Precalculus Enhanced with Graphing Utilities 4th Edition Solution Manual

Name: Precalculus Enhanced with Graphing Utilities 4th Edition Solution Manual
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Author: Samuel Clark

Solve your textbook questions with ease using Precalculus Enhanced with Graphing Utilities 4th Edition Solution Manual, a comprehensive and easy-to-follow guide.

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1
Chapter 1
Graphs
Section 1.1
1. 0
2. ( )5 3 8 8− − = =
3. 2 2
3 4 25 5+ = =
4. 2 2 2
11 60 121 3600 3721 61+ = + = =
Since the sum of the squares of two of the sides
of the triangle equals the square of the third side,
the triangle is a right triangle.
5. x-coordinate, or abscissa; y-coordinate, or
ordinate.
6. quadrants
7. midpoint
8. False; the distance between two points is never
negative.
9. False; points that lie in Quadrant IV will have a
positive x-coordinate and a negative y-
coordinate. The point ( )1, 4− lies in Quadrant II.
10. True; 1 2 1 2
,
2 2
x x y y
M + +⎛ ⎞
= ⎜ ⎟
⎝ ⎠
11. (a) Quadrant II
(b) Positive x-axis
(c) Quadrant III
(d) Quadrant I
(e) Negative y-axis
(f) Quadrant IV
12. (a) Quadrant I
(b) Quadrant III
(c) Quadrant II
(d) Quadrant I
(e) Positive y-axis
(f) Negative x-axis
13. The points will be on a vertical line that is two
units to the right of the y-axis.
14. The points will be on a horizontal line that is
three units above the x-axis.

Chapter 1: Graphs ISM: Precalculus EGU
2
15. ( )1, 4− ; Quadrant II
16. (3, 4); Quadrant I
17. (3, 1); Quadrant I
18. ( )6, 4− − ; Quadrant III
19. min 11
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25. min 6
max 6
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min 4
max 4
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max 2
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ISM: Precalculus EGU Chapter 1: Graphs
3
30. min 22
max 10
scl 2
min 4
max 8
scl 1
X
X
X
Y
Y
Y
= −
= −
=
=
=
=
31. 2 2
1 2( , ) (2 0) (1 0) 4 1 5d P P = − + − = + =
32. 2 2
1 2( , ) ( 2 0) (1 0) 4 1 5d P P = − − + − = + =
33. 2 2
1 2( , ) ( 2 1) (2 1) 9 1 10d P P = − − + − = + =
34. ( )2 2
1 2( , ) 2 ( 1) (2 1)
9 1 10
d P P = − − + −
= + =
35. ( )( ) ( )
2 22 2
1 2( , ) (5 3) 4 4 2 8
4 64 68 2 17
d P P = − + − − = +
= + = =
36. ( )( ) ( ) ( )
2 2 2 2
1 2( , ) 2 1 4 0 3 4
9 16 25 5
d P P = − − + − = +
= + = =
37. ( )2 2
1 2
2 2
( , ) 6 ( 3) (0 2)
9 ( 2)
81 4
85
d P P = − − + −
= + −
= +
=
38. ( ) ( )2 2 2 2
1 2( , ) 4 2 2 ( 3) 2 5
4 25 29
d P P = − + − − = +
= + =
39. ( )22 2 2
1 2( , ) (6 4) 4 ( 3) 2 7
4 49 53
d P P = − + − − = +
= + =
40. ( ) ( )2 2
1 2
2 2
( , ) 6 ( 4) 2 ( 3)
10 5 100 25
125 5 5
d P P = − − + − −
= + = +
= =
41. ( )2 2
1 2
2 2
( , ) 2.3 ( 0.2 (1.1 0.3)
(2.5) (0.8) 6.25 0.64
6.89 2.625
d P P = − − + −
= + = +
= ≈
42. ( )2 2
1 2
2 2
( , ) 0.3 1.2 (1.1 2.3)
( 1.5) ( 1.2) 2.25 1.44
3.69 1.92
d P P = − − + −
= − + − = +
= ≈
43. 2 2 2 2
1 2( , ) (0 ) (0 )d P P a b a b= − + − = +
44. 2 2 2 2
1 2
2
( , ) (0 ) (0 )
2 2
d P P a a a a
a a
= − + − = +
= =
45. ( ) ( )1 21,3 ; 5,15P P= =
( ) ( ) ( )
( ) ( )
2 2
1 2
2 2
, 5 1 15 3
4 12
16 144
160 4 10
d P P = − + −
= +
= +
= =
46. ( ) ( )1 28, 4 ; 2,3P P= − − =
( ) ( )( ) ( )( )
( ) ( )
2 2
1 2
2 2
, 2 8 3 4
10 7
100 49
149
d P P = − − + − −
= +
= +
=
47. ( ) ( )1 24, 6 ; 4, 8P P= − = −
( ) ( )( ) ( )
( ) ( )
2 2
1 2
2 2
, 4 4 8 6
8 14
64 196
260 2 65
d P P = − − + − −
= + −
= +
= =
48. ( ) ( )1 20, 6 ; 3, 8P P= = −
( ) ( ) ( )
( ) ( )
2 2
1 2
2 2
, 3 0 8 6
3 14
9 196
205
d P P = − + −

