This document contains the solutions to odd-numbered exercises from Chapter P of a calculus textbook. It provides answers and work for 43 problems involving graphing functions, finding intercepts, determining symmetries, and other skills related to functions and their graphs. The problems progress from simple linear functions to more complex expressions involving square roots, cubes, and other operations.

1. PA R T I
C H A P T E R P
Preparation for Calculus
Section P.1 Graphs and Models . . . . . . . . . . . . . . . . . . . . . . 2
Section P.2 Linear Models and Rates of Change . . . . . . . . . . . . . 7
Section P.3 Functions and Their Graphs . . . . . . . . . . . . . . . . . 14
Section P.4 Fitting Models to Data . . . . . . . . . . . . . . . . . . . . 18
Review Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

2. C H A P T E R P
Preparation for Calculus
Section P.1 Graphs and Models
Solutions to Odd-Numbered Exercises
1
1. y ϭ Ϫ 2 x ϩ 2 3. y ϭ 4 Ϫ x2
x-intercept: ͑4, 0͒ x-intercepts: ͑2, 0͒, ͑Ϫ2, 0͒
y-intercept: ͑0, 2͒ y-intercept: ͑0, 4͒
Matches graph (b) Matches graph (a)
5. y ϭ 3x ϩ 1
2 7. y ϭ 4 Ϫ x2
x Ϫ4 Ϫ2 0 2 4 x Ϫ3 Ϫ2 0 2 3
y Ϫ5 Ϫ2 1 4 7 y Ϫ5 0 4 0 Ϫ5
y y
8 6
(4, 7)
6 (0, 4)
4 (2, 4)
2
2 (−2, 0)
(0, 1) (2, 0)
x x
−8 −6 −4 2 4 6 8 −6 −4 4 6
(− 2, −2) −2
(−4, − 5) −4
(− 3, − 5) −4 (3, − 5)
−6
−8 −6
9. y ϭ x ϩ 2Խ Խ 11. y ϭ Ίx Ϫ 4
x Ϫ5 Ϫ4 Ϫ3 Ϫ2 Ϫ1 0 1 x 0 1 4 9 16
y 3 2 1 0 1 2 3 y Ϫ4 Ϫ3 Ϫ2 Ϫ1 0
y y
10
6 8
6
4 4
(−5, 3)
(1, 3) 2 (4, − 2) (16, 0)
(−4, 2) 2 (0, 2) x
−2 2 12 14 16 18
(−3, 1) (−1, 1) (9, − 1)
x −4 (1, − 3)
−6 −4 (− 2, 0) 2 −6 (0, − 4)
−8
−2
− 10
2

3. Section P.1 Graphs and Models 3
13. 15. 5
Xmin = -3
(− 4.00, 3)
Xmax = 5 (2, 1.73)
Xscl = 1 −6 6
Ymin = -3
Ymax = 5
−3
Yscl = 1
(a) ͑2, y͒ ϭ ͑2, 1.73͒ ͑ y ϭ Ί5 Ϫ 2 ϭ Ί3 Ϸ 1.73͒
Note that y ϭ 4 when x ϭ 0.
(b) ͑x, 3͒ ϭ ͑Ϫ4, 3͒ ͑ 3 ϭ Ί5 Ϫ ͑Ϫ4͒ ͒
17. y ϭ x2 ϩ x Ϫ 2 19. y ϭ x2Ί25 Ϫ x2
y-intercept: y ϭ 02 ϩ 0 Ϫ 2 y-intercept: y ϭ 02Ί25 Ϫ 02
y ϭ Ϫ2; ͑0, Ϫ2͒ y ϭ 0; ͑0, 0͒
x-intercepts: 0 ϭ x2 ϩ x Ϫ 2 x-intercepts: 0 ϭ x2Ί25 Ϫ x2
0 ϭ ͑x ϩ 2͒͑x Ϫ 1͒ 0 ϭ x2Ί͑5 Ϫ x͒͑5 ϩ x͒
x ϭ Ϫ2, 1; ͑Ϫ2, 0͒, ͑1, 0͒ x ϭ 0, ± 5; ͑0, 0͒; ͑± 5, 0͒
3͑2 Ϫ Ίx ͒
21. y ϭ 23. x2y Ϫ x2 ϩ 4y ϭ 0
x
y-intercept:
y-intercept: None. x cannot equal 0.
02͑y͒ Ϫ 02 ϩ 4y ϭ 0
3͑2 Ϫ Ίx͒
x-intercepts: 0ϭ
x y ϭ 0; ͑0, 0͒
0 ϭ 2 Ϫ Ίx x-intercept:
x ϭ 4; ͑4, 0͒ x2͑0͒ Ϫ x2 ϩ 4͑0͒ ϭ 0
x ϭ 0; ͑0, 0͒
25. Symmetric with respect to the y-axis since 27. Symmetric with respect to the x-axis since
y ϭ ͑Ϫx͒ Ϫ 2 ϭ x Ϫ 2.
2 2
͑Ϫy͒2 ϭ y2 ϭ x3 Ϫ 4x.
29. Symmetric with respect to the origin since 31. y ϭ 4 Ϫ Ίx ϩ 3
͑Ϫx͒͑Ϫy͒ ϭ xy ϭ 4. No symmetry with respect to either axis or the origin.
33. Symmetric with respect to the origin since Խ Խ
35. y ϭ x3 ϩ x is symmetric with respect to the y-axis
Ϫx
Ϫy ϭ
͑Ϫx͒2 ϩ 1
Խ Խ Խ
since y ϭ ͑Ϫx͒3 ϩ ͑Ϫx͒ ϭ Ϫ ͑x3 ϩ x͒ ϭ x3 ϩ x . Խ Խ Խ
x
yϭ .
x2 ϩ 1
37. y ϭ Ϫ3x ϩ 2 y
Intercepts:
͑ 2 , 0͒, ͑0, 2͒
3
2 (0, 2)
1 2
Symmetry: none 3, 0
x
1 2 3
1

