50

Chapter 2,
continued
2.2 Exercises (pp. 86–88)
6. Let (x1, y1) 5 (7, 3) and (x2, y2) 5 (21, 7).
y2 2 y1
723
4
1
m5}
5}
5}
5 2}2
x2 2 x1
21 2 7
28
Skill Practice
1. The slope m of a nonvertical line is the ratio of vertical
7. (24, 3), (2, 26)
change to horizontal change.
3
29
2. If the slopes of both lines are equal then the two lines are
parallel. If the product of the slopes of the lines equals
21 then the slopes are negative reciprocals of each other
and the lines are perpendicular.
Because m < 0, the line falls.
8. (7, 1), (7, 21)
21 2 1
21 2 (24)
3
3. m 5 } 5 }
422
2
22
m5}
5}
727
0
Because m is undefined, the line is vertical.
Because m > 0, the line rises.
9. (3, 22), (5, 22)
329
26
1
4. m 5 } 5 } 5 }
24 2 8
212
2
22 2 (22)
0
m5}
5 }2 5 0
523
Because m > 0, the line rises.
Because m 5 0, the line is horizontal.
25
5
24 2 1
5. m 5 } 5 } 5 2}
825
3
3
10. (5, 6), (1, 24)
24 2 6
Because m < 0, the line falls.
5
210
m5}
5}
5 }2
125
24
22 2 (22)
0
6. m 5 } 5 } 5 0
6
3 2 (23)
Because m > 0, the line rises.
Because m 5 0, the line is horizontal.
11. Line 1: through (22, 8) and (2, 24)
28
24 2 4
7. m 5 } 5 } 5 24
2
1 2 (21)
Line 2: through (25, 1) and (22, 2)
24 2 8
212
5}
5 23
m1 5 }
4
2 2 (22)
221
Because m < 0, the line falls.
25 2 5
210
8. m 5 } 5 }
0
26 2 (26)
1
5 }3
m2 5 }
22 2 (25)
Because m is undefined, the line is vertical.
1
3
Because m1m2 5 23 + } 5 21, m1 and m2 are negative
reciprocals of each other. So, the lines are perpendicular.
Because m > 0, the line rises.
12. Line 1: through (24, 22) and (1, 7)
326
23
3
10. m 5 } 5 } 5 }
24
4
27 2 (23)
Line 2: through (21, 24) and( 3, 5)
7 2 (22)
9
5 2 (24)
9
5 }5
m1 5 }
1 2 (24)
Because m > 0, the line rises.
924
5
11. m 5 } 5 }
424
0
5 }4
m2 5 }
3 2 (21)
9
3 2 (24)
7
9. m 5 } 5 }
4
21 2 (25)
9
Because m1m2 5 }5 + }4 Þ 21 and m1 Þ m2, the lines are
neither perpendicular nor parallel.
change in diameter
13. Average rate of change 5 }}
change in time
251 in. 2 248 in.
5 }}
2005 2 1965
Because m is undefined, the line is vertical.
325
22
12. m 5 } 5 } 5 21
725
2
Because m < 0, the line falls.
23 2 (23)
0
13. m 5 } 5 } 5 0
420
4
Because m 5 0, the line is horizontal.
24 2 (21)
23
3
14. m 5 } 5 } 5 }
21 2 1
22
2
3 in.
5}
40 years
5 0.075 inch per year
Find the number of years from 2005 to 2105, multiply
this number by the average rate of change to find the
total increase in diameter during the period 2005 - 2105.
Number of years 5 2105 2 2005 5 100
Increase in diameter 5 (100 years)1 0.075 }
year 2 5 7.5
inch
inches. In 2105, the diameter of the sequoia will be
about 251 1 7.5 5 258.5 inches.
Because m > 0, the line rises.
15. The error is found in the run of the slope. The run should
be x2 2 x1.
(24, 23), (2, 21)
21 2 (23)
2
1
5 }6 5 }3
m5}
2 2 (24)
16. The error is found in the rise and the run. The rise should
be y2 2 y1, and the run should be x2 2 x1.
124
23
1
m5}
5}
5 2}2
6
5 2 (21)
50
Algebra 2
Worked-Out Solution Key
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
26 2 3
m5}
5}
5 2}2
6
2 2 (24)
Chapter 2,
continued
1 2 (24)
5
17. A; m 5 } 5 }
522
3
24. (2, 12), (5, 30)
Because m > 0, the line rises.
18. Line 1: through (3, 21) and (6, 24)
x is measured in hours and y is
measured in dollars.
change in dollars
Average rate of change 5 }}
change in hours
Line 2: through (24, 5) and (22, 7)
30 dollars 2 12 dollars
5 }}
5 hours 2 2 hours
24 2 (21)
23
5}
5 21
m1 5 }
623
3
18 dollars
5}
3 hours
725
2
5 }2 5 1
m2 5 }
22 2 (24)
Because m1m2 5 21 + 1 5 21, m1 and m2 are negative
reciprocals of each other. So, the lines are perpendicular.