Chapter 1: Graphs ISM: Precalculus EGU
4
49. ( 2,5), (1,3), ( 1, 0)A B C= − = = −
( )
( )
( )
2 2
2 2
2 2
2 2
2 2
2 2
( , ) 1 ( 2) (3 5)
3 ( 2) 9 4
13
( , ) 1 1 (0 3)
( 2) ( 3) 4 9
13
( , ) 1 ( 2) (0 5)
1 ( 5) 1 25
26
d A B
d B C
d A C
= − − + −
= + − = +
=
= − − + −
= − + − = +
=
= − − − + −
= + − = +
=
Verifying that ∆ ABC is a right triangle by the
Pythagorean Theorem:
[ ] [ ] [ ]
( ) ( ) ( )
2 2 2
2 2 2
( , ) ( , ) ( , )
13 13 26
13 13 26
26 26
d A B d B C d A C+ =
+ =
+ =
=
The area of a triangle is 1
2
A bh= ⋅ . In this
problem,
[ ] [ ]1 ( , ) ( , )
2
1 13 13
2
1 13
2
13 square units2
A d A B d B C= ⋅⋅
= ⋅⋅
= ⋅
=
50. ( 2, 5), (12, 3), (10, 11)A B C= − = = −
( )
( )
( )
2 2
2 2
2 2
2 2
2 2
2 2
( , ) 12 ( 2) (3 5)
14 ( 2)
196 4 200
10 2
( , ) 10 12 ( 11 3)
( 2) ( 14)
4 196 200
10 2
( , ) 10 ( 2) ( 11 5)
12 ( 16)
144 256 400
20
d A B
d B C
d A C
= − − + −
= + −
= + =
=
= − + − −
= − + −
= + =
=
= − − + − −
= + −
= + =
=
Verifying that ∆ ABC is a right triangle by the
Pythagorean Theorem:
[ ] [ ] [ ]
( ) ( ) ( )
2 2 2
2 2 2
( , ) ( , ) ( , )
10 2 10 2 20
200 200 400
400 400
d A B d B C d A C+ =
+ =
+ =
=
The area of a triangle is 1
2
A bh= . In this
problem,
[ ] [ ]1 ( , ) ( , )
2
1 10 2 10 2
2
1 100 2
2
100 square units
A d A B d B C= ⋅ ⋅
= ⋅ ⋅
= ⋅ ⋅
=