4. 4 Chapter P Preparation for Calculus
x
39. y ϭ Ϫ4 41. y ϭ 1 Ϫ x2 43. y ϭ ͑x ϩ 3͒2
2
Intercepts: Intercepts:
Intercepts:
͑1, 0͒, ͑Ϫ1, 0͒, ͑0, 1͒ ͑Ϫ3, 0͒, ͑0, 9͒
͑8, 0͒, ͑0, Ϫ4͒
Symmetry: y-axis Symmetry: none
Symmetry: none
y
y y
2 12
2
(8, 0) 10
x
(0, 1)
(0, 9)
2 2 4 8 10 8
2 ( 1, 0) (1, 0)
x
−2 2
(0, 4)
6 1
2
8 2 x
− 10 − 8 − 6 (− 3, 0) 2 4
10 −2
45. y ϭ x3 ϩ 2 47. y ϭ xΊx ϩ 2 49. x ϭ y3
Intercepts: Intercepts: Intercepts: ͑0, 0͒
͑ Ϫ Ί2, 0͒, ͑0, 2͒
3
͑0, 0͒, ͑Ϫ2, 0͒ Symmetry: origin
Symmetry: none Symmetry: none y
4
y Domain: x ≥ Ϫ2 3
5 2
y
4 (0, 0)
x
3 6 −4 −3 −2 −1 1 2 3 4
5
3
( 2, 0) −2
(0, 2) 4
1 −3
3
−4
x 2
3 2 1 2 3
1 (− 2, 0) 1
(0, 0)
x
−4 −3 −1 1 2 3 4
−2
1
51. y ϭ
x
y 53. y ϭ 6 Ϫ x ԽԽ y
3 8
Intercepts: none Intercepts: 6 (0, 6)
2
4
Symmetry: origin 1 ͑0, 6͒, ͑Ϫ6, 0͒, ͑6, 0͒ (− 6, 0) 2
(6, 0)
x x
1 2 3 Symmetry: y-axis −8 −4 −2 2 4 6 8
−2
−4
−6
−8
55. y2 Ϫ x ϭ 9 57. x ϩ 3y2 ϭ 6
y2 ϭ x ϩ 9 3y2 ϭ 6 Ϫ x
y ϭ ± Ίx ϩ 9
Ί2 Ϫ 3
4
x
(0, 3) yϭ± 3
Intercepts: (−9, 0) (0, 2 )
− 11 1
͑0, 3͒, ͑0, Ϫ3͒, ͑Ϫ9, 0͒ Intercepts: −1
(6, 0)
8
Symmetry: x-axis
(0, − 3)
͑6, 0͒, ͑0, Ί2͒, ͑0, Ϫ Ί2͒ (0, − 2 )
−4
−3
Symmetry: x-axis

5. Section P.1 Graphs and Models 5
59. y ϭ ͑x ϩ 2͒͑x Ϫ 4͒͑x Ϫ 6͒ (other answers possible) 61. Some possible equations:
yϭx
y ϭ x3
y ϭ 3x3 Ϫ x
y ϭ Ίx
3
63. xϩyϭ2⇒yϭ2Ϫx 65. xϩyϭ7⇒yϭ7Ϫx
2x Ϫ y ϭ 1 ⇒ y ϭ 2x Ϫ 1 3x Ϫ 11
3x Ϫ 2y ϭ 11 ⇒ y ϭ
2
2 Ϫ x ϭ 2x Ϫ 1
3x Ϫ 11
3 ϭ 3x 7Ϫxϭ
2
1ϭx
14 Ϫ 2x ϭ 3x Ϫ 11
The corresponding y-value is y ϭ 1.
Ϫ5x ϭ Ϫ25
Point of intersection: ͑1, 1͒ xϭ5
The corresponding y-value is y ϭ 2.
Point of intersection: ͑5, 2͒
67. x2 ϩ y ϭ 6 ⇒ y ϭ 6 Ϫ x2 69. x2 ϩ y 2 ϭ 5 ⇒ y 2 ϭ 5 Ϫ x 2
xϩyϭ4⇒yϭ4Ϫx xϪyϭ1⇒yϭxϪ1
6 Ϫ x2 ϭ 4 Ϫ x 5 Ϫ x2 ϭ ͑x Ϫ 1͒2
0 ϭ x2 Ϫ x Ϫ 2 5 Ϫ x2 ϭ x2 Ϫ 2x ϩ 1
0 ϭ ͑x Ϫ 2͒͑x ϩ 1͒ 0 ϭ 2x2 Ϫ 2x Ϫ 4 ϭ 2͑x ϩ 1͒͑x Ϫ 2͒
x ϭ 2, Ϫ1 x ϭ Ϫ1 or x ϭ 2
The corresponding y-values are y ϭ 2 (for x ϭ 2) The corresponding y-values are y ϭ Ϫ2 and y ϭ 1.
and y ϭ 5 (for x ϭ Ϫ1).
Points of intersection: ͑Ϫ1, Ϫ2͒, ͑2, 1͒
Points of intersection: ͑2, 2͒, ͑Ϫ1, 5͒
71. y ϭ x3 73. y ϭ x3 Ϫ 2x2 ϩ x Ϫ 1
yϭx y ϭ Ϫx2 ϩ 3x Ϫ 1
x3 ϭ x x3 Ϫ 2x2 ϩ x Ϫ 1 ϭ Ϫx2 ϩ 3x Ϫ 1
x3 Ϫ x ϭ 0 x3 Ϫ x2 Ϫ 2x ϭ 0
x͑x ϩ 1͒͑x Ϫ 1͒ ϭ 0 x͑x Ϫ 2͒͑x ϩ 1͒ ϭ 0
x ϭ 0, x ϭ Ϫ1, or x ϭ 1 x ϭ Ϫ1, 0, 2
The corresponding y-values are y ϭ 0, y ϭ Ϫ1, and ͑Ϫ1, Ϫ5͒, ͑0, Ϫ1͒, ͑2, 1͒
y ϭ 1.
4
y = x 3 − 2x 2 + x − 1
Points of intersection: ͑0, 0͒, ͑Ϫ1, Ϫ1͒, ͑1, 1͒
(2, 1)
−4 6
(0, −1)
(−1, −5)
−8
y = −x 2 + 3 x − 1