19. Line 1: through (1, 5) and (3, 22)
Line 2: through (23, 2) and (4, 0)
22 2 5
5 6 dollars per hour
25. (0, 11), (3, 50)
x is measured in gallons and y is
measured in miles.
change in miles
Average rate of change 5 }}
change in gallons
50 miles 2 11 miles
27
5 }}
3 gallons 2 0 gallons
5}
m1 5 }
321
2
022
39 miles
22
5}
3 gallons
5}
m2 5 }
7
4 2 (23)
27
5 13 miles per gallon
22
+}
5 1 Þ 21 and m1 Þ m2,
Because m1m2 5 }
2
7
the lines are neither perpendicular nor parallel.
26. (3, 10), (5, 18)
20. Line 1: through (21, 4) and (2, 5)
Line 2: through (26, 2) and (0, 4)
524
1
422
2
x is measured in seconds and y is
measured in feet.
change in feet
Average rate of change 5 }}
change in seconds
18 feet 2 10 feet
5 }3
m1 5 }
2 2 (21)
5 }}
5 seconds 2 3 seconds
1
8 feet
5 }6 5 }3
m2 5 }
0 2 (26)
Because m1 5 m2 (and the lines are different), you can
conclude that the lines are parallel.
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
21. Line 1: through (5, 8) and (7, 2)
Line 2: through (27, 22) and (24, 21)
228
26
5}
5 23
m1 5 }
725
2
5}
2 seconds
5 4 feet per second
27. (1, 8), (7, 20)
x is measured in seconds and y is
measured in meters.
change in meters
Average rate of change 5 }}
change in seconds
20 meters 2 8 meters
5 }}
7 seconds 2 1 second
21 2 (22)
1
5 }3
m2 5 }
24 2 (27)
12 meters
5}
6 seconds
1
Because m1m2 5 23 + }3 5 21, m1 and m2 are negative
reciprocals of each other. So, the lines are perpendicular.
22. Line 1: through (23, 2) and (5, 0)
Line 2: through (21, 24) and (3, 23)
022
22
1
5}
5 2}4
m1 5 }
8
5 2 (23)
23 2 (24)
1
5 }4
m2 5 }
3 2 (21)
1 1
1
Þ 21 and m1 Þ m2 ,
Because m1m2 5 2}4 + }4 5 2}
16
the lines are neither perpendicular nor parallel.
23. Line 1: through (1, 24) and (4, 22)
Line 2: through (8, 1) and (14, 5)
22 2 (24)
2
5 }3
m1 5 }
421
521
4
2
5 }6 5 }3
m2 5 }
14 2 8
5 2 meters per second
28. If lines *1 and/or *2 are vertical, their slopes would be
undefined.
1
2
3
7
29. Let (x1, y1) 5 21, } and (x2, y2) 5 0, } .
2
2
7
3
}2}
2
2
}
1
2
4
}
2
}
m 5 0 2 (21) 5 1 5 2
1
2
3
5
30. Let (x1, y1) 5 2}, 22 and (x2, y2) 5 }, 23 .
4
4
23 2 (22)
21
1
}
1
2
1
5}
5 2}2
m5}
8
3
5
} 2 2}
4
4
2
4
5
1 5
31. Let (x1, y1) 5 2}, } and (x2, y2) 5 }, 3 .
2 2
2
1
5
3 2 }2
2
1
2
}
1
2
1
m5}
5 }6 5 }6
1
5
1 22
} 2 2}
2
}
2
Because m1 5 m2 (and the lines are different), you can
conclude that the lines are parallel.
Algebra 2
Worked-Out Solution Key
51
continued
32. Let (x1, y1) 5 (24.2, 0.1) and (x2, y2) 5 (23.2, 0.1).
37. (2, 23) and (k, 7); m 5 22
7 2 (23)
0.1 2 0.1
0
m 5 }}
5 }1 5 0
23.2 2 (24.2)
10
m5}
5}
5 22
k22
k22
33. Let (x1, y1) 5 (20.3, 2.2) and (x2, y2) 5 (1.7, 20.8).
20.8 2 2.2
10
22
}5k22
3
23
m5}
5}
5 2}2
2
1.7 2 (20.3)
25 5 k 2 2
23 5 k
34. Let (x1, y1) 5 (3.5, 22) and (x2, y2) 5 (4.5, 0.5).
0.5 2 (22)
7 2 (23)
10
5}
5 22
Check: m 5 }
25
23 2 2
2.5
m5}
5}
5 2.5
1
4.5 2 (3.5)
38. (0, k) and (3, 4); m 5 1
35. No; no
42k
When finding the slope of a line, it does not matter which
points are picked on the line.
