0 5 10

-10

-5

0

5

10

5. a.

-10 -5 0 5 10

-10

-5

0

5

10 b.

-10 -5 0 5 10

-10

0

10

20

2.2 EQUATIONS OF LINES 59

6. a.

-10 -5 0 5 10

-10

-5

0

5

10 b.

-10 0 10 20

-10

-5

0

5

10

7. a.

-10 -5 0 5 10

-10

-5

0

5

10 b.

-10 0 10 20

-10

0

10

20

30

8. a.

-10 -5 0 5 10

-10

-5

0

5

10 b.

-10 0 10 20 30

-10

0

10

20

30

40

9.

-10 0 10 20 30 40

-10

0

10

20

30

10.

-10 0 10 20 30

-40

-20

0

20

60 2 FUNCTIONS AND THEIR GRAPHS

11.

-10 -5 0 5

-40

-30

-20

-10

0

10

12.

-5 0 5 10 15 20 25

-40

-30

-20

-10

0

10

Excel

1. 3.2x + 2.1y – 6.72 = 0

-15

-10

-5

0

5

10

15

20

-15 -10 -5 0 5 10 15

x

y

2. 2.3x – 4.1y – 9.43 = 0

-10

-8

-6

-4

-2

0

2

4

-15 -10 -5 0 5 10 15

x

y

3. 1.6x + 5.1y = 8.16

-2

-1

0

1

2

3

4

5

6

-15 -10 -5 0 5 10 15

x

y

4. -3.2x + 2.1y = 6.72

-15

-10

-5

0

5

10

15

20

-15 -10 -5 0 5 10 15

x

y

5. 12.1x + 4.1y = 49.61

-30

-20

-10

0

10

20

30

40

50

-15 -10 -5 0 5 10 15

x

y

6. 4.1x – 15.2y = 62.32

0

2

4

6

8

10

-20 -10 0 10 20 30 x

y

2.3 FUNCTIONS AND THEIR GRAPHS 61

7. 20x + 16y = 300

-10

0

10

20

30

40

-20 -10 0 10 20 30

x

y

8. 32.2x + 21y = 676.2

-20

-10

0

10

20

30

40

50

60

-20 -10 0 10 20 30 40

x

y

9. 20x + 30y = 600

-10

-5

0

5

10

15

20

25

30

-20 0 20 40 60

x

y

10. 30x – 20y = 600

-50

-40

-30

-20

-10

0

10

20

-20 -10 0 10 20 30 40

x

y

11. 22.4x + 16.1y = 352

-10

0

10

20

30

40

-20 -10 0 10 20 30

x

y

12. 18.2x – 15.1y = 274.8

-35

-30

-25

-20

-15

-10

-5

0

5

10

-20 -10 0 10 20 30

x

y

2.3 Functions and Their Graphs

Concept Questions page 96

1. a. A function is a rule that associates with each element in a set A exactly one element in a set B.

b. The domain of a function f is the set of all elements x in a set A such that f (x) is an element in B. The range of

f is the set of all elements f (x) such that x is an element in its domain.

c. An independent variable is a variable in the domain of a function f . The dependent variable is y = f (x).

62 2 FUNCTIONS AND THEIR GRAPHS

2. a. The graph of a function f is the set of all ordered pairs (x, y) such that y = f (x), x being an element in the

domain of f .

[ ]

Domain

x

y

y=f(x)

0

b. Use the vertical line test to determine if every vertical line intersects the curve in at most one point. If so, then

the curve is the graph of a function.

3. a. Yes, every vertical line intersects the curve in at most one point.

b. No, a vertical line intersects the curve at more than one point.

c. No, a vertical line intersects the curve at more than one point.

d. Yes, every vertical line intersects the curve in at most one point.

4. The domain is [1, 5) and the range is K

1

2 , 2

s

âˆª (2, 4].

Exercises page 97

1. f (x) = 5x + 6. Therefore f (3) = 5 (3) + 6 = 21, f (âˆ’3) = 5 (âˆ’3) + 6 = âˆ’9, f (a) = 5 (a) + 6 = 5a + 6,

f (âˆ’a) = 5 (âˆ’a) + 6 = âˆ’5a + 6, and f (a + 3) = 5 (a + 3) + 6 = 5a + 15 + 6 = 5a + 21.

2. f (x) = 4x âˆ’ 3. Therefore, f (4) = 4 (4) âˆ’ 3 = 16 âˆ’ 3 = 13, f

r

1

4

s

= 4

r

1

4

s

âˆ’ 3 = 1 âˆ’ 3 = âˆ’2,

f (0) = 4 (0) âˆ’ 3 = âˆ’3, f (a) = 4 (a) âˆ’ 3 = 4a âˆ’ 3, and f (a + 1) = 4 (a + 1) âˆ’ 3 = 4a + 1.