6. 6 Chapter P Preparation for Calculus
75. 5.5Ίx ϩ 10,000 ϭ 3.29x
͑ 5.5Ίx ͒ 2 ϭ ͑3.29x Ϫ 10,000͒2
30.25x ϭ 10.8241x2 Ϫ 65,800x ϩ 100,000,000
0 ϭ 10.8241x2 Ϫ 65,830.25x ϩ 100,000,000 Use the Quadratic Formula.
x Ϸ 3133 units
The other root, x Ϸ 2949, does not satisfy the equation R ϭ C.
This problem can also be solved by using a graphing utility and finding the intersection of the graphs of C and R.
77. (a) Using a graphing utility, you obtain (b) 250
y ϭ Ϫ0.0153t2 ϩ 4.9971t ϩ 34.9405
(c) For the year 2004, t ϭ 34 and
−5
y Ϸ 187.2 CPI. 35
− 50
79. 400
0 100
0
If the diameter is doubled, the resistance is changed by approximately a factor of ͑1͞4͒. For instance, y͑20͒ Ϸ 26.555 and
y͑40͒ Ϸ 6.36125.
81. False; x-axis symmetry means that if ͑1, Ϫ2͒ is on the graph, then ͑1, 2͒ is also on the graph.
83. True; the x-intercepts are
΂Ϫb ± Ίb2 Ϫ 4ac
2a
,0 .΃
85. Distance to the origin ϭ K ϫ Distance to ͑2, 0͒
Ίx2 ϩ y2 ϭ KΊ͑x Ϫ 2͒2 ϩ y2, K 1
x2 ϩ y 2 ϭ K 2͑x2 Ϫ 4x ϩ 4 ϩ y2͒
͑1 Ϫ K 2͒ x 2 ϩ ͑1 Ϫ K 2͒y 2 ϩ 4K 2x Ϫ 4K 2 ϭ 0
Note: This is the equation of a circle!

7. Section P.2 Linear Models and Rates of Change 7
Section P.2 Linear Models and Rates of Change
1. m ϭ 1 3. m ϭ 0 5. m ϭ Ϫ12
2 Ϫ ͑Ϫ4͒ 5Ϫ1
7. y
9. m ϭ 11. m ϭ
5Ϫ3 2Ϫ2
5 m=1
4 6 4
ϭ ϭ3 ϭ
3 (2, 3) 2 0
2
y
m = −3 undefined
1 m is 2
undefined 3
x y
2 (5, 2)
1 3 4 5
−1 m = −2
1
6
x
−1 1 2 3 5 6 7 5 (2, 5)
4
−2
3
−3
2
−4 (3, − 4)
1 (2, 1)
−5
x
−2 −1 1 3 4 5 6
−1
−2
2͞3 Ϫ 1͞6 y
13. m ϭ
Ϫ1͞2 Ϫ ͑Ϫ3͞4͒ 3
2
1͞2 (− 1 , 2 )
ϭ ϭ2 2 3 (− 3 , 1 )
4 6
1͞4 x
−3 −2 1 2 3
−1
−2
−3
15. Since the slope is 0, the line is horizontal and its equation is y ϭ 1. Therefore, three additional points are ͑0, 1͒, ͑1, 1͒,
and ͑3, 1͒.
17. The equation of this line is
y Ϫ 7 ϭ Ϫ3͑x Ϫ 1͒
y ϭ Ϫ3x ϩ 10 .
Therefore, three additional points are ͑0, 10͒, ͑2, 4͒, and ͑3, 1͒.
19. Given a line L, you can use any two distinct points to calculate its slope. Since a line is straight, the ratio of the change in
y-values to the change in x-values will always be the same. See Section P.2 Exercise 93 for a proof.

8. 8 Chapter P Preparation for Calculus
21. (a) (b) The slopes of the line segments are
Population (in millions)
270
255.0 Ϫ 252.1
ϭ 2.9
260 2Ϫ1
257.7 Ϫ 255.0
250
ϭ 2.7
3Ϫ2
1 2 3 4 5 6 7 8 9 260.3 Ϫ 257.7
Year (0 ↔ 1990) ϭ 2.6
4Ϫ3
262.8 Ϫ 260.3
ϭ 2.5
5Ϫ4
265.2 Ϫ 262.8
ϭ 2.4
6Ϫ5
267.7 Ϫ 265.2
ϭ 2.5
7Ϫ6
270.3 Ϫ 267.7
ϭ 2.6
8Ϫ7
The population increased most rapidly from 1991 to 1992.
͑m ϭ 2.9͒
23. x ϩ 5y ϭ 20 25. x ϭ 4
yϭ Ϫ1 x
5 ϩ4 The line is vertical. Therefore, the slope is undefined and
1 there is no y-intercept.
Therefore, the slope is m ϭ Ϫ5 and the y-intercept is
͑0, 4͒.
27. y ϭ 3x ϩ 3
4 29. y ϭ 2x
3 31. y ϩ 2 ϭ 3͑x Ϫ 3͒
4y ϭ 3x ϩ 12 3y ϭ 2x y ϩ 2 ϭ 3x Ϫ 9
0 ϭ 3x Ϫ 4y ϩ 12 2x Ϫ 3y ϭ 0 y ϭ 3x Ϫ 11
y y y Ϫ 3x ϩ 11 ϭ 0
5 4
y
4 3
(0, 3) 3
2 2
2
1
1 x
(0, 0)
x −2 −1 1 2 3 4 5 6
x −1
−4 −3 −2 −1 1 1 2 3 4
−1 −2 (3, − 2)
−3
−4
−5
6Ϫ0 1 Ϫ ͑Ϫ3͒ 8Ϫ0 8
33. m ϭ ϭ3 35. m ϭ ϭ2 37. m ϭ ϭϪ
2Ϫ0 2Ϫ0 2Ϫ5 3
y Ϫ 0 ϭ 3͑x Ϫ 0͒ 8
y Ϫ 1 ϭ 2͑x Ϫ 2͒ y Ϫ 0 ϭ Ϫ ͑x Ϫ 5͒
3
y ϭ 3x y Ϫ 1 ϭ 2x Ϫ 4 8 40
yϭϪ xϩ
y 0 ϭ 2x Ϫ y Ϫ 3 3 3
8
y
3y ϩ 8x Ϫ 40 ϭ 0
6 (2, 6)
4 2 y
2 1 (2, 1)
(0, 0) 9
x x
−8 −6 −4 −2 2 4 6 8 8 (2, 8)
−2 −1 2 3 4 5 7
−1
6
−2 5
−3 4
−8 (0, −3) 3
2
−5 1 (5, 0)
x
−1 1 2 3 4 6 7 8 9
−2