42k
m5}
5}
51
320
3
42k53
Sample answer:
022
m5}
5}
5 2}2
4
1 2 (23)
Q(21, 1) to T(5, 22)
5}
5 2}2
m5}
6
5 2 (21)
R (1, 0) to S (3, 21)
5}
5 2}2
m5}
321
2
22 2 1
21 2 0
k51
1
22
P (23, 2) to R (1, 0)
1
23
39. (24, 2k) and (k, 25); m 5 21
1
21
3
421
5 }3 5 1
Check: m 5 }
320
25 2 2k
25 2 2k
5}
5 21
m5}
k 2 (24)
k14
It also does not make a difference which point is 1 x1 y1 2
and which point is 1 x2 y2 2.
25 2 2k 5 2k 2 4
25 5 k 2 4
from before
P (23, 2) to R (1, 0)
21 5 k
1
25 2 2(21)
25 1 2
23
Check: m 5 }
5}
5}
5 21
21 1 4
3
21 2 (24)
m 5 2}2
1
Q(21, 1) to T(5, 22) m 5 2}2
R (1, 0) to S (3, 21)
40. (22, k) and (2k, 2); m 5 20.25
22k
1
22k
m5}
5}
5 20.25
2k 2 (22)
2k 1 2
m 5 2}2
2 2 k 5 20.25(2k 1 2)
reverse
220
2
m5}
5}
5 2}2
23 2 1
24
T(5, 22) to Q(21, 1)
1 2 (22)
3
1
5}
5 2}2
m5}
21 2 5
26
S (3, 21) to R(1, 0)
5}
5 2}2
m5}
123
22
0 2 (21)
1
36. Sample answer:
2 2 k 5 20.5k 2 0.5
1
R (1, 0) to P (23, 2)
2 5 0.5k 2 0.5
2.5 5 0.5k
55k
1
225
23
23
5}
5}
5 20.25
Check: m 5 }
10 1 2
12
2(5) 2 (22)
Problem Solving
y
41.
(0, 3)
5 28 ft
run 5 48 ft
24
21
rise
2
rise
x
28
7
Slope 5 }
5}
run 5 }
48
12
1 (1, 21)
42.
rise 5 400 ft
The points (21, 7) and (1, 21) also lie on the line.
run 5 685 ft
rise
400
80
Slope 5 }
5}
run 5 }
685
137
52
Algebra 2
Worked-Out Solution Key
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
Chapter 2,
Chapter 2,
continued
43.
b.
rise 5 195 ft
rise 5 80 ft
run 5 3000 ft
rise
195
run 5 240 ft
13
Slope 5 }
5}
run 5 }
3000
200
a2 1 b2 5 c2
80 1 240 2 5 c 2
6400 1 57,600 5 c 2
64,000 5 c 2
c ø 253
The slide is about 253 feet long.
2
13
5 0.065 5 6.5%
grade 5 }
200
44. The rise of each ramp of the three-section ramp is
5.25 ft
3
} 5 1.75 ft. The slope of each ramp of the three-
rise
1.75
1
section ramp is: slope 5 }
5}
.
run 5 }
28
16
1 16
c. The slope will be increased from } to } because
3 47
the run will be decreased and the vertical distance
remains the same.
If there were only a single-section ramp, the slope would
3(1.75)
rise
5.25
3
be: slope 5 }
5}
5}
.
run 5 }
28
28
16
The slope of a single-section ramp would be 3 times
as steep as the three-section ramp. The benefit of the
three-section ramp is that people can walk at a slope that
doesn’t require so much work. It is also easier for those
who use a wheelchair.
36 miles
49. Average highway rate 5 36 miles per gallon 5 }
1 gallon
Average city rate 5 24 miles per gallon
24 miles
36 miles 1 36 miles
1 gallon 1 1.5 gallons
change in gallons
214 gallons 2 400 gallons
5 }}
30 days 2 0 days
5 28.8 miles per gallon
Mixed Review for TAKS
50. B;
* 5 w 1 40
2186 gallons
5}
30 days
560 5 2w 1 2(w 1 40)
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
change in centimeters
46. Average rate of change 5 }}
change in years
560 5 4w 1 80
* 5 120 1 40
* 5 160
3.6 cm
5}
80 years
The length of the playground is 160 feet.
51. J;
c 5 185 1 45h
rise
15 ft
3
47. a. slope 5 }
5 }8
run 5 }
40 ft
3
8
1
4
Because } > }, the roof satisfies the building code.
c. Let r represent the minimum rise which satisfies the
3
r
code, then }
5}
l r 5 10. Therefore the rise
40
12
exceeds the code minimum 15 2 10 5 5 ft.
w 1 40
120 5 w
15.5 cm 2 11.9 cm
5 }}
110 years 2 30 years
3 ft
1
b. minimum pitch 5 } 5 }
4
12 ft
w
P 5 2w 1 2*
5 26.2 gallons per day
5 0.045 centimeters per year
72 miles
2.5 gallons
}} 5 }
45. A;
Average rate of change 5 }}
change in days
36 miles
5 }
5}
1 gallon
1.5 gallons
Lesson 2.3
2.3 Guided Practice (pp. 90–92)
1.
y
2
21
x
48. a. Let x represent the horizontal distance you cover when
descending the slide.
rise
80
1
slope 5 }
5 }3
run 5 }
x
The graphs of y 5 22x and y 5 x both have a y-intercept
of 0, but the graph of y 5 22x has a slope of 22 instead
of 1.
x 5 80(3) 5 240
You cover a horizontal distance of 240 feet when
descending the slide.