ISM: Precalculus EGU Chapter 1: Graphs
5
51. ( 5,3), (6, 0), (5,5)A B C= − = =
( )
( )
( )
2 2
2 2
2 2
2 2
2 2
2 2
( , ) 6 ( 5) (0 3)
11 ( 3) 121 9
130
( , ) 5 6 (5 0)
( 1) 5 1 25
26
( , ) 5 ( 5) (5 3)
10 2 100 4
104
2 26
d A B
d B C
d A C
= − − + −
= + − = +
=
= − + −
= − + = +
=
= − − + −
= + = +
=
=
Verifying that ∆ ABC is a right triangle by the
Pythagorean Theorem:
[ ] [ ] [ ]
( ) ( ) ( )
2 2 2
2 2 2
( , ) ( , ) ( , )
104 26 130
104 26 130
130 130
d A C d B C d A B+ =
+ =
+ =
=
The area of a triangle is 1
2
A bh= . In this
problem,
[ ] [ ]1 ( , ) ( , )
2
1 104 26
2
1 2 26 26
2
1 2 26
2
26 square units
A d A C d B C= ⋅ ⋅
= ⋅ ⋅
= ⋅ ⋅
= ⋅ ⋅
=
52. ( 6, 3), (3, 5), ( 1, 5)A B C= − = − = −
( )
( )
( )
2 2
2 2
2 2
2 2
2 2
2 2
( , ) 3 ( 6) ( 5 3)
9 ( 8) 81 64
145
( , ) 1 3 (5 ( 5))
( 4) 10 16 100
116
( , ) 1 ( 6) (5 3)
5 2 25 4
29
d A B
d B C
d A C
= − − + − −
= + − = +
=
= − − + − −
= − + = +
=
= − − − + −
= + = +
=
Verifying that ∆ ABC is a right triangle by the
Pythagorean Theorem:
[ ] [ ] [ ]
( ) ( ) ( )
2 2 2
2 2 2
( , ) ( , ) ( , )
29 116 145
29 116 145
145 145
d A C d B C d A B+ =
+ =
+ =
=
The area of a triangle is 1
2
A bh= . In this
problem,
[ ] [ ]1 ( , ) ( , )
2
1 29 116
2
1 29 2 29
2
1 2 29
2
29 square units
A d A C d B C= ⋅ ⋅
= ⋅ ⋅
= ⋅ ⋅
= ⋅ ⋅
=

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Chapter 1: Graphs ISM: Precalculus EGU
6
53. (4, 3), (0, 3), (4, 2)A B C= − = − =
( )
( ) ( )
( )
22
2 2
2 2
2 2
22
2 2
( , ) (0 4) 3 ( 3)
( 4) 0 16 0
16
4
( , ) 4 0 2 ( 3)
4 5 16 25
41
( , ) (4 4) 2 ( 3)
0 5 0 25
25
5
d A B
d B C
d A C
= − + − − −
= − + = +
=
=
= − + − −
= + = +
=
= − + − −
= + = +
=
=
Verifying that ∆ ABC is a right triangle by the
Pythagorean Theorem:
[ ] [ ] [ ]
( )
2 2 2
2
2 2
( , ) ( , ) ( , )
4 5 41
16 25 41
41 41
d A B d A C d B C+ =
+ =
+ =
=
The area of a triangle is 1
2
A bh= . In this
problem,
[ ] [ ]1 ( , ) ( , )
2
1 4 5
2
10 square units
A d A B d A C= ⋅ ⋅
= ⋅ ⋅
=
54. (4, 3), (4, 1), (2, 1)A B C= − = =
( )
( ) ( )
( )
22
2 2
2 2
2 2
22
2 2
( , ) (4 4) 1 ( 3)
0 4
0 16
16
4
( , ) 2 4 1 1
( 2) 0 4 0
4
2
( , ) (2 4) 1 ( 3)
( 2) 4 4 16
20
2 5
d A B
d B C
d A C
= − + − −
= +
= +
=
=
= − + −
= − + = +
=
=
= − + − −
= − + = +
=
=
Verifying that ∆ ABC is a right triangle by the
Pythagorean Theorem:
[ ] [ ] [ ]
( )
2 2 2
2
2 2
( , ) ( , ) ( , )
4 2 2 5
16 4 20
20 20
d A B d B C d A C+ =
+ =
+ =
=
The area of a triangle is 1
2
A bh= . In this
problem,
[ ] [ ]1 ( , ) ( , )
2
1 4 2
2
4 square units
A d A B d B C= ⋅ ⋅
= ⋅ ⋅
=