9. Section P.2 Linear Models and Rates of Change 9
8Ϫ1 7͞2 Ϫ 3͞4 11͞4 11
39. m ϭ Undefined. 41. m ϭ ϭ ϭ 43. xϭ3
5Ϫ5 1͞2 Ϫ 0 1͞2 2
xϪ3ϭ0
Vertical line x ϭ 5 3 11
yϪ ϭ ͑x Ϫ 0͒
4 2 y
y
11 3 2
9
8 (5, 8) yϭ xϩ
7
2 4 1
6
5 22x Ϫ 4y ϩ 3 ϭ 0 (3, 0)
x
4 1 2 4
3
2 y −1
1 (5, 1)
x 4 −2
−1 1 2 3 4 6 7 8 9
3
(1 , 7)
2 2
−2
2
1 ( 0, 3 )
4
x
−4 −3 −2 −1 1 2 3 4
x y x y
45. ϩ ϭ1 47. ϩ ϭ1
2 3 a a
3x ϩ 2y Ϫ 6 ϭ 0 1 2
ϩ ϭ1
a a
3
ϭ1
a
aϭ3⇒xϩyϭ3
xϩyϪ3ϭ0
49. y ϭ Ϫ3 51. y ϭ Ϫ2x ϩ 1
yϩ3ϭ0 y
3
y
2
1
x
−3 −2 −1 1 2 3 4 5
x
−2 −2 −1 1 2
−1
−4
−5
−6
53. y Ϫ 2 ϭ 3͑x Ϫ 1͒
2 55. 2x Ϫ y Ϫ 3 ϭ 0
yϭ 3
2x ϩ 1
2 y ϭ 2x Ϫ 3
2y Ϫ 3x Ϫ 1 ϭ 0 y
y 1
4 x
2 1 2 3
3
1
2
1 2
x 3
−4 −3 −2 1 2 3 4
−2
−3
−4

10. 10 Chapter P Preparation for Calculus
57. 10 10
− 10 10 − 15 15
− 10 − 10
The lines do not appear perpendicular. The lines appear perpendicular.
The lines are perpendicular because their slopes 1 and Ϫ1 are negative reciprocals of each other.
You must use a square setting in order for perpendicular lines to appear perpendicular.
59. 4x Ϫ 2y ϭ 3 61. 5x Ϫ 3y ϭ 0
y ϭ 2x Ϫ 2
3
y ϭ 5x
3
mϭ2 mϭ5
3
(a) y Ϫ 1 ϭ 2͑x Ϫ 2͒ (a) y Ϫ 7 ϭ 5 ͑x Ϫ 3 ͒
8 3 4
y Ϫ 1 ϭ 2x Ϫ 4 24y Ϫ 21 ϭ 40x Ϫ 30
2x Ϫ y Ϫ 3 ϭ 0 24y Ϫ 40x ϩ 9 ϭ 0
(b) yϪ1ϭ Ϫ 2 ͑x
1
Ϫ 2͒ (b) y Ϫ 7 ϭ Ϫ 3 ͑x Ϫ 3 ͒
8 5 4
2y Ϫ 2 ϭ Ϫx ϩ 2 40y Ϫ 35 ϭ Ϫ24x ϩ 18
x ϩ 2y Ϫ 4 ϭ 0 40y ϩ 24x Ϫ 53 ϭ 0
63. (a) x ϭ 2 ⇒ x Ϫ 2 ϭ 0
(b) y ϭ 5 ⇒ y Ϫ 5 ϭ 0
65. The slope is 125. Hence, V ϭ 125͑t Ϫ 1͒ ϩ 2540
ϭ 125t ϩ 2415
67. The slope is Ϫ2000. Hence, V ϭ Ϫ2000͑t Ϫ 1͒ ϩ 20,400
ϭ Ϫ2000t ϩ 22,400
69. 5
(2, 4)
−3 6
(0, 0)
−1
You can use the graphing utility to determine that the points of intersection are ͑0, 0͒ and ͑2, 4͒. Analytically,
x2 ϭ 4x Ϫ x2
2x2 Ϫ 4x ϭ 0
2x͑x Ϫ 2͒ ϭ 0
x ϭ 0 ⇒ y ϭ 0 ⇒ ͑0, 0͒
x ϭ 2 ⇒ y ϭ 4 ⇒ ͑2, 4͒.
The slope of the line joining ͑0, 0͒ and ͑2, 4͒ is m ϭ ͑4 Ϫ 0͒͑͞2 Ϫ 0͒ ϭ 2. Hence, an equation of the line is
y Ϫ 0 ϭ 2͑x Ϫ 0͒
y ϭ 2x.

