SOLUTIONS MANUAL
ALGEBRA &
TRIGONOMETRY
FIFTH EDITION
R
U
SE
PRECALCULUS: A RIGHT
IS
TRIANGLE APPROACH
O
FIFTH EDITION
N
N
O
Judith A. Beecher
C
Judith A. Penna
ED
Marvin L. Bittinger
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,Chapter 1
Graphs, Functions, and Models
4. y
Exercise Set 1.1
4 (1, 4)
1. Point A is located 5 units to the left of the y-axis and 2
(5, 0) (4, 0)
4 units up from the x-axis, so its coordinates are (−5, 4). 4 2 2 4 x
Point B is located 2 units to the right of the y-axis and 2
(4, 2)
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2 units down from the x-axis, so its coordinates are (2, −2). 4 (2, 4)
Point C is located 0 units to the right or left of the y-axis
and 5 units down from the x-axis, so its coordinates are
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(0, −5). 5. To graph (−5, 1) we move from the origin 5 units to the
left of the y-axis. Then we move 1 unit up from the x-axis.
Point D is located 3 units to the right of the y-axis and
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5 units up from the x-axis, so its coordinates are (3, 5). To graph (5, 1) we move from the origin 5 units to the right
of the y-axis. Then we move 1 unit up from the x-axis.
Point E is located 5 units to the left of the y-axis and
4 units down from the x-axis, so its coordinates are To graph (2, 3) we move from the origin 2 units to the right
(−5, −4). of the y-axis. Then we move 3 units up from the x-axis.
Point F is located 3 units to the right of the y-axis and
0 units up or down from the x-axis, so its coordinates are
(3, 0). IS To graph (2, −1) we move from the origin 2 units to the
right of the y-axis. Then we move 1 unit down from the
x-axis.
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To graph (0, 1) we do not move to the right or the left of
2. G: (2, 1); H: (0, 0); I: (4, −3); J: (−4, 0); K: (−2, 3); the y-axis since the first coordinate is 0. From the origin
L: (0, 5) we move 1 unit up.
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3. To graph (4, 0) we move from the origin 4 units to the right
y
of the y-axis. Since the second coordinate is 0, we do not
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move up or down from the x-axis.
4
To graph (−3, −5) we move from the origin 3 units to the (2, 3)
2
left of the y-axis. Then we move 5 units down from the
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(5, 1) (0, 1) (5, 1)
x-axis. 4 2 4 x
To graph (−1, 4) we move from the origin 1 unit to the left 2 (2, 1)
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of the y-axis. Then we move 4 units up from the x-axis. 4
To graph (0, 2) we do not move to the right or the left of
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the y-axis since the first coordinate is 0. From the origin y
6.
we move 2 units up.
To graph (2, −2) we move from the origin 2 units to the 4
right of the y-axis. Then we move 2 units down from the (5, 2)
2
x-axis. (5, 0) (4, 0)
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4 2 2 4 x
y 2
4 (4, 3)
(1, 4) 4 (1, 5)
2 (0, 2)
(4, 0) 7. The first coordinate represents the year and the second co-
4 2 2 4 x ordinate represents the number of Sprint Cup Series races
2 (2, 2)
in which Tony Stewart finished in the top five. The or-
(3, 5) 4 dered pairs are (2008, 10), (2009, 15), (2010, 9), (2011, 9),
(2012, 12), and (2013, 5).
8. The first coordinate represents the year and the second
coordinate represents the percent of Marines who are
women. The ordered pairs are (1960, 1%), (1970, 0.9%),
(1980, 3.6%), (1990, 4.9%), (2000, 6.1%), and (2011, 6.8%).
Copyright
c 2016 Pearson Education, Inc.
,2 Chapter 1: Graphs, Functions, and Models
9. To determine whether (−1, −9) is a solution, substitute 12. For (1.5, 2.6): x2 + y 2 = 9
−1 for x and −9 for y.
(1.5)2 + (2.6)2 ? 9
y = 7x − 2
2.25 + 6.76
−9 ? 7(−1) − 2 9.01 9 FALSE
−7 − 2 (1.5, 2.6) is not a solution.
−9 −9 TRUE For (−3, 0): x2 + y 2 = 9
The equation −9 = −9 is true, so (−1, −9) is a solution. (−3)2 + 02 ? 9
To determine whether (0, 2) is a solution, substitute 0 for
9+0
x and 2 for y.
9 9 TRUE
y = 7x − 2
(−3, 0) is a solution.