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ISM: Precalculus EGU Chapter 1: Graphs
7
55. The coordinates of the midpoint are:
1 2 1 2
( , ) ,
2 2
4 43 5 ,
2 2
8 0
,
2 2
(4, 0)
x x y y
x y + +⎛ ⎞
= ⎜ ⎟
⎝ ⎠
− ++⎛ ⎞
= ⎜ ⎟
⎝ ⎠
⎛ ⎞
= ⎜ ⎟
⎝ ⎠
=
56. The coordinates of the midpoint are:
( )
1 2 1 2
( , ) ,
2 2
2 2 0 4
,
2 2
0 4
,
2 2
0, 2
x x y y
x y + +⎛ ⎞
= ⎜ ⎟
⎝ ⎠
− + +⎛ ⎞
= ⎜ ⎟
⎝ ⎠
⎛ ⎞
= ⎜ ⎟
⎝ ⎠
=
57. The coordinates of the midpoint are:
1 2 1 2
( , ) ,
2 2
3 6 2 0
,
2 2
3 2
,
2 2
3 ,1
2
x x y y
x y + +⎛ ⎞
= ⎜ ⎟
⎝ ⎠
− + +⎛ ⎞
= ⎜ ⎟
⎝ ⎠
⎛ ⎞
= ⎜ ⎟
⎝ ⎠
⎛ ⎞
= ⎜ ⎟
⎝ ⎠
58. The coordinates of the midpoint are:
1 2 1 2
( , ) ,
2 2
2 4 3 2
,
2 2
6 1
,
2 2
1
3, 2
x x y y
x y + +⎛ ⎞
= ⎜ ⎟
⎝ ⎠
+ − +⎛ ⎞
= ⎜ ⎟
⎝ ⎠
−⎛ ⎞
= ⎜ ⎟
⎝ ⎠
⎛ ⎞
= −⎜ ⎟
⎝ ⎠
59. The coordinates of the midpoint are:
1 2 1 2
( , ) ,
2 2
4 6 3 1
,
2 2
210 ,
2 2
(5, 1)
x x y y
x y + +⎛ ⎞
= ⎜ ⎟
⎝ ⎠
+ − +⎛ ⎞
= ⎜ ⎟
⎝ ⎠
−⎛ ⎞
= ⎜ ⎟
⎝ ⎠
= −
60. The coordinates of the midpoint are:
1 2 1 2
( , ) ,
2 2
4 2 3 2
,
2 2
2 1
,
2 2
1
1, 2
x x y y
x y + +⎛ ⎞
= ⎜ ⎟
⎝ ⎠
− + − +⎛ ⎞
= ⎜ ⎟
⎝ ⎠
− −⎛ ⎞
= ⎜ ⎟
⎝ ⎠
⎛ ⎞
= − −⎜ ⎟
⎝ ⎠
61. The coordinates of the midpoint are:
1 2 1 2
( , ) ,
2 2
0.2 2.3 0.3 1.1
,
2 2
2.1 1.4
,
2 2
(1.05, 0.7)
x x y y
x y + +⎛ ⎞
= ⎜ ⎟
⎝ ⎠
− + +⎛ ⎞
= ⎜ ⎟
⎝ ⎠
⎛ ⎞
= ⎜ ⎟
⎝ ⎠
=
62. The coordinates of the midpoint are:
( )
1 2 1 2
( , ) ,
2 2
1.2 ( 0.3) 2.3 1.1
,
2 2
0.9 3.4
,
2 2
0.45, 1.7
x x y y
x y + +⎛ ⎞
= ⎜

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Chapter 1: Graphs ISM: Precalculus EGU
8
64. The coordinates of the midpoint are:
1 2 1 2
( , ) ,
2 2
0 0
,
2 2
,
2 2
x x y y
x y
a a
a a
+ +⎛ ⎞
= ⎜ ⎟
⎝ ⎠
+ +⎛ ⎞
= ⎜ ⎟
⎝ ⎠
⎛ ⎞
= ⎜ ⎟
⎝ ⎠
65. Consider points of the form ( )2, y that are a
distance of 5 units from the point ( )2, 1− − .
( ) ( )
( ) ( )
( ) ( )
2 2
2 1 2 1
2 2
2 2
2
2
2 2 1
4 1
16 1 2
2 17
d x x y y
y
y
y y
y y
= − + −
= − − + − −
= − + − −
= + + +
= + +
( )
( )( )
2
2
2 2
2
2
5 2 17
5 2 17
25 2 17
0 2 8
0 4 2
y y
y y
y y
y y
y y
= + +
= + +
= + +
= + −
= + −
4 0
4
y
y
+ =
= −
or 2 0
2
y
y
− =
=
Thus, the points ( )2, 4− and ( )2, 2 are a
distance of 5 units from the point ( )2, 1− − .
66. Consider points of the form ( ), 3x − that are a
distance of 13 units from the point ( )1, 2 .
( ) ( )
( ) ( )( )
( )
2 2
2 1 2 1
22
22
2
2
1 2 3
2 1 5
2 1 25
2 26
d x x y y
x
x x
x x
x x
= − + −
= − + − −
= − + +
= − + +
= − +
( )
( )( )
2
2
2 2
2
2
13 2 26
13 2 26
169 2 26
0 2 143
0 13 11
x x
x x
x x
x x
x x
= − +
= − +
= − +
= − −
= − +
13 0
13
x
x
− =
=
or 11 0
11
x
x
+ =
= −
Thus, the points ( )13, 3− and ( )11, 3− − are a
distance of 13 units from the point ( )1, 2 .
67. Points on the x-axis have a y-coordinate of 0.
Thus, we consider points of the form ( ), 0x that
are a distance of 5 units from the point ( )4, 3− .
( ) ( )
( ) ( )
( )
2 2
2 1 2 1
2 2
22
2
2
4 3 0
16 8 3
16 8