11. Section P.2 Linear Models and Rates of Change 11
1Ϫ0
71. m1 ϭ ϭ Ϫ1
Ϫ2 Ϫ ͑Ϫ1͒
Ϫ2 Ϫ 0 2
m2 ϭ ϭϪ
2 Ϫ ͑Ϫ1͒ 3
m1 m2
The points are not collinear.
y
73. Equations of perpendicular bisectors:
aϪb aϩb
yϪ
c
2
ϭ
c
xϪ
2 ΂ ΃ (b, c)
aϩb bϪa
yϪ
c
2
ϭ
Ϫc
xϪ
2 ΂ ΃ ( b − a , 2c )
2
( a + b , 2c )
2
x
(− a, 0) (a, 0)
Letting x ϭ 0 in either equation gives the point of inter-
section:
΂0, Ϫa ϩ b2 ϩ c2
΃
2
.
2c
This point lies on the third perpendicular bisector, x ϭ 0.
75. Equations of altitudes: y
aϪb
yϭ ͑x ϩ a͒
c
(b, c)
xϭb
aϩb (a, 0)
yϭϪ ͑x Ϫ a͒ (− a, 0)
x
c
Solving simultaneously, the point of intersection is
Ϫ b2
΂b, a ΃
2
.
c
77. Find the equation of the line through the points ͑0, 32͒ and ͑100, 212͒.
m ϭ 180 ϭ 9
100 5
F Ϫ 32 ϭ 5 ͑C Ϫ 0͒
9
F ϭ 9 C ϩ 32
5
5F Ϫ 9C Ϫ 160 ϭ 0
For F ϭ 72Њ, C Ϸ 22.2Њ.
79. (a) W1 ϭ 0.75x ϩ 12.50 (b) 50
W2 ϭ 1.30x ϩ 9.20
(c) Both jobs pay $17 per hour if 6 units are produced. (6, 17)
For someone who can produce more than 6 units per 0 30
hour, the second offer would pay more. For a worker 0
who produces less than 6 units per hour, the first offer
pays more. Using a graphing utility, the point of intersection is
approximately ͑6, 17͒. Analytically,
0.75x ϩ 12.50 ϭ 1.30x ϩ 9.20
3.3 ϭ 0.55x ⇒ x ϭ 6
y ϭ 0.75͑6͒ ϩ 12.50 ϭ 17.

12. 12 Chapter P Preparation for Calculus
81. (a) Two points are ͑50, 580͒ and ͑47, 625͒. The slope is (b) 50
625 Ϫ 580
mϭ ϭ Ϫ15.
47 Ϫ 50
p Ϫ 580 ϭ Ϫ15͑x Ϫ 50͒ 0
0
1500
p ϭ Ϫ15x ϩ 750 ϩ 580 ϭ Ϫ15x ϩ 1330
If p ϭ 655, x ϭ 15 ͑1330 Ϫ 655͒ ϭ 45 units.
1
or x ϭ 15 ͑1330 Ϫ p͒
1
(c) If p ϭ 595, x ϭ 15 ͑1330 Ϫ 595͒ ϭ 49 units.
1
83. 4x ϩ 3y Ϫ 10 ϭ 0 ⇒ d ϭ Խ4͑0͒ ϩ 3͑0͒ Ϫ 10Խ ϭ 10 ϭ 2
Ί42 ϩ 32 5
85. x Ϫ y Ϫ 2 ϭ 0 ⇒ d ϭ Խ1͑Ϫ2͒ ϩ ͑Ϫ1͒͑1͒ Ϫ 2Խ ϭ 5
ϭ
5Ί2
Ί1 ϩ 1
2 2 Ί2 2
87. A point on the line x ϩ y ϭ 1 is ͑0, 1͒. The distance from the point ͑0, 1͒ to x ϩ y Ϫ 5 ϭ 0 is
dϭ Խ1͑0͒ ϩ 1͑1͒ Ϫ 5Խ ϭ Խ1 Ϫ 5Խ ϭ 4
ϭ 2Ί2.
Ί12 ϩ 12 Ί2 Ί2
89. If A ϭ 0, then By ϩ C ϭ 0 is the horizontal line y ϭ ϪC͞B. The distance to ͑x1, y1͒ is
Խ
d ϭ y1 Ϫ
B Խ
΂ϪC ΃ ϭ Խ ԽBԽ Խ ϭ Խ ΊA ϩ B Խ.
By ϩ C 1 Ax ϩ By ϩ C 1
2
1
2
If B ϭ 0, then Ax ϩ C ϭ 0 is the vertical line x ϭ ϪC͞A. The distance to ͑x1, y1͒ is
Խ
d ϭ x1 Ϫ
A Խ
΂ϪC ΃ ϭ Խ ԽAԽ Խ ϭ Խ ΊA ϩ B Խ.
Ax ϩ C 1 Ax ϩ By ϩ C 1
2
1
2
(Note that A and B cannot both be zero.)
The slope of the line Ax ϩ By ϩ C ϭ 0 is ϪA͞B. The equation of the line through ͑x1, y1͒ perpendicular
to Ax ϩ By ϩ C ϭ 0 is:
B
y Ϫ y1 ϭ ͑x Ϫ x1͒
A
Ay Ϫ Ay1 ϭ Bx Ϫ Bx1
Bx1 Ϫ Ay1 ϭ Bx Ϫ Ay
The point of intersection of these two lines is:
Ax ϩ By ϭ ϪC ⇒ A2x ϩ ABy ϭ ϪAC (1)
Bx Ϫ Ay ϭ Bx1 Ϫ Ay1 ⇒ B 2x Ϫ ABy ϭ B2x 1 Ϫ ABy1 (2)
͑A2 ϩ B2͒x ϭ ϪAC ϩ B2x1 Ϫ ABy1 (By adding equations (1) and (2))
ϪAC ϩ B2x1 Ϫ ABy1
xϭ
A2 ϩ B2
Ax ϩ By ϭ ϪC ⇒ ABx ϩ B2y ϭ ϪBC (3)
Bx Ϫ Ay ϭ Bx1 Ϫ Ay1⇒ ϪABx ϩ A2 y ϭ ϪABx1 ϩ A2 y1 (4)
͑A2 ϩ B2͒y ϭ ϪBC Ϫ ABx1 ϩ A2y1 (By adding equations (3) and (4))
ϪBC Ϫ ABx1 ϩ A2y1
yϭ
A2 ϩ B2
—CONTINUED—