Algebra 2
Worked-Out Solution Key
53
Chapter 2,
2.
continued
Step 3: Estimate the body length of the calf at age 10
months by starting at 10 on the x-axis and moving up
until you reach the graph. Then move left to the y-axis.
At age 10 months, the body length of the calf is about
108 inches.
y
1
x
21
11. 2x 1 5y 5 10
x-intercept: 2x 1 5(0) 5 10
The graphs of y 5 x 2 2 and y 5 x both have a slope
of 1, but the graph of y 5 x 2 2 has a y-intercept of 22
instead of 0.
3.
x55
The x-intercept is 5. So, plot the point (5, 0).
y
y-intercept: 2(0) 1 5y 5 10
y52
1
The y-intercept is 2. So, plot the point (0, 2).
x
21
y
The graphs of y 5 4x and y 5 x both have a y-intercept
of 0, but the graph of y 5 4x has a slope of 4 instead
of 1.
4.
5.
y
1
x
21
12. 3x 2 2y 5 12
y
x-intercept: 3x 2 2(0) 5 12
1
x54
1
x
21
21
x
The x-intercept is 4. So, plot the point (4, 0).
y-intercept: 3(0) 2 2y 5 12
1
7.
y 5 26
y
The y-intercept is 26. So, plot the point (0, 26).
x
21
1
8.
9.
y
x
21
1
21
y
x
y
13. x 5 1
1
x
21
2
21
x
The graph of x 5 1 is the vertical line that passes through
the point (1, 0).
y
10. Step 1: Graph the equation y 5 6x 1 48.
1
Body length (in.)
y
150
120
14. y 5 24
90
The graph of y 5 24 is the horizontal line that passes
through the point (0, 24).
60
30
0
1
21
0
4
8
12
16
Age (months)
x
Step 2: Interpret the slope and y-intercept. The slope 6
represents the calf ’s rate of growth in inches per month.
The y-intercept 48 represents a newborn calf’s body
length in inches.
54
x
21
Algebra 2
Worked-Out Solution Key
y
x
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
6.
y
Chapter 2,
continued
2.3 Exercises (pp. 93–96)
8.
f(x)
Skill Practice
1
1. The linear equation y 5 2x 1 5 is written in
x
21
slope-intercept form.
2. Identify the x-intercept: let y 5 0, solve for x, then plot
the point.
Identify the y-intercept: let x 5 0, solve for y, then plot
the point.
Draw a line through the two points.
3.
The graph of y 5 23x 1 2 has a slope of 23 and a
y-intercept of 2 instead of having a slope of 1 and a
y-intercept of 0.
9. y 5 2x 2 3
y
The y-intercept is 23, so plot the point (0, 23).
1
x
21
The slope is 21, so plot a second point by starting at
(0, 23) and then moving down 1 unit and right 1 unit.
The second point is (1, 24).
The graphs of y 5 3x and y 5 x both have a y-intercept of
0, but the graph of y 5 3x has a slope of 3 instead of 1.
4.
1
y
x
21
y
1
x
21
10. y 5 x 2 6
The graphs of y 5 2x and y 5 x both have a y-intercept
of 0, but the graph of y 5 x has a slope of 21 instead
of 1.
5.
y
The y-intercept is 26, so plot the point (0, 26).
The slope is 1, so plot a second point by starting at
(0, 26) and then moving up 1 unit and right 1 unit. The
second point is (1, 25).
1
y
x
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
21
1
21
x
The graphs of y 5 x 1 5 and y 5 x both have a slope
of 1, but the graph of y 5 x 1 5 has a y-intercept of 5
instead of 0.
6.
1
x
21
11. y 5 2x 1 6
The y-intercept is 6, so plot the point (0, 6).
y
The slope is 2, so plot a second point by starting at (0, 6)
and then moving up 2 units and right 1 unit. The second
point is (1, 8).
y
The graphs of y 5 x 2 2 and y 5 x both have a slope of
1, but the graph of y 5 x 2 2 has a y-intercept of 22
instead of 0.
7.
f(x)
1
1
21
21
x
x
The graph of y 5 2x 2 1 has a slope of 2 and a
y-intercept of 21 instead of having a slope of 1 and a
y-intercept of 0.
Algebra 2
Worked-Out Solution Key
55
Chapter 2,
continued
5
16. f (x) 5 2} x 1 1
4
12. y 5 3x 2 4
The y-intercept is 24, so plot the point (0, 24).
The slope is 3, so plot a second point by starting at
(0, 24) and then moving up 3 units and right 1 unit. The
second point is (1, 21).
y
The y-intercept is 1, so plot the point (0, 1).
5
The slope is 2}4, so plot a second point by starting at
(0, 1) and then moving down 5 units and right 4 units.