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ISM: Precalculus EGU Chapter 1: Graphs
9
68. Points on the y-axis have an x-coordinate of 0.
Thus, we consider points of the form ( )0, y that
are a distance of 5 units from the point ( )4, 4 .
( ) ( )
( ) ( )
2 2
2 1 2 1
2 2
2 2
2
2
4 0 4
4 16 8
16 16 8
8 32
d x x y y
y
y y
y y
y y
= − + −
= − + −
= + − +
= + − +
= − +
( )
( )( )
2
2
2 2
2
2
5 8 32
5 8 32
25 8 32
0 8 7
0 7 1
y y
y y
y y
y y
y y
= − +
= − +
= − +
= − +
= − −
7 0
7
y
y
− =
=
or 1 0
1
y
y
− =
=
Thus, the points ( )0, 7 and ( )0,1 are on the y-
axis and a distance of 5 units from the point
( )4, 4 .
69. The midpoint of AB is:
( )
0 6 0 0
,
2 2
3, 0
D + +⎛ ⎞
= ⎜ ⎟
⎝ ⎠
=
The midpoint of AC is:
( )
0 4 0 4
,
2 2
2, 2
E + +⎛ ⎞
= ⎜ ⎟
⎝ ⎠
=
The midpoint of BC is:
( )
6 4 0 4
,
2 2
5, 2
F + +⎛ ⎞
= ⎜ ⎟
⎝ ⎠
=
( )2 2
2 2
( , ) 0 4 (3 4)
( 4) ( 1) 16 1
17
d C D = − + −
= − + − = +
=
( )2 2
2 2
( , ) 2 6 (2 0)
( 4) 2 16 4
20 2 5
d B E = − + −
= − + = +
= =
2 2
2 2
( , ) (2 0) (5 0)
2 5
4 25
29
d A F = − + −
= +
= +
=
70. Let 1 2(0, 0), (0, 4), ( , )P P P x y= = =
( )
( )
( )
2 2
1 2
2 2
1
2 2
2 2
2 2
2
2 2
2 2
, (0 0) (4 0)
16 4
, ( 0) ( 0)
4
16
, ( 0) ( 4)
( 4) 4
( 4) 16
d P P
d P P x y
x y
x y
d P P x y
x y
x y
= − + −
= =
= − + −
= + =
→ + =
= − + −
= + − =
→ + − =
Therefore,
( )22
2 2
4
8 16
8 16
2
y y
y y y
y
y
= −
= − +
=
=
which gives
2 2
2
2 16
12
2 3
x
x
x
+ =
=
= ±
Two triangles are possible. The third vertex is
( ) ( )2 3, 2 or 2 3, 2− .

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Chapter 1: Graphs ISM: Precalculus EGU
10
71. Let ( )1 0, 0P = , ( )2 0,P s= , ( )3 , 0P s= , and
( )4 ,P s s= .
(0,s)
(s,0)
(0,0)
(s,s)
X
Y
The points 1P and 4P are endpoints of one
diagonal and the points 2P and 3P are the
endpoints of the other diagonal.
1,4
0 0
, ,
2 2 2 2
s s s s
M + +⎛ ⎞ ⎛ ⎞
= =⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
2,3
0 0
, ,
2 2 2 2
s s s s
M + +⎛ ⎞ ⎛ ⎞
= =⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
The midpoints of the diagonals are the same.
Therefore, the diagonals of a square intersect at
their midpoints.
72. Let ( )1 0, 0P = , ( )2 , 0P a= , and
3
3
,
2 2
a a
P ⎛ ⎞
= ⎜ ⎟⎜ ⎟
⎝ ⎠
. To show that these vertices
form an equilateral triangle, we need to show
that the distance between any pair of points is the
same constant value.
( ) ( ) ( )
( ) ( )
2 2
1 2 2 1 2 1
2 2
2
,
0 0 0
d P P x x y y
a
a a
= − + −
= − + −
= =
( ) ( ) ( )2 2
2 3 2 1 2 1
22
2 2 2 2
,
3 0
2 2
3 4
4 4 4
d P P x x y y
a a
a
a a a a a