13. Section P.2 Linear Models and Rates of Change 13
89. —CONTINUED—
΂ϪAC ϩ Bϩx BϪ ABy , ϪBC Ϫ ABxB ϩ A y ΃ point of intersection
2 2
1 1 1 1
A 2 A ϩ 2 2 2
The distance between ͑x1, y1͒ and this point gives us the distance between ͑x1, y1͒ and the line Ax ϩ By ϩ C ϭ 0.
Ί΄ ϪAC ϩ Bϩx BϪ ABy Ϫ x ΅ ϩ ΄ ϪBC Ϫ ABxB ϩ A y ΅
2 2 2 2
dϭ 1 1 1 1
Ϫ y1
A A ϩ 2 2 1 2 2
ϭ Ί΄
ϪAC Ϫ ABy Ϫ A x
΅ ϩ ΄ ϪBC Ϫ ABxB Ϫ B y ΅
2 2 2 2
1 1 1 1
A ϩB A ϩ 2 2 2 2
ϭΊ΄ ϪA͑C Aϩ ϩ Bϩ Ax ͒΅ ϩ ΄ ϪB͑C Aϩ ϩ Bϩ By ͒΅
By Ax
2
1
2
1
2
2
1
2
1
2
ϭΊ
͑A ϩ B ͒͑C ϩ Ax ϩ By ͒
2 2
1 1
2
͑A ϩ B ͒ 2 2 2
ϭ ԽAx1 ϩ By1 ϩ CԽ
ΊA2 ϩ B2
91. For simplicity, let the vertices of the rhombus be ͑0, 0͒,
y
͑a, 0͒, ͑b, c͒, and ͑a ϩ b, c͒, as shown in the figure. The
slopes of the diagonals are then (b, c) (a + b , c )
c c
m1 ϭ and m2 ϭ .
aϩb bϪa
x
(0, 0) (a, 0)
Since the sides of the Rhombus are equal, a2 ϭ b2 ϩ c2,
and we have
c c c2 c2
m1m2 ϭ
aϩb
и b Ϫ a ϭ b2 Ϫ a2 ϭ Ϫc2 ϭ Ϫ1.
Therefore, the diagonals are perpendicular.
93. Consider the figure below in which the four points are 95. True.
collinear. Since the triangles are similar, the result imme-
a c a
diately follows. ax ϩ by ϭ c1 ⇒ y ϭ Ϫ x ϩ 1 ⇒ m1 ϭ Ϫ
b b b
y2‫ ء‬Ϫ y1‫ ء‬y2 Ϫ y1
ϭ b c b
x2‫ ء‬Ϫ x1‫ ء‬x2 Ϫ x1 bx Ϫ ay ϭ c2 ⇒ y ϭ x Ϫ 2 ⇒ m2 ϭ
a a a
y
1
m2 ϭ Ϫ
(x 2 , y2 ) (x * , y* )
2 2
m1
(x1, y1 )
(x *, y* )
1 1
x