The second point is (4, 24).
y
1
1
x
21
x
21
3
17. f (x) 5 }x 2 3
2
13. y 5 4x 2 1
The y-intercept is 21, so plot the point (0, 21).
The slope is 4, so plot a second point on the line by
starting at (0, 21) and then moving up 4 units and right
1 unit. The second point is (1, 3).
The y-intercept is 23, so plot the point (0, 23).
3
The slope is }2, so plot a second point by starting at
(0, 23) and then moving up 3 units and right 2 units.
The second point is (2, 0).
y
y
1
1
x
21
5
18. f (x) 5 }x 1 4
3
The y-intercept is 22, so plot the point (0, 22).
2
The slope is }3 , so plot a second point by starting
at (0, 22) and then moving up 2 units and right 3 units.
The second point is (3, 0).
The y-intercept is 4, so plot the point (0, 4).
5
The slope is }3, so plot a second point by starting at (0, 4)
and then moving up 5 units and right 3 units. The second
point is (3, 9).
y
y
x
21
1
x
21
1
15. f (x) 5 2} x 2 1
2
19. f (x) 5 21.5x 1 2
The y-intercept is 21, so plot the point (0, 21).
1
2
The slope is 2}, so plot a second point by starting at
(0, 21) and then moving down 1 unit and right 2 units.
The second point is (2, 22).
The y-intercept is 2, so plot the point (0, 2).
The slope is 21.5, so plot a second point by starting at
(0, 2) and then moving down 1.5 units and right 1 unit.
The second point is (1, 0.5).
y
y
1
1
22
56
Algebra 2
Worked-Out Solution Key
21
x
x
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
2
14. y 5 } x 2 2
3
1
x
21
Chapter 2,
continued
20. f (x) 5 3x 2 1.5
26. 3x 2 4y 5 212
The y-intercept is 21.5, so plot the point (0, 21.5).
x-intercept: 3x 2 4(0) 5 212
The slope is 3, so plot a second point by starting at
(0, 21.5) and then moving up 3 units and right 1 unit. The
second point is (1, 1.5).
The x-intercept is 24.
x 5 24
y-intercept: 3(0) 2 4y 5 212
y
y53
y-intercept is 3.
27. 2x 2 y 5 10
1
x
21
x-intercept: 2x 2 0 5 10
x55
21. The y-intercept was incorrectly plotted at (0, 2) instead of
(0, 3), and the slope was incorrectly graphed as 3 instead
of 2.
The x-intercept is 5.
y-intercept: 2(0) 2 y 5 10
y 5 210
The y-intercept is 210.
y
28. 4x 2 5y 5 20
x-intercept: 4x 2 5(0) 5 20
1
1
x55
x
The x-intercept is 5.
1
22. The slope was incorrectly graphed as } instead of 4.
4
y
y-intercept: 4(0) 2 5y 5 20
y 5 24
The y-intercept is 24.
1
29. 26x 1 8y 5 236
x
21
x-intercept: 26x 1 8(0) 5 236
x56
The x-intercept is 6.
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
23. C; 4x 2 3y 5 18
y-intercept: 26(0) 1 8y 5 236
23y 5 18 2 4x
4
y 5 }3 x 2 6
24. x 2 y 5 4
x-intercept: x 2 (0) 5 4
x54
The x-intercept is 4.
y-intercept: (0) 2 y 5 4
y 5 24
The y-intercept is 24.
25. x 1 5y 5 215
x-intercept: x 1 5(0) 5 215
x 5 215
The x-intercept is 215.
y-intercept: 0 1 5y 5 215
y 5 23
The y-intercept is 23.
y 5 24.5
The y-intercept is 24.5.
30. C;
5x 2 6y 5 30
5x 2 6(0) 5 30
x56
The x-intercept is 6.
31. x 1 4y 5 8
y
(0, 2)
x-intercept: x 1 4(0) 5 8
1
x58
y-intercept: 0 1 4y 5 8
(8, 0)
x
22
y52
32. 2x 2 6y 5 212
y
x-intercept: 2x 2 6(0) 5 212
1
x 5 26
y-intercept: 2(0) 2 6y 5 212
(26, 0)
21
(0, 2)
x
y52
Algebra 2
Worked-Out Solution Key
57
Chapter 2,
continued
33. The graph of x 5 4 is the vertical line that passes through
the point (4, 0).
5
41. The graph of x 5 2} is the vertical line that passes
2
5
through the point 2}2, 0 .
1
y
2
y
1
(4, 0)
1
x
21
(
5
22 ,
)
x
21
0
34. The graph of y 5 22 is the horizontal line that passes
through the point (0, 22).