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ISM: Precalculus EGU Chapter 1: Graphs
11
73. 2 2
1 2
2 2
( , ) ( 4 2) (1 1)
( 6) 0
36
6
d P P = − − + −
= − +
=
=
( )2 2
2 3
2 2
( , ) 4 ( 4) ( 3 1)
0 ( 4)
16
4
d P P = − − − + − −
= + −
=
=
2 2
1 3
2 2
( , ) ( 4 2) ( 3 1)
( 6) ( 4)
36 16
52
2 13
d P P = − − + − −
= − + −
= +
=
=
Since [ ] [ ] [ ]2 22
1 2 2 3 1 3( , ) ( , ) ( , )d P P d P P d P P+ = ,
the triangle is a right triangle.
74. ( )2 2
1 2
2 2
( , ) 6 ( 1) (2 4)
7 ( 2)
49 4
53
d P P = − − + −
= + −
= +
=
( )2 2
2 3
2 2
( , ) 4 6 ( 5 2)
( 2) ( 7)
4 49
53
d P P = − + − −
= − + −
= +
=
( )2 2
1 3
2 2
( , ) 4 ( 1) ( 5 4)
5 ( 9)
25 81
106
d P P = − − + − −
= + −
= +
=
Since [ ] [ ] [ ]2 22
1 2 2 3 1 3( , ) ( , ) ( , )d P P d P P d P P+ = ,
the triangle is a right triangle.
Since ( ) ( )1 2 2 3, ,d P P d P P= , the triangle is
isosceles.
Therefore, the triangle is an isosceles right
triangle.
75. ( ) ( )2 2
1 2
2 2
( , ) 0 ( 2) 7 ( 1)
2 8 4 64 68
2 17
d P P = − − + − −
= + = + =
=
( )2 2
2 3
2 2
( , ) 3 0 (2 7)
3 ( 5) 9 25
34
d P P = − + −
= + − = +
=
( ) ( )2 2
1

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Chapter 1: Graphs ISM: Precalculus EGU
12
77. Using the Pythagorean Theorem:
2 2 2
2
2
90 90
8100 8100
16200
16200 90 2 127.28 feet
d
d
d
d
+ =
+ =
=
= = ≈
90
9090
90
d
78. Using the Pythagorean Theorem:
2 2 2
2 2
60 60
3600 3600 7200
7200 60 2 84.85 feet
d
d d
d
+ =
+ = → =
= = ≈
60
6060
60
d
79. a. First: (90, 0), Second: (90, 90)
Third: (0, 90)
(0,0)
(0,90)
(90,0)
(90,90)
X
Y
b. Using the distance formula:
2 2
2 2
(310 90) (15 90)
220 ( 75)
54025 232.4 feet
d = − + −
= + −
= ≈
c. Using the distance formula:
2 2
2 2
(300 0) (300 90)
300 210
134100 366.2 feet
d = − + −
= +
= ≈
80. a. First: (60, 0), Second: (60, 60)
Third: (0, 60)
(0,0)
(0,60)
(60,0)
(60,60)
x
y
b. Using the distance formula:
2 2
2 2
(180 60) (20 60)
120 ( 40)
16000 126.5 feet
d = − + −
= + −
= ≈
c. Using the distance formula:
2 2
2 2
(220 0) (220 60)
220 160
74000 272.0 feet
d = − + −
= +
= ≈
81. The Neon heading east moves a distance 30t
after t hours. The truck heading south moves a
distance 40t after t hours. Their distance apart
after t hours is:
2 2
2 2
2
(30 ) (40 )
900 1600
2500
50
d t t
t t
t
t
= +
= +
=
=
d
40t
30t