2 ? 7 · 0 − 2 1 4
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13. To determine whether − , −
0−2 is a solution, substitute
2 5
2 −2 FALSE 1 4
− for a and − for b.
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The equation 2 = −2 is false, so (0, 2) is not a solution. 2 5
2a + 5b = 3
1 1 4
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10. For , 8 : y = −4x + 10
2 2 − +5 − ? 3
2 5
1
8 ? −4 · + 10 −1 − 4
2
−5 3 FALSE
−2 + 10
1
8 8
, 8 is a solution.
TRUE
IStion.
3
1 4
The equation −5 = 3 is false, so − , −
2 5
is not a solu-
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2 To determine whether 0, is a solution, substitute 0 for
5
For (−1, 6): y = −4x + 10 3
a and for b.
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5
6 ? −4(−1) + 10
2a + 5b = 3
4 + 10
3
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6 14 FALSE 2·0+5· ? 3
5
(−1, 6) is not a solution.
0+3
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2 3
11. To determine whether , is a solution, substitute
2 3 3 TRUE
3 4 3 3
3
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for x and for y. The equation 3 = 3 is true, so 0, is a solution.
4 5
6x − 4y = 1 3
14. For 0, : 3m + 4n = 6
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2 3 2
6· −4· ? 1 3
3 4 3·0+4· ? 6
2
4−3
0+6
1 1 TRUE
6 6 TRUE
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2 3
The equation 1 = 1 is true, so , is a solution. 3
3 4 0, is a solution.
3 2
To determine whether 1, is a solution, substitute 1 for 2
2
3 For ,1 : 3m + 4n = 6
x and for y. 3
2 2
6x − 4y = 1 3·
+4·1 ? 6
3
3
6·1−4· ? 1 2+4
2
6 6 TRUE
6−6 2
0 1 FALSE The equation 6 = 6 is true, so
3
, 1 is a solution.
3
The equation 0 = 1 is false, so 1, is not a solution.
2
Copyright
c 2016 Pearson Education, Inc.
, Exercise Set 1.1 3
15. To determine whether (−0.75, 2.75) is a solution, substi- y
(0, 5)
tute −0.75 for x and 2.75 for y.
4
x2 − y 2 = 3 5x 3y 15
2
(3, 0)
(−0.75)2 − (2.75)2 ? 3
4 2 2 4
x
0.5625 − 7.5625 2
−7 3 FALSE 4
The equation −7 = 3 is false, so (−0.75, 2.75) is not a
solution.
18. y
To determine whether (2, −1) is a solution, substitute 2
for x and −1 for y. 4
x2 − y 2 = 3 2
(4, 0)
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22 − (−1)2 ? 3 4 2 2 4 x
2
4−1 (0, 2)
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4
3 3 TRUE
The equation 3 = 3 is true, so (2, −1) is a solution. 2x 4y 8
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2
16. For (2, −4): 5x + 2y = 70 19. Graph 2x + y = 4.
5 · 2 + 2(−4)2 ? 70 To find the x-intercept we replace y with 0 and solve for
10 + 2 · 16 x.
10 + 32 2x + 0 = 4
(2, −4) is not a solution.
42 70 FALSE
IS 2x = 4
x=2
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For (4, −5): 5x + 2y 2 = 70 The x-intercept is (2, 0).
5 · 4 + 2(−5) ? 70
2 To find the y-intercept we replace x with 0 and solve for
y.
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20 + 2 · 25
2·0+y = 4
20 + 50
70 70 TRUE y=4
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(4, −5) is a solution. The y-intercept is (0, 4).
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17. Graph 5x − 3y = −15. We plot the intercepts and draw the line that contains
them. We could find a third point as a check that the
To find the x-intercept we replace y with 0 and solve for intercepts were found correctly.
x.
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5x − 3 · 0 = −15 y
5x = −15 2x y 4 4 (0, 4)
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x = −3 2
(2, 0)
The x-intercept is (−3, 0).
4 2 2 4 x
To find the y-intercept we replace x with 0 and solve for 2
y. 4
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5 · 0 − 3y = −15
−3y = −15
20. y
y=5
The y-intercept is (0, 5). 6 (0, 6)
We plot the intercepts and draw the line that contains 4
them. We could find a third point as a check that the 2
(2, 0)
intercepts were found correctly.
4 2 2 4 x
2
3x y 6
Copyright
c 2016 Pearson Education, Inc.
