Chapter 11 Test
18.
2
14fx x 
Since
1a
is positive, the parabola opens upward.
The vertex of the parabola is
,1,4hk 
and the
2
01 4 14 5y  
19.
2
23fx x x
Since
1a
is positive, the parabola opens upward.
The x–coordinate of the vertex of the parabola is
22
1
2212
b
a

 
and the
y–coordinate of the vertex of the parabola is
1
b
ff



The x–intercepts are –1 and 3. Set x = 0 and solve
for y to obtain the y–intercept.
2
0203 3y 
downward and has a maximum at
64 64 2.
2 2 16 32
b
xa
  

The ball reaches its maximum height in two
The baseball reaches a maximum height of 69 feet
after 2 seconds.
21.
2
016 645tt  
Solve using the quadratic formula.
16 64 5abc 


2
64 64 4 16 5
216
x 
Chapter 11 Quadratic Equations and Functions
22.
2
46 360fx x x 
23. Let 25.ux
2
2
25 42530
430
310
xx
uu
uu
 


24. Let
2
.ux
42
13 36 0
xx

Chapter 11 Test
25. Let
1/3
.ux
2/3 1/3
2
1/3 1/3
980
980
xx
xx


26.
2
12 0xx 
Solve the related quadratic equation.
2
12 0
430
xx
xx
 

40or 30
43
xx
xx
 

The boundary points are 3 and 4.
Interval Test Value Test Conclusion
,3  −4
2
44120
80, false

,3  does not belong to the solution set.
Chapter 11 Quadratic Equations and Functions
27. 21
3
3
x
x
Express the inequality so that one side is zero.
21
30
3
x
x

Find the values of x that make the numerator and denominator zero.
10 0 and 3 0
10 3
10
xx
xx
x
 
 
The boundary points are 3 and 10. Exclude 3 from the solution set, since this would make the denominator zero.
Interval Test Value Test Conclusion
,3 0
20 1 3
03
13, true
3

,3 belongs to the solution set.
Cumulative Review Exercises (Chapters 1 – 11)
1.
91132
99136
xx
xx
 
 
Cumulative Review
2.
34 7
29
xy
xy


Multiply the second equation by 2 and add the result
to the first equation.
34 7
xy

3. 39
23 16
52 15
xy z
xyz
xyz
 


Multiply the second equation by 3 and add to the
first equation to eliminate z.
We now have a system of two equations in two
variables.
78 39
3 1
xy
xy


3 1
3(1) 1
31
4
4
xy
y
y
y
y




4.
71892
218 2
220
220
22
10
xx
x
x
x
x
 




conditions. Therefore, the solution set is
,4 
.
6.
248or 73
24 10
xx
xx
 

Chapter 11 Quadratic Equations and Functions
8.
242
3
x

9.
2
46 24
33 9
xx x


46 24
33 33
xx xx

 
10.
431xx 

22
431
431
43231
xx
xx
xx x
 

   
Cumulative Review
11.
2
2
254
2450
xx
xx


Apply the quadratic formula:
2 4 5abc
2


12.
21
33
11
33
2
560
560
xx
xx


Substitute
1
3
x
back in for u.
13.
2
260xx
Solve the related quadratic equation.
2
Chapter 11 Quadratic Equations and Functions
The boundary points are
2
and
3
2
.
2
Interval Test Value Test Conclusion
3
, 2 , 2 does not belong to the solution set.
23 3 60
90,False
   

14.
36
36
6
3
12
xy
yx
x
y
yx




15.
11
2
fx x
This is a linear function with slope
1
2
m
and y-intercept
1b
.
Cumulative Review
16. 32 6xy
First, graph the equation 32 6xy
as a dashed
line.
32 6
236
xy
yx


17.
2
232fx x  
Since
2a
is negative, the parabola opens
downward. The vertex of the parabola is
,3,2hk
. Replace
fx
with 0 to find
x-intercepts.
2
2
02032
23 2
29 2 18 2 16
f 
 
    
16 24
xy
 
19.
26
32 4 6 34
74
54 20
20
xy xy x y
xy




20.
2222
22
8911 745
8911745
516
xxyy xxyy
xxyyxxyy
xxyy
  
 
 
21.
2
2
31256 1525
6135
xx xxx
xx


Chapter 11 Quadratic Equations and Functions
23.
2
32
5710
x
xxx

24.
2
22
2
2
2
99
11
33
11
9
3
33 3
3
x
xx
x
xx
x
xx
xx x
xx x





26.
232
22
510 50
25 2
52
xy x y x y
xxy
xy x


29.
422
2
81 1 9 1 9 1
xxx
 

22
339
xyx xyy
  
32.

2
2
2
315 2
315 2
213
fgx fx gx
xx x
xx x
xx




2
552513
fg

22
22 2
()3()15 315
23315315
h
ah ah a a
h
aahhah aa


Cumulative Review
37.
39
39
xy
yx

 
The line whose equation we want to find has a slope
38. Let x = the computer’s original price.
0.30 434
0.70 434
xx
x

3
x

Disregard
13
3
because the width of a rectangle
cannot be negative. If
4x
, then
invested at 14%.
41. Because I varies inversely as R, we have the
following for a constant k: