Chapter 11 Review Exercises
34. a.
2
( ) 0.125 2.3 27gx x x
2
(35) 0.125(35) 2.3(35) 27 261g
b.
2
2
2
( ) 0.125 0.8 99
267 0.125 0.8 99
0 0.125 0.8 168
fx x x
xx
xx
35. a.
2
(35) 0.125(35) 2.3(35) 27 261g
This value is shown in the graph by the point
(35, 261).
36.
2
0161403tt
Apply the Pythagorean Theorem.
16 140 3abc
140 19,600 192
32
37.
2
14fx x
2
14
x
Apply the square root property.
Axis of symmetry: 1x .
38.
2
42fx x
Since 1a is positive, the parabola opens upward.
The vertex of the parabola is
,4,2hk and
Chapter 11 Quadratic Equations and Functions
Axis of symmetry: 4x .
39.
2
23fx x x
Since 1a is negative, the parabola opens
Apply the zero product principle.
30or 10
31
xx
xx
Axis of symmetry: 1x.
40.
2
246fx x x
Since 2a is positive, the parabola opens upward.
The x–coordinate of the vertex of the parabola is
44
1
2224
b
a
and the
y–coordinate of the vertex of the parabola is
2
1
2
b
ff
a
for y to obtain the y–intercept.
2
20 40 6y
20 0 6
006 6
y
y
Chapter 11 Review Exercises
41.
2
0.02 1fx x x
Since
0.02a
is negative, the function opens
downward and has a maximum at
11
25.
2 2 0.02 0.04
b
xa
42.
2
16 400 40st t t
Since
16a
is negative, the function opens
downward and has a maximum at
43.
2
104.5 1501.5 6016fx x x
Since ais positive, the function opens upward and
has a minimum at
1501.5 7.2.
2 2 104.5
b
xa
44. Maximize the area using
.
A
lw
2
1000 2
2 1000
A
xx x
A
xx x
Since
2a
is negative, the function opens
downward and has a maximum at
The area is maximized at 125,000 square yards
when the width is 250 yards and the length is
1000 2 250 500
yards.
The numbers which minimize the product are 7 and
7
. The minimum product is
7 7 49.
46. Let
2
.ux
42
680
xx
Chapter 11 Quadratic Equations and Functions
47. Let
.ux
2
2
780
780
780
xx
xx
uu
Check:
1718 0
48. Let
2
2.ux x
2
22
214 215
xx xx
Replace
2
by 2 .uxx
First, consider u = 15.
2
2
215
2150
xx
xx
49. Let
1
.ux
21
56 0
xx
87
uu
Replace
1
by .ux
11
8or 7
xx
Chapter 11 Review Exercises
50. Let
1
3
.ux
21
33
12 0
xx
Replace
1
3
by .ux
51. Let
1
4
.ux
11
24
3100
xx
16
x
Disregard –5 because the fourth root of x cannot be a negative number.
We must check 16, because both sides of the equation were raised to an even power.
Check 16.x
11
24
16 3 16 10 0
Chapter 11 Quadratic Equations and Functions
52.
2
2530
xx
Solve the related quadratic equation.
2
The boundary points are
1
3 and .
2
Interval Test Value Test Conclusion
,3
−4
2
24 54 30 false
,3
does not belong to the solution set.
53.
2
2940
xx
Solve the related quadratic equation.
The boundary points are
1
2
4 and .
Interval Test Value Test Conclusion
,4
−5
2
25 95 40 true
,4
belongs to the solution set.
1
4, 2
−1
2
21 91 40 false
1
4, 2
does not belong to the solution set.
Chapter 11 Review Exercises
54.
32
23
xxx
Solve the related polynomial equation.
32
23
xx x
The boundary points are
3
, 0, and 1.
32
Interval Test Value Test Conclusion
4
, 3 , 3 does not belong
424 34 False
to the solution set.
The solution set is
3, 0 1,
.
55.
60
2
x
x
Find the values of x that make the numerator and denominator zero.
The boundary points are –2 and 6.
Interval Test Value Test Conclusion
,2
−3
36 0 true
32
,2
belongs to the solution set.
Chapter 11 Quadratic Equations and Functions
56.
35
4
x
x
Express the inequality so that one side is zero.
350
x
4
x
The boundary points are
23
4 and 4
. Exclude 4 from the solution set, since this would make the denominator zero.
Interval Test Value Test Conclusion
,4
0
03 5
04
35, true
,4
belongs to the solution set.
Chapter 11 Review Exercises
57.
2
16 48
st t t
To find when the height is more than 32 feet above the ground, solve the inequality
2
16 48 32.
tt
Solve the related quadratic equation.
The boundary points are 1 and 2.
Interval Test Value Test Conclusion
0,1
0.5
2
16 0.5 48 0.5 32
20 32, false
0,1
does not belong to the solution set.
58. a.
2
15 15
0 0 30 0 200 0 0 200 0 0 200 200
88
H
The heart rate is 200 beats per minute immediately following the workout.
Chapter 11 Quadratic Equations and Functions
The boundary points are 4 and 12.
Interval Test Value Test Conclusion
0, 4
1
2
15 1 30 1 200 110
8
7
171 110, true
8
0, 4
belongs to the solution set.
Chapter 11 Test
1.
2
2
2
250
25
5
x
x
x
2.
2
320
320
x
x
Chapter 11 Test
4.
2
2_____
5
xx
Since b =
2
5
, add
222
11
12 1
225
25 5
b
.
2
2
21 1
525 5
xx x
6. Use the Pythagorean Theorem.
222
2
2
50 50
2500 2500
5000
5000
x
x
x
x
8.54
≈
8.
()
26
512
midpoint ,
22
78
,
22
7,4
2
−+−
−+
=
−
=
=−
2
2
44418
16 32 16
bac
Since the discriminant is negative, there are two
imaginary solutions which are complex conjugates.
11.
2
2
295
xx
Chapter 11 Quadratic Equations and Functions
12.
2
850
xx
Solve using the quadratic formula.
1 8 5
abc
The solution set is
411 .
13.
2
2
2250
225
x
x
Apply the square root principle.
225
25
x
xi
The solution set is
25.
i
14.
2
2650
2 6 5
xx
ab c
15. Because the solution set is
3, 7 , we have
3or 7
30 70
xx
xx
100 0
x
17. a. 2014 is 11 years after 2003.
2
2
( ) 2.4 0.7 29
(11) 2.4(11) 0.7(11) 29
327.1
327
fx x x
f
According to the function, in 2014 there were
327 “Bicycle Friendly” communities. This
underestimates the number shown in the graph
by 5.
b.
2
2
2
() 2.4 0.7 29
1206 2.4 0.7 29
0 2.4 0.7 1177
fx x x
xx
xx