1
1
42. } x 1 2y 5 22
2
1
x-intercept: }2 x 1 2(0) 5 22
y
x
21
(24, 0)
1
y
(0, 21)
24
x
x 5 24
(0, 22)
1
y-intercept: }2 (0) 1 2y 5 22
35. 5x 2 y 5 3
1
x-intercept: 5x 2 0 5 3
y
y 5 21
( , 0)
x
3
5
21
3
x 5 }5
43.
x 5 22
36. 3x 1 4y 5 12
y51
(0, 3)
x-intercept: 3x 1 4(0) 5 12
x
21
y-intercept: 6y 2 3(0) 5 6
y
44. 23 1 x 5 0
x53
1
(4, 0)
21
The graph of x 5 3 is the vertical line that passes through
the point (3, 0).
x
y53
37. 25x 1 10y 5 20
y
y
x-intercept: 25x 1 10(0) 5 20
x 5 24
1
(24, 0)
1
(0, 2)
x
21
1
x
y52
38. 2x 2 y 5 6
x-intercept: 2x 2 0 5 6
1
(26, 0)
y 1 7 5 22x
1
y 1 2x 5 7
x
21
x 5 26
45.
y
y
21
x-intercept: 0 1 2x 5 27
7
y-intercept: 20 2 y 5 6
x 5 2}2
y56
y-intercept: y 1 2(0) 5 27
(0, 26)
y 5 27
46. 4y 5 16
39. The graph of y 5 1.5 is the horizontal line that passes
y54
through the point (0, 1.5).
The graph of y 5 4 is the horizontal line that passes
through the point (0, 4).
y
y
1 (0, 1.5)
21
x
1
40. 2.5x 2 5y 5 215
y
x-intercept: 2.5x 2 5(0) 5 215
(0, 3)
x 5 26
y-intercept: 2.5(0) 2 5y 5 215
y53
58
Algebra 2
Worked-Out Solution Key
21
1
(26, 0)
21
x
x
x
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
x54
y-intercept: 25(0) 1 10y 5 20
2
x-intercept: 6(0) 2 3x 5 6
y 5 23
y-intercept: 3(0) 1 4y 5 12
y
6y 2 3x 5 6
(0, 23)
y-intercept: 5(0) 2 y 5 3
6y 5 3x 1 6
Chapter 2,
47.
continued
8y 5 22x 1 20
53. 2y 2 5 5 0
y
8y 1 2x 5 20
5
y 5 }2
x-intercept: 8(0) 1 2x 5 20
5
1
x 5 10
x
21
y-intercept: 8y 1 2(0) 5 20
5
The graph of y 5 }2 is the horizontal line that passes
5
through the point 1 0, }2 2.
y 5 }2
48.
y
1
y
4x 5 2}2 y 2 1
1
4
1
4x 1 }2 y 5 21
1
x-intercept: 4x 1 }2 (0) 5 21
54. 5y 5 7.5 2 2.5x
1
x 5 2}4
1
y-intercept: 24(0) 1 }2 y 5 21
x
21
y
5y 1 2.5x 5 7.5
x-intercept: 5(0) 1 2.5x 5 7.5
x53
y 5 22
49.
x
21
1
x
1
y-intercept: 5y 1 2.5(0) 5 7.5
24x 5 8y 1 12
1
24x 2 8y 5 12
y
y 5 1.5
1
x-intercept: 24x 2 8(0) 5 12
x
55. Sample answer:
A line that has an x-intercept but no y-intercept is a
vertical line. One example is the line x 5 3.
x 5 23
y-intercept: 24(0) 2 8y 5 12
y
3
y 5 2}2
1
50. 3.5x 5 10.5
x
1
x53
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
The graph of x 5 3 is the vertical line that passes through
the point (3, 0).
y
A line that has a y-intercept but no x-intercept is a
horizontal line. One example is the line y 5 2
1
y
x
21
1
x
1
51. y 2 5.5x 5 6
y 5 5.5x 1 6
The y-intercept is 6, so plot the point (0, 6).
The slope is 5.5, so plot a second point by starting at
(0, 6) and then moving up 5.5 units and right 1 unit.
The second point is (1, 11.5).
56. Sample answer:
For positive values of m, as m increases, the steepness
of the line increases. For negative values of m, as m
decreases, the steepness of the line increases. You can
conclude that the further away m gets from zero, the
steeper the line will be.
y
6
y
22
x
y 5 23x
52. 14 2 3x 5 7y
y
1
14
y 5 4x
x
22
3x 1 7y 5 14
x-intercept: 3x 1 7(0) 5 14
3
y5
x
1
x
2
1
y 5 24 x
21
x5}
3
y-intercept: 3(0) 1 7y 5 14
y52
Algebra 2
Worked-Out Solution Key
59
Chapter 2,
continued
57. Ax 1 By 5 C
61. C(g) 5 3g 1 1.5
The y-intercept, 1.5, represents the cost to rent shoes,
$1.50. The slope, 3, represents the cost per game, $3.
x-intercept: Ax 1 B(0) 5 C
C
x5}
A
2
Cost
y-intercept: A(0) 1 By 5 C
C
y5}
B
The point 1 0, }
is on the line.
B2
C
C(A)
B(21)(C)
Number of games
A
5 } 5 2}
.
B
62. To determine a reasonable domain, find the minimum
58. Sample answer:
Two points on the line are (0, b) and (1, m 1 b).
m1b2b
m
5}
5 m.