14. 14 Chapter P Preparation for Calculus
Section P.3 Functions and Their Graphs
1. (a) f ͑0͒ ϭ 2͑0͒ Ϫ 3 ϭ Ϫ3 3. (a) g͑0͒ ϭ 3 Ϫ 02 ϭ 3
(b) g͑Ί3͒ ϭ 3 Ϫ ͑Ί3͒ ϭ 3 Ϫ 3 ϭ 0
2
(b) f ͑Ϫ3͒ ϭ 2͑Ϫ3͒ Ϫ 3 ϭ Ϫ9
(c) f ͑b͒ ϭ 2b Ϫ 3 (c) g͑Ϫ2͒ ϭ 3 Ϫ ͑Ϫ2͒2 ϭ 3 Ϫ 4 ϭ Ϫ1
(d) f ͑x Ϫ 1͒ ϭ 2͑x Ϫ 1͒ Ϫ 3 ϭ 2x Ϫ 5 (d) g͑t Ϫ 1͒ ϭ 3 Ϫ ͑t Ϫ 1͒2 ϭ Ϫt2 ϩ 2t ϩ 2
5. (a) f ͑0͒ ϭ cos͑2͑0͒͒ ϭ cos 0 ϭ 1 ΂ ␲΃ ϭ cos΂2΂Ϫ ␲΃΃ ϭ cos΂Ϫ ␲΃ ϭ 0
(b) f Ϫ
4 4 2
(c) f ΂␲΃ ϭ cos΂2΂␲΃΃ ϭ cos23␲ ϭ Ϫ 1
3 3 2
f ͑x ϩ ⌬x͒ Ϫ f ͑x͒ ͑x ϩ ⌬x͒3 Ϫ x3 x3 ϩ 3x2⌬x ϩ 3x͑⌬x͒2 ϩ ͑⌬x͒3 Ϫ x3
7. ϭ ϭ ϭ 3x2 ϩ 3x⌬x ϩ ͑⌬x͒2, ⌬x 0
⌬x ⌬x ⌬x
f ͑x͒ Ϫ f ͑2͒ ͑1͞Ίx Ϫ 1 Ϫ 1͒
9. ϭ
xϪ2 xϪ2
1 Ϫ Ίx Ϫ 1 1 ϩ Ίx Ϫ 1 2Ϫx Ϫ1
ϭ и ϭ ϭ
͑x Ϫ 2͒Ίx Ϫ 1 1 ϩ Ίx Ϫ 1 ͑x Ϫ 2͒Ίx Ϫ 1͑1 ϩ Ίx Ϫ 1͒ Ίx Ϫ 1͑1 ϩ Ίx Ϫ 1͒
,x 2
␲t
11. h͑x͒ ϭ Ϫ Ίx ϩ 3 13. f ͑t͒ ϭ sec
4
Domain: x ϩ 3 ≥ 0 ⇒ ͓Ϫ3, ϱ͒
␲t ͑2k ϩ 1͒␲
⇒t 4k ϩ 2
Range: ͑Ϫ ϱ, 0͔ 4 2
Domain: all t 4k ϩ 2, k an integer
Range: ͑Ϫ ϱ, Ϫ1͔, ͓1, ϱ͒
1
15. f ͑x͒ ϭ
x
Domain: ͑Ϫ ϱ, 0͒, ͑0, ϱ͒
Range: ͑Ϫ ϱ, 0͒, ͑0, ϱ͒
17. f ͑x͒ ϭ Ά2x ϩ 1, xx < 0
2x ϩ 2, ≥ 0
19. f ͑x͒ ϭ ΆϪxԽ ϩ 1, xx < 1
Խx
ϩ 1, ≥ 1
(a) f ͑Ϫ1͒ ϭ 2͑Ϫ1͒ ϩ 1 ϭ Ϫ1 Խ Խ
(a) f ͑Ϫ3͒ ϭ Ϫ3 ϩ 1 ϭ 4
(b) f ͑0͒ ϭ 2͑0͒ ϩ 2 ϭ 2 (b) f ͑1͒ ϭ Ϫ1 ϩ 1 ϭ 0
(c) f ͑2͒ ϭ 2͑2͒ ϩ 2 ϭ 6 (c) f ͑3͒ ϭ Ϫ3 ϩ 1 ϭ Ϫ2
(d) f ͑t ϩ 1͒ ϭ 2͑t ϩ 1͒ ϭ 2t ϩ 4
2 2 2
(d) f ͑b ϩ 1͒ ϭ Ϫ ͑b ϩ 1͒ ϩ 1 ϭ Ϫb
2 2 2
(Note: t2 ϩ 1 ≥ 0 for all t) Domain: ͑Ϫ ϱ, ϱ͒
Domain: ͑Ϫ ϱ, ϱ͒ Range: ͑Ϫ ϱ, 0͔ ʜ ͓1, ϱ͒
Range: ͑Ϫ ϱ, 1͒, ͓2, ϱ͒

15. Section P.3 Functions and Their Graphs 15
21. f ͑x͒ ϭ 4 Ϫ x y
23. h͑x͒ ϭ Ίx Ϫ 1 y
Domain: ͑Ϫ ϱ, ϱ͒ 8 Domain: ͓1, ϱ͒ 2
6
Range: ͑Ϫ ϱ, ϱ͒ Range: ͓0, ϱ͒ 1
x
2 1 2 3
x
4 2 2 4
25. f ͑x͒ ϭ Ί9 Ϫ x2 y
27. g͑t͒ ϭ 2 sin ␲ t y
4
Domain: ͓Ϫ3, 3͔ Domain: ͑Ϫ ϱ, ϱ͒
2
2
1
Range: ͓0, 3͔ Range: ͓Ϫ2, 2͔
x t
4 2 2 4 2 3
2 −1
29. x Ϫ y 2 ϭ 0 ⇒ y ϭ ± Ίx 31. y is a function of x. Vertical lines intersect the graph
at most once.
y is not a function of x. Some vertical lines intersect
the graph twice.
33. x2 ϩ y2 ϭ 4 ⇒ y ϭ ± Ί4 Ϫ x2 35. y2 ϭ x2 Ϫ 1 ⇒ y ϭ ± Ίx2 Ϫ 1
y is not a function of x since there are two values of y for y is not a function of x since there are two values of y for
some x. some x.
ԽԽ Խ
37. f ͑x͒ ϭ x ϩ x Ϫ 2 Խ
If x < 0, then f ͑x͒ ϭ Ϫx Ϫ ͑x Ϫ 2͒ ϭ Ϫ2x ϩ 2 ϭ 2͑1 Ϫ x͒.
If 0 ≤ x < 2, then f ͑x͒ ϭ x Ϫ ͑x Ϫ 2͒ ϭ 2.
If x ≥ 2, then f ͑x͒ ϭ x ϩ ͑x Ϫ 2͒ ϭ 2x Ϫ 2 ϭ 2͑x Ϫ 1͒.
Thus,
Ά
2͑1 Ϫ x͒, x < 0
f ͑x͒ ϭ 2, 0 ≤ x < 2.
2͑x Ϫ 1͒, x ≥ 2.
39. The function is g͑x͒ ϭ cx2. Since ͑1, Ϫ2͒ satisfies the 41. The function is r͑x͒ ϭ c͞x, since it must be undefined at
equation, c ϭ Ϫ2. Thus, g͑x͒ ϭ Ϫ2x2. x ϭ 0. Since ͑1, 32͒ satisfies the equation, c ϭ 32. Thus,
r͑x͒ ϭ 32͞x.
43. (a) For each time t, there corresponds a depth d. 45. d
(b) Domain: 0 ≤ t ≤ 5
27
Range: 0 ≤ d ≤ 30 18
(c) d
9
30
25 t
t1 t2 t3
20
15
10
5
t
1 2 3 4 5 6