3. g (x) = 3×2 âˆ’ 6x + 3, so g (0) = 3 (0) âˆ’ 6 (0) + 3 = 3, g (âˆ’1) = 3 (âˆ’1)

2 âˆ’ 6 (âˆ’1) + 3 = 3 + 6 + 3 = 12,

g (a) = 3 (a)

2 âˆ’ 6 (a) + 3 = 3a2 âˆ’ 6a + 3, g (âˆ’a) = 3 (âˆ’a)

2 âˆ’ 6 (âˆ’a) + 3 = 3a2 + 6a + 3, and

g (x + 1) = 3 (x + 1)

2 âˆ’ 6 (x + 1) + 3 = 3

b

x2 + 2x + 1

c

âˆ’ 6x âˆ’ 6 + 3 = 3×2 + 6x + 3 âˆ’ 6x âˆ’ 3 = 3×2.

4. h (x) = x3 âˆ’ x2 + x + 1, so h (âˆ’5) = (âˆ’5)

3 âˆ’ (âˆ’5)

2 + (âˆ’5) + 1 = âˆ’125 âˆ’ 25 âˆ’ 5 + 1 = âˆ’154,

h (0) = (0)

3 âˆ’ (0)

2 + 0 + 1 = 1, h (a) = a3 âˆ’ (a)

2 + a + 1 = a3 âˆ’ a2 + a + 1, and

h (âˆ’a) = (âˆ’a)

3 âˆ’ (âˆ’a)

2 + (âˆ’a) + 1 = âˆ’a3 âˆ’ a2 âˆ’ a + 1.

5. f (x) = 2x + 5, so f (a + h) = 2 (a + h) + 5 = 2a + 2h + 5, f (âˆ’a) = 2 (âˆ’a) + 5 = âˆ’2a + 5,

f

b

a2

c = 2

b

a2

c

+ 5 = 2a2 + 5, f (a âˆ’ 2h) = 2 (a âˆ’ 2h) + 5 = 2a âˆ’ 4h + 5, and

f (2a âˆ’ h) = 2 (2a âˆ’ h) + 5 = 4a âˆ’ 2h + 5

6. g (x) = âˆ’x2 + 2x, g (a + h) = âˆ’ (a + h)

2 + 2 (a + h) = âˆ’a2 âˆ’ 2ah âˆ’ h2 + 2a + 2h,

g (âˆ’a) = âˆ’ (âˆ’a)

2 + 2 (âˆ’a) = âˆ’a2 âˆ’ 2a = âˆ’a (a + 2), g

bâˆša

c

= âˆ’ bâˆša

c2 + 2

bâˆša

c

= âˆ’a + 2

âˆša,

a + g (a) = a âˆ’ a2 + 2a = âˆ’a2 + 3a = âˆ’a (a âˆ’ 3), and

1

g (a) = 1

âˆ’a2 + 2a = âˆ’ 1

a (a âˆ’ 2)

.

2.3 FUNCTIONS AND THEIR GRAPHS 63

7. s (t) = 2t

t2 âˆ’ 1

. Therefore, s (4) = 2 (4)

(4)

2 âˆ’ 1 = 8

15, s (0) = 2 (0)

02 âˆ’ 1 = 0,

s (a) = 2 (a)

a2 âˆ’ 1 = 2a

a2 âˆ’ 1

, s (2 + a) = 2 (2 + a)

(2 + a)

2 âˆ’ 1 = 2 (2 + a)

a2 + 4a + 4 âˆ’ 1 = 2 (2 + a)

a2 + 4a + 3

, and

s (t + 1) = 2 (t + 1)

(t + 1)

2 âˆ’ 1 = 2 (t + 1)

t2 + 2t + 1 âˆ’ 1 = 2 (t + 1)

t (t + 2)

.

8. g (u) = (3u âˆ’ 2)

3/2. Therefore, g (1) = [3 (1) âˆ’ 2]

3/2 = (1)

3/2 = 1, g (6) = [3 (6) âˆ’ 2]

3/2 = 163/2 = 43 = 64,

g

r

11

3

s

=

K

3

r

11

3

s

âˆ’ 2

L3/2

= (9)

3/2 = 27, and g (u + 1) = [3 (u + 1) âˆ’ 2]

3/2 = (3u + 1)

3/2.

9. f (t) = 2t2

âˆšt âˆ’ 1

. Therefore, f (2) = 2

b

22

c

âˆš2 âˆ’ 1 = 8, f (a) = 2a2

âˆša âˆ’ 1

, f (x + 1) = 2 (x + 1)

2

âˆš(x + 1) âˆ’ 1 = 2 (x + 1)

2

âˆšx ,

and f (x âˆ’ 1) = 2 (x âˆ’ 1)

2

âˆš(x âˆ’ 1) âˆ’ 1 = 2 (x âˆ’ 1)

2

âˆšx âˆ’ 2 .