Using the slope formula gives }
1
120
Problem Solving
59. To find the total cost of the membership after 9 months,
Cost
start at 9 on the x-axis and move up until you reach the
graph. Then move left to the y-axis. The total cost of the
membership after 9 months is about $480.
y
600
525
450
375
300
225
150
75
0
0 1 2 3 4 5 6 7 8 x
Months
60. y 5 5x 1 35
Cost (dollars)
The slope, 5, represents how much it costs to stay per
night, $5. The y-intercept, 35, represents the annual
membership fee, $35, without staying any nights.
y
80
70
60
50
40
30
20
10
0
60
and maximum values of w. The minimum value of w is 0
because the number of weeks cannot be negative. To find
the maximum value of w, let M(w) 5 0 and solve for w.
m(w) 5 230w 1 300
0 5 230w 1 300
30w 5 300
w 5 10
So, a reasonable domain is 0awa10.
To determine a reasonable range, substitute the minimum
and maximum values for w into the equation.
w 5 0: m(w) 5 230(0) 1 300
m(w) 5 300
w 5 10: m(w) 5 230(10) 1 300
m(w) 5 0
So, a reasonable range is 0am(w)a300.
The slope, 230, represents the number of minutes per
week you use the card. So, you use 30 minutes per week.
M(w)
300
200
100
0
0
2
4
6
8
Number of weeks
10 w
63. The y-intercept will be 30, the amount of money on
0 1 2 3 4 5 6 7 8 9 x
Number of nights
Algebra 2
Worked-Out Solution Key
the card ($30) before any smoothies were purchased.
The line will fall from left to right because the more
smoothies you buy, the less the amount on the card
will be.
64. a. Domain: 0axa60; Range: 0aya20
The y-intercept represents the amount of time it would
take to go 1800 yards down the river if there were no
drifting time, 20 minutes.
The x-intercept represents the amount of time it would
take to go 1800 yards down the river if there were no
paddling time, 60 minutes.
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
The slope is
C
B
}
C
02}
A
}20
Number of minutes
1
C(g)
36
32
28
24
20
16
12
8
4
0
0 1 2 3 4 5 6 7 8 g
C
The point }
, 0 is on the line.
A
Chapter 2,
continued
b. If you paddle for 5 minutes, let y 5 5.
30x 1 90(5) 5 1800
30x 1 450 5 1800
30x 5 1350
x 5 45
When y 5 5, x 5 45. So, the total trip time is
x 1 y 5 50 minutes.
c. If you paddle and drift equal amounts of time, then
x 5 y.
30x 1 90x 5 1800
120x 5 1800
x 5 15
When x 5 15 and y 5 15, the total trip time is
x 1 y 5 30 minutes.
66. Sample answer:
Let a 5 10: 5s 1 7(10) 5 150
5s 1 70 5 150
5s 5 80
s 5 16
The honor society could buy 16 science museum tickets
and 10 art museum tickets.
Let a 5 5: 5s 1 7(5) 5 150
5s 1 35 5 150
5s 5 115
s 5 23
The honor society could buy 23 science museum tickets
and 5 art museum tickets.
a
20
16
Art museum tickets
Paddling time (min.)
y
20
12
8
4
16
12
8
4
0
0
14
28
42
56
Drifting time (min.)
x
0
65. Sample answer:
67. a.
Let r 5 0: 6(0) 1 3.5w 5 14
3.5w 5 14
6
12
18
24
30 s
Science museum tickets
t (min)
h (ft)
w54
You could walk 4 hours and run 0 hours.
b.
Height (feet)
Let w 5 0: 6r 1 3.5(0) 5 14
6r 5 14
1
r 5 2 }3
1
You could walk 0 hours and run 2 }3 hours.
Let w 5 1: 6r 1 3.5(1) 5 14
6r 5 10.5
3
You could walk 1 hour and run 1 }4 hours.
w
h
900
800
700
600
500
400
300
200
100
0
0
0
1
2
3
4
5
200
350
500
650
800
950
1
2
3
4
t
Time (minutes)
3
r 5 1.75 5 1}4
c. Balloon’s
height
5
Ascent
Initial
1
+ Time
rate
height
h 5 200 1 150 + t
h 5 200 1 150t
Walking time (hours)
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
0
4
68. a. The graphs of y 5 1400 2 50x and
y 5 1200 2 50x have the same slope but
y 5 1400 2 50x has a y-intercept of 1400 and
y 5 1200 2 50x has a y-intercept of 1200. The
graphs are parallel.
3
2
1
0
0
1
2
r
Running time (hours)
Algebra 2
Worked-Out Solution Key
61
Chapter 2,
continued
b. The y-intercepts represent how many words each
research paper contains, the x-intercepts represent how
long it will take each person to type the research paper,
and the slopes represent the rate of typing for each
person. Your paper has 1400 words, it will take you 28
minutes to finish, and you type at a rate of 50 words
per minute.Your friend’s paper has 1200 words, it will
take him/her 24 minutes to finish, and he/she types at a
rate of 50 words per minute.
c. Although you both type at the same rate, your friend
will finish first, because your friend’s paper is shorter.