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ISM: Precalculus EGU Chapter 1: Graphs
13
82. 15 miles 5280 ft 1 hr 22 ft/sec
1 hr 1 mile 3600 sec
⋅ ⋅ =
( )22
2
100 22
10000 484
d t
t
= +
= +
100
22t
d
Section 1.2
1. ( )
( )
2 3 1 7
2 3 6
3 3
6
x
x
x
x
+ − = −
+ = −
+ = −
= −
The solution set is { }6− .
2.
( )( )
2
2 9 5 0
2 1 5 0
x x
x x
− − =
+ − =
2 1 0
2 1
1
2
x
x
x
+ =
= −
= −
or 5 0
5
x
x
− =
=
The solution set is { }1 ,5
2
− .
3. intercepts
4. zeros; roots
5. y-axis
6. 4
7. ( )3, 4−
8. True
9. False; we solve for y in terms of x.
10. False; a graph can be symmetric with respect to
both coordinate axes (in such cases it will also be
symmetric with respect to the origin).
For example: 2 2 1x y+ =
11.
12.
13.
14.

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Chapter 1: Graphs ISM: Precalculus EGU
14
15.
5
−5
y
5−5
a = (5, 2)
b = ( 5, 2)− −
c = ( 5, 2)−
(5, −2)
16.
17.
18.
19.
20.
21. 4
y x x= −
4
0 0 0
0 0
= −
=
4
1 1 1
1 0
= −
≠
4
0 ( 1) 1
0 1 1
= − − −
≠ − −
(0, 0) is on the graph of the equation.
22. 3 2y x x= −
3
0 0 2 0
0 0
= −
=
3
1 1 2 1
1 1
= −
≠ −
3
1 1 2 1
1 1
− = −
− = −
(0, 0) and (1, –1) are on the graph of the
equation.
23. 2 2 9y x= +
2 2
3 0 9
9 9
= +
=
2 2
0 3 9
0 18
= +
≠
2 2
0 ( 3) 9
0 18
= − +
≠
(0, 3) is on the graph of the equation.
24. 3 1y x= +
3
2 1 1
8 2
= +
≠
3
1 0 1
1 1
= +
=
3
0 1 1
0 0
= − +
=
(0, 1) and (–1, 0) are on the graph of the
equation.

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ISM: Precalculus EGU Chapter 1: Graphs
15
25. 2 2 4x y+ =
2 2
0 2 4
4 4
+ =
=
2 2
( 2) 2 4
8 4
− + =
≠
2 2
2 2 4
4 4
+ =
=
( )(0, 2) and 2, 2 are on the graph of the
equation.
26. 2 2
4 4x y+ =
2 2
0 4 1 4
4 4
+ ⋅ =
=
2 2
2 4 0 4
4 4
+ ⋅ =
=
( )22 1
2
2 4 4
5 4
+ =
≠
(0, 1) and (2, 0) are on the graph of the equation.
27. a. Intercepts: ( )1, 0− and ( )1, 0
b. Symmetric with respect to the x-axis, y-axis,
and the origin.
28. a. Intercepts: ( )0,1
b. Not symmetric to x-axis, y-axis, or origin
29. a. Intercepts: ( )2 0,
π− , ( )0,1 , and ( )2 , 0
π
b. Symmetric with respect to the y-axis.
30. a. Intercepts: ( )2, 0− , ( )0, 3− , and ( )2, 0
b. Symmetric with respect to the y-axis.
31. a. Intercepts: ( )0, 0
b. Symmetric with respect to the x-axis.
32. a. Intercepts: ( )2, 0− , ( )0, 2 , ( )0, 2− , and
( )2, 0
b. Symmetric with respect to the x-axis, y-axis,
and the origin.
33. a. Intercepts: ( )2, 0− , ( )0, 0 , and ( )2, 0
b. Symmetric with respect to the origin.
34. a. Intercepts: ( )4, 0− , ( )0, 0 , and ( )4, 0
b. Symmetric with respect to the origin.
35.
−10
y
x
5−5
(0, 9)−
10
(2, 5)−( 2, 5)−−
36. y
x
5
−5
(5, 3)
(5, 3)−
( 4, 0)−
−5 5
(0, 2)−
(0, 2)
37.
−5
y
x
5
π−π
, 2
2
π⎛ ⎞
− −⎜ ⎟
⎝ ⎠
, 2
2
π⎛ ⎞
⎜ ⎟
⎝ ⎠
(π, 0)(−π, 0)
38.

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