solution manual for College Algebra Graphs and Models 7th edition Marvin L. Bittinger
Chapter 1
Graphs, Functions, and Models
Check Your Understanding Section 1.1
1. The point (−5,0) is on an axis, so it is not in any quadrant.
The statement is false.
2. The ordered pair (1,−6) is located 1 unit right of the origin
and 6 units below it. The ordered pair (−6,1) is located 6
units left of the origin and 1 unit above it. Thus, (1,−6)
and (−6,1) do not name the same point. The statement
is false.
3. True; the first coordinate of a point is also called the
abscissa.
4. True; the point (−2,7) is 2 units left of the origin and
7 units above it.
5. True; the second coordinate of a point is also called the
ordinate.
6. False; the point (0,−3) is on the y-axis.
Exercise Set 1.1
1. Point A is located 5 units to the left of the y-axis and
4 units up from the x-axis, so its coordinates are (−5,4).
Point B is located 2 units to the right of the y-axis and
2 units down from the x-axis, so its coordinates are (2,−2).
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
(0,−5).
Point D is located 3 units to the right of the y-axis and
5 units up from the x-axis, so its coordinates are (3,5).
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
(−5,−4).
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).
3. To graph (4,0) we move from the origin 4 units to the right
of the y-axis. Since the second coordinate is 0, we do not
move up or down from the x-axis.
To graph (−3,−5) we move from the origin 3 units to the
left of the y-axis. Then we move 5 units down from the
x-axis.
To graph (−1,4) we move from the origin 1 unit to the
left of the y-axis. Then we move 4 units up from the
x-axis.
To graph (0,2) we do not move to the right or the left of
the y-axis since the first coordinate is 0. From the origin
we move 2 units up.
To graph (2,−2) we move from the origin 2 units to the
right of the y-axis. Then we move 2 units down from the
x-axis.
4
2
2
4
4 2 2 4
x
y
( 1, 4)
( 3, 5)
(0, 2) (4, 0)
(2, 2)
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.
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.
To graph (2,3) we move from the origin 2 units to the right
of the y-axis. Then we move 3 units up from the x-axis.
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.
To graph (0,1) we do not move to the right or the left of
the y-axis since the first coordinate is 0. From the origin
we move 1 unit up.
4
2
2
4
4 2 4
x
y
( 5, 1)
(2, 3)
(0, 1) (5, 1)
(2, 1)
7. The first coordinate represents the year and the corre-
sponding second coordinate represents the number of cities
served by Southwest Airlines. The ordered pairs are
(1971, 3), (1981, 15), (1991, 32), (2001, 59), (2011, 72),
and (2021, 121).
9. To determine whether (−1,−9) is a solution, substitute
−1 for x and −9 for y.
y = 7x − 2
−9 ? 7(−1) − 2
?
? −7 − 2
−9
?
−9 TRUE
The equation −9 = −9 is true, so (−1,−9) is a solution.
To determine whether (0,2) is a solution, substitute 0 for
x and 2 for y.
y = 7x − 2
2 ? 7 · 0 − 2
?
? 0 − 2
2
?
−2 FALSE
The equation 2 = −2 is false, so (0,2) is not a solution.
11. To determine whether
? 2
3 ,
3
4
?
is a solution, substitute
2
3
for x and
3
4
for y.
6x − 4y = 1
6 ·
2
3
− 4 ·
3
4
? 1
?
4 − 3
?
1
?
1 TRUE
The equation 1 = 1 is true, so
? 2
3 ,
3
4
?
is a solution.
To determine whether
?
1,
3
2
?
is a solution, substitute 1 for
x and
3
2
for y.
6x − 4y = 1
6 · 1 − 4 ·
3
2
? 1
?
6 − 6
?
0
?
1 FALSE
The equation 0 = 1 is false, so
?
1,
3
2
?
is not a solution.
13. To determine whether
?
−
1
2 ,−
4
5
?
is a solution, substitute
− 1
2
for a and − 4
5
for b.
2a + 5b = 3
2
?
−
1
2
?
+ 5
?
−
4
5
?
? 3
?
−1 − 4
?
−5
?
3 FALSE
The equation −5 = 3 is false, so
?
−
1
2 ,−
4
5
?
is not a solu-
tion.
To determine whether
?
0,
3
5
?
is a solution, substitute 0 for
a and
3
5
for b.
2a + 5b = 3
2 · 0 + 5 ·
3
5
? 3
?
0 + 3
?
3
?
3 TRUE
The equation 3 = 3 is true, so
?
0,
3
5
?
is a solution.
15. To determine whether (−0.75,2.75) is a solution, substi-
tute −0.75 for x and 2.75 for y.
x 2 − y 2 = 3
(−0.75) 2 − (2.75) 2 ? 3
0.5625 − 7.5625
?
−7
?
3 FALSE
The equation −7 = 3 is false, so (−0.75,2.75) is not a
solution.
To determine whether (2,−1) is a solution, substitute 2
for x and −1 for y.
x 2 − y 2 = 3
2 2 − (−1) 2 ? 3
4 − 1
?
3
?
3 TRUE
The equation 3 = 3 is true, so (2,−1) is a solution.
17. Graph 5x − 3y = −15.
To find the x-intercept we replace y with 0 and solve for
x.
5x − 3 · 0 = −15
5x = −15
x = −3
The x-intercept is (−3,0).
To find the y-intercept we replace x with 0 and solve for
y.
5 · 0 − 3y = −15
−3y = −15
y = 5
The y-intercept is (0,5).
We plot the intercepts and draw the line that contains
them. We could find a third point as a check that the
intercepts were found correctly.
y
x
4 2 2 4
4