It will only take him/her 24 minutes to finish, while it
takes you 28 minutes to finish.
y
The interior angles measure 358, 358, and 1108.
71. F;
1
1
V 5 }3 :r 2h 5 }3 :(1.252)(4) ø 6.54 in.3
The volume of the cone is about 6.5 cubic inches.
Quiz 2.1–2.3 (p. 96)
3. The relation is not a function because the input 0 is
mapped onto both 4 and 5.
4. Line 1: let (x1, y1) 5 (23, 27) and (x2, y2) 5 (1, 9)
0
6
12
18
Minutes
24
x
Line 2: let (x1, y1)5 (21, 24) and (x2, y2) 5 (0, 22)
22 2 (24)
2
5 }1 5 2
m2 5 }
0 2 (21)
Because m1m2 5 4(2) 5 8 Þ 21 and m1 Þ m2, the lines
are neither perpendicular nor parallel.
5. Line 1: let (x1, y1) 5 (2, 7) and (x2, y2) 5 (21, 22)
22 2 7
Start by making a table of values.
2
16
Line 2: let (x1, y1) 5 (3, 26) and (x2, y2) 5 (26, 23)
b. 3x 1 4y 5 25
1
9 2 (27)
m1 5 }
5}
54
4
1 2 (23)
3
4
5
6
7
8
6.25 5.5 4.75 4 3.25 2.5 1.75 1 0.25
From the table, it appears that (3, 4) and (7, 1) are the
only solutions where x and y are whole numbers.
Check:
(3, 4): 3(3) 1 4(4) 0 25
9 1 16 0 25
29
m1 5 }
5}
53
21 2 2
23
(7, 1): 3(7) 1 4(1) 0 25
21 1 4 0 25
25 5 25 25 5 25 23 2 (26)
3
1
5}
5 2}3
m2 5 }
26 2 3
29
1
Because m1m2 5 3 + 2}3 5 21, m1 and m2 are negative
reciprocals of each other. So, the lines are perpendicular.
6. y 5 25x 1 3
The y-intercept is 3, so plot the point (0, 3).
The slope is 25, so plot a second point on the line by
starting at (0, 3) and then moving down 5 units and right
1 unit.
c. Not all the solutions from part (b) represent
y
combinations of rectangles that can actually cover
the grid. (3, 4) works, but (7, 1) does not work.
(3, 4)
1
(7, 1)
21
not possible
62
Algebra 2
Worked-Out Solution Key
x
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
Words left
x 5 35
mappped onto exactly one output value.
600
rectangle is 3, the area of the 4 by 1 rectangle is 4,
and the area of the 5 by 5 rectangle is 25. Let x 5 the
number of 3 by 1 rectangles you need and y 5 the
number of 4 by 1 rectangles you need. Therefore,
3x 1 4y 5 25. x and y must be whole numbers
because you cannot use partial rectangles.
y
2x 5 70
2. The relation is a function because each input value is
900
69. a. Area 5 length 3 width. The area of the 3 by 1
0
1108 1 2x8 5 1808
mapped onto exactly one output value.
300
x
70. D;
1. The relation is a function because each input value is
1200
0
Mixed Review for TAKS
Chapter 2,
continued
7. x 5 10
2. 3x 2 y 5 4
The graph of x 5 10 is the vertical line that passes
through the point (10, 0).
2y 5 4 2 3x
y 5 3x 2 4
y
2
x
22
8. 4x 1 3y 5 224
x-intercept: 4x 1 3(0) 5 224
3. 3x 2 6y 5 218
x 5 26
26y 5 23x 2 18
y-intercept: 4(0) 1 3y 5 224
1
y 5 }2 x 1 3
y 5 28
1
y
21
x
4.
8x 5 5y 1 16
8x 2 16 5 5y
16
8
}x 2 } 5 y
5
5
9. Because she rowed a total of 3333 miles, a reasonable
range is 0ada3333.
To determine a reasonable domain, find the values of
t when d is at its minimum and maximum values.
d 5 0: 0 5 41t
d 5 3333: 3333 5 41t
81.3 ø t
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
05t
So, a reasonable domain is 0ata82.
To estimate how long it took Tori Murden to row 1000
miles, start at 1000 on the vertical axis and move right
until you reach the graph. Then move down to the
horizontal axis. It took Tori Murden about 24 days to
row 1000 miles.
4x 5 25y 2 240
4x 1 240 5 25y
240
25
4
25
}x 1 } 5 y
16
d
0
50
t
22
5.
28
0
290
Graphing Calculator Activity 2.3 (p. 97)
30
6. 1.25x 1 4.2y 5 28.7
1. y 1 14 5 17 2 2x
4.2y 5 28.7 2 1.25x
y 5 3 2 2x
41
1.25
2}
x
y5}
6
4.2
16
0
28
260
0
60
Algebra 2
Worked-Out Solution Key
63