16. 16 Chapter P Preparation for Calculus
y y
47. (a) The graph is shifted (b) The graph is shifted
3 units to the left. 4 1 unit to the right. 4
2
x x
−6 −4 −2 2 4 −2 2 4 6 8
−2 −2
−4 −4
−6 −6
y y
(c) The graph is shifted (d) The graph is shifted
2 units upward. 6 4 units downward. x
−4 −2 2 4 6
4 −2
2 −4
x −6
−4 −2 2 4 6
−2 −8
y y
(e) The graph is stretched (f) The graph is stretched
vertically by a factor of 3. x vertically by a factor 4
−4 −2 4 6
−2
of 1.
4 2
−4 x
−4 −2 2 4 6
−6
−8
− 10 −6
49. (a) y ϭ Ίx ϩ 2 (b) y ϭ Ϫ Ίx (c) y ϭ Ίx Ϫ 2
y y y
4
4 1
3
3 x
1 2 3 4 2
1 1
2
x
2 1 2 3 4 5 6
1 −1
3 −2
x
1 2 3 4
Vertical shift 2 units upward Reflection about the x-axis Horizontal shift 2 units to the
right
51. (a) T͑4͒ ϭ 16Њ, T͑15͒ Ϸ 23Њ
(b) If H͑t͒ ϭ T͑t Ϫ 1͒, then the program would turn on (and off) one hour later.
(c) If H͑t͒ ϭ T͑t͒ Ϫ 1, then the overall temperature would be reduced 1 degree.
3
53. f ͑x͒ ϭ x2, g͑x͒ ϭ Ίx 55. f ͑x͒ ϭ , g͑x͒ ϭ x2 Ϫ 1
x
͑ f Њ g͒͑x͒ ϭ f ͑g͑x͒͒ ϭ f ͑ Ίx ͒ ϭ ͑ Ίx ͒ ϭ x, x ≥ 0
2
3
͑ f Њ g͒͑x͒ ϭ f ͑g͑x͒͒ ϭ f ͑x2 Ϫ 1͒ ϭ
Domain: ͓0, ϱ͒ x2 Ϫ 1
͑g Њ f ͒͑x͒ ϭ g͑ f ͑x͒͒ ϭ g͑x2͒ ϭ Ίx2 ϭ ԽxԽ Domain: all x ±1
9 Ϫ x2
Domain: ͑Ϫ ϱ, ϱ͒
΂3΃ ϭ ΂3΃
2
9
͑g Њ f ͒͑x͒ ϭ g͑ f ͑x͒͒ ϭ g Ϫ1ϭ 2
Ϫ1ϭ
x x x x2
No. Their domains are different. ͑ f Њ g͒ ϭ ͑g Њ f ͒ for x ≥ 0.
Domain: all x 0
No, f Њ g g Њ f.

17. Section P.3 Functions and Their Graphs 17
57. ͑A Њ r͒͑t͒ ϭ A͑r͑t͒͒ ϭ A͑0.6t͒ ϭ ␲͑0.6t͒2 ϭ 0.36␲t 2 59. f ͑Ϫx͒ ϭ ͑Ϫx͒2͑4 Ϫ ͑Ϫx͒2͒ ϭ x2͑4 Ϫ x2͒ ϭ f ͑x͒
͑A Њ r͒͑t͒ represents the area of the circle at time t. Even
61. f ͑Ϫx͒ ϭ ͑Ϫx͒ cos͑Ϫx͒ ϭ Ϫx cos x ϭ Ϫf ͑x͒
Odd
63. (a) If f is even, then ͑ 2 , 4͒ is on the graph. (b) If f is odd, then ͑ 2 , Ϫ4͒ is on the graph.
3 3
65. f ͑Ϫx͒ ϭ a2nϩ1͑Ϫx͒2nϩ1 ϩ . . . ϩ a3͑Ϫx͒3 ϩ a1͑Ϫx͒
ϭ Ϫ ͓a2nϩ1x2nϩ1 ϩ . . . ϩ a3x3 ϩ a1x͔
ϭ Ϫf ͑x͒
Odd
67. Let F ͑x͒ ϭ f ͑x͒g͑x͒ where f and g are even. Then
F ͑Ϫx͒ ϭ f ͑Ϫx͒g͑Ϫx͒ ϭ f ͑x͒g͑x͒ ϭ F ͑x͒.
Thus, F ͑x͒ is even. Let F ͑x͒ ϭ f ͑x͒g͑x͒ where f and g are odd. Then
F ͑Ϫx͒ ϭ f ͑Ϫx͒g͑Ϫx͒ ϭ ͓Ϫf ͑x͔͓͒Ϫg͑x͔͒ ϭ f ͑x͒g͑x͒ ϭ F ͑x͒.
Thus, F ͑x͒ is even.
69. f ͑x͒ ϭ x2 ϩ 1 and g͑x͒ ϭ x4 are even. f ͑x͒ ϭ x3 Ϫ x is odd and g͑x͒ ϭ x2 is even.
f ͑x͒g͑x͒ ϭ ͑x2 ϩ 1͒͑x4͒ ϭ x6 ϩ x4 is even. f ͑x͒g͑x͒ ϭ ͑x3 Ϫ x͒͑x2͒ ϭ x5 Ϫ x3 is odd.
5 4
−6 6
−4 4
−1 −4
71. (a) x length and width volume V (b) 1200
1 24 Ϫ 2͑1͒ 484
2 24 Ϫ 2͑2͒ 800
0 7
3 24 Ϫ 2͑3͒ 972 0
4 24 Ϫ 2͑4͒ 1024 Yes, V is a function of x.
5 24 Ϫ 2͑5͒ 980 (d) 1100
6 24 Ϫ 2͑6͒ 864
The maximum volume appears to be 1024 cm3.
−1 12
(c) V ϭ x͑24 Ϫ 2x͒2 ϭ 4x͑12 Ϫ x͒2 − 100
Domain: 0 < x < 12 Maximum volume is V ϭ 1024 cm3 for box having
dimensions 4 ϫ 16 ϫ 16 cm.
73. False; let f ͑x͒ ϭ x2. 75. True, the function is even.
Then f ͑Ϫ3͒ ϭ f ͑3͒ ϭ 9, but Ϫ3 3.