10. f (x) = 2 + 2

âˆš5 âˆ’ x. Therefore, f (âˆ’4) = 2 + 2

âˆš5 âˆ’ (âˆ’4) = 2 + 2

âˆš9 = 2 + 2 (3) = 8,

f (1) = 2 + 2

âˆš5 âˆ’ 1 = 2 + 2

âˆš

4 = 2 + 4 = 6, f

r

11

4

s

= 2 + 2

r

5 âˆ’ 11

4

s1/2

= 2 + 2

r

9

4

s1/2

= 2 + 2

r

3

2

s

= 5,

and f (x + 5) = 2 + 2

âˆš5 âˆ’ (x + 5) = 2 + 2

âˆšâˆ’x.

11. For x = âˆ’3 â‰¤ 0, we calculate f (âˆ’3) = (âˆ’3)

2 + 1 = 9 + 1 = 10. For x = 0 â‰¤ 0, we calculate

f (0) = (0)

2 + 1 = 1. For x = 1 > 0, we calculate f (1) = âˆš

1 = 1.

12. For x = âˆ’2 < 2, g (âˆ’2) = âˆ’1

2 (âˆ’2) + 1 = 1 + 1 = 2. For x = 0 < 2, g (0) = âˆ’1

2 (0) + 1 = 0 + 1 = 1. For

x = 2 â‰¥ 2, g (2) = âˆš2 âˆ’ 2 = 0. For x = 4 â‰¥ 2, g (4) = âˆš4 âˆ’ 2 = âˆš

2.

13. For x = âˆ’1 < 1, f (âˆ’1) = âˆ’1

2 (âˆ’1)

2 + 3 = 5

2 . For x = 0 < 1, f (0) = âˆ’1

2 (0)

2 + 3 = 3. For x = 1 â‰¥ 1,

f (1) = 2

b

12

c

+ 1 = 3. For x = 2 â‰¥ 1, f (2) = 2

b

22

c

+ 1 = 9.

14. For x = 0 â‰¤ 1, f (0) = 2 + âˆš1 âˆ’ 0 = 2 + 1 = 3. For x = 1 â‰¤ 1, f (1) = 2 + âˆš1 âˆ’ 1 = 2 + 0 = 2. For

x = 2 > 1, f (2) = 1

1 âˆ’ 2 = 1

âˆ’1 = âˆ’1.

15. a. f (0) = âˆ’2.

b. (i) f (x) = 3 when x â‰ˆ 2.

(ii) f (x) = 0 when x = 1.

c. [0, 6]

d. [âˆ’2, 6]

16. a. f (7) = 3.

b. x = 4 and x = 6.

c. x = 2; 0.

d. [âˆ’1, 9]; [âˆ’2, 6].

17. g (2) = âˆš

22 âˆ’ 1 = âˆš3, so the point r

2,

âˆš3

s

lies on the graph of g.

18. f (3) = 3 + 1

âˆš

32 + 7

+ 2 = 4

âˆš16

+ 2 = 4

4 + 2 = 3, so the point (3, 3) lies on the graph of f .

19. f (âˆ’2) = |âˆ’2 âˆ’ 1|

âˆ’2 + 1 = |âˆ’3|

âˆ’1 = âˆ’3, so the point (âˆ’2, âˆ’3) does lie on the graph of f .

64 2 FUNCTIONS AND THEIR GRAPHS

20. h (âˆ’3) = |âˆ’3 + 1|

(âˆ’3)

3 + 1 = 2

âˆ’27 + 1 = âˆ’ 2

26 = âˆ’ 1

13, so the point r

âˆ’3, âˆ’ 1

13s

does lie on the graph of h.

21. Because the point (1, 5) lies on the graph of f it satisfies the equation defining f . Thus,

f (1) = 2 (1)

2 âˆ’ 4 (1) + c = 5, or c = 7.

22. Because the point (2, 4) lies on the graph of f it satisfies the equation defining f . Thus,

f (2) = 2

S

9 âˆ’ (2)

2 + c = 4, or c = 4 âˆ’ 2

âˆš5.

23. Because f (x) is a real number for any value of x, the domain of f is (âˆ’âˆž,âˆž).

24. Because f (x) is a real number for any value of x, the domain of f is (âˆ’âˆž,âˆž).

25. f (x) is not defined at x = 0 and so the domain of f is (âˆ’âˆž, 0) âˆª (0,âˆž).

26. g (x) is not defined at x = 1 and so the domain of g is (âˆ’âˆž, 1) âˆª (1,âˆž).

27. f (x) is a real number for all values of x. Note that x2 + 1 â‰¥ 1 for all x. Therefore, the domain of f is (âˆ’âˆž,âˆž).

28. Because the square root of a number is defined for all real numbers greater than or equal to zero, we have x âˆ’ 5 â‰¥ 0

or x â‰¥ 5, and the domain is [5,âˆž).

29. Because the square root of a number is defined for all real numbers greater than or equal to zero, we have 5 âˆ’ x â‰¥ 0,

or âˆ’x â‰¥ âˆ’5 and so x â‰¤ 5. (Recall that multiplying by âˆ’1 reverses the sign of an inequality.) Therefore, the domain

of f is (âˆ’âˆž, 5].

30. Because 2×2 + 3 is always greater than zero, the domain of g is (âˆ’âˆž,âˆž).

31. The denominator of f is zero when x2 âˆ’ 4 = 0, or x = Â±2. Therefore, the domain of f is

(âˆ’âˆž, âˆ’2) âˆª (âˆ’2, 2) âˆª (2,âˆž).

32. The denominator of f is equal to zero when x2 + x âˆ’ 2 = (x + 2) (x âˆ’ 1) = 0; that is, when x = âˆ’2 or x = 1.

Therefore, the domain of f is (âˆ’âˆž, âˆ’2) âˆª (âˆ’2, 1) âˆª (1,âˆž).

33. f is defined when x + 3 â‰¥ 0, that is, when x â‰¥ âˆ’3. Therefore, the domain of f is [âˆ’3,âˆž).

34. g is defined when x âˆ’ 1 â‰¥ 0; that is when x â‰¥ 1. Therefore, the domain of g is [1,âˆž).

35. The numerator is defined when 1 âˆ’ x â‰¥ 0, âˆ’x â‰¥ âˆ’1 or x â‰¤ 1. Furthermore, the denominator is zero when x = Â±2.

Therefore, the domain is the set of all real numbers in (âˆ’âˆž, âˆ’2) âˆª (âˆ’2, 1].

36. The numerator is defined when x âˆ’ 1 â‰¥ 0, or x â‰¥ 1, and the denominator is zero when x = âˆ’2 and when x = 3. So

the domain is [1, 3) âˆª (3,âˆž).

2.3 FUNCTIONS AND THEIR GRAPHS 65

37. a. The domain of f is the set of all real numbers.

b. f (x) = x2 âˆ’ x âˆ’ 6, so

f (âˆ’3) = (âˆ’3)

2 âˆ’ (âˆ’3) âˆ’ 6 = 9 + 3 âˆ’ 6 = 6,

f (âˆ’2) = (âˆ’2)

2 âˆ’ (âˆ’2) âˆ’ 6 = 4 + 2 âˆ’ 6 = 0,

f (âˆ’1) = (âˆ’1)

2 âˆ’ (âˆ’1) âˆ’ 6 = 1 + 1 âˆ’ 6 = âˆ’4,

f (0) = (0)

2 âˆ’ (0) âˆ’ 6 = âˆ’6,

c.

0

10

_4 _2 2 4 x

y

5

_5

f

r

1

2

s

=

r

1

2

s2

âˆ’

r

1

2

s

âˆ’ 6 = 1

4 âˆ’ 2

4 âˆ’ 24

4 = âˆ’25

4 , f (1) = (1)

2 âˆ’ 1 âˆ’ 6 = âˆ’6,

f (2) = (2)

2 âˆ’ 2 âˆ’ 6 = 4 âˆ’ 2 âˆ’ 6 = âˆ’4, and f (3) = (3)

2 âˆ’ 3 âˆ’ 6 = 9 âˆ’ 3 âˆ’ 6 = 0.

38. f (x) = 2×2 + x âˆ’ 3.

a. Because f (x) is a real number for all values of x, the domain of f

is (âˆ’âˆž,âˆž).

b.

x âˆ’3 âˆ’2 âˆ’1 âˆ’1

2 0 1 2 3

y 12 3 âˆ’2 âˆ’3 âˆ’3 0 7 18

c.

_4

0

4

8

y

_3 _2 _1 1 2 x

39. f (x) = 2×2 + 1 has domain (âˆ’âˆž,âˆž) and range

[1,âˆž).

0

10

20

_3 _2 _1 1 2 x

y

40. f (x) = 9 âˆ’ x2 has domain (âˆ’âˆž,âˆž) and range

(âˆ’âˆž, 9].

_8

_4

0

4

8

_4 _2 2 x

y

41. f (x) = 2 + âˆšx has domain [0,âˆž) and range

[2,âˆž).

0

2

4

y

4 8 12 x

42. g (x) = 4 âˆ’ âˆšx has domain [0,âˆž)