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CHAPTER 4 TEST C
Directions: Show all work.
Name: ___________________________
Section: __________________________
1. Which one of the following graphs represents y  6 x  x 2 ?
a.
b.
c.
d.
e.
f.
 5 
2. Which of the following equations has intercepts at  2, 0  ,  0,1 and   , 0  ?
 2 
1
1
1
a. y    2 x  5  x  2  b. y    5 x  2  x  2  c. y    2 x  5  x  2 
10
4
10
d. y 
1
 2 x  5 x  2 
10
e. y 
1
 2 x  5 x  2 
10
y
f.
1
 5 x  2  x  2 
4
3. Which one of the following gives an equation for the parabola shown?
a. y  3  x  2   2
2
b. y  3  x  2   2
2
c.
y  2  x  2   2
2
d. y  2  x  2   2
2
4. The parabola y  ax 2  bx  c whose coefficients fit the system of equations at right
contains which one of the following sets of points?
a b  c 1
a.
1, 1 ,  1, 2 , 5,3
Chapter 4/Form C
b.
 1, 1 , 1, 2 ,  3,5
4a  2b  c  1
9a  3b  c  5
c.  1,1 , 1, 2 ,  5,3
d.
1,1 ,  2, 1 , 3,5
e.
 1,1 ,  2, 1 ,  3,5
f.
 1,1 ,  4, 1 , 9,5
5. Which one of the following equations results from combining the equations in the
y  3x 2  2 x  8
system
?
x  3y  6
a. 9 x 2  5 x  24  0
b. 9 x 2  6 x  24  0
c. 3x 2  2 x  6  0
d. 9 x 2  5 x  18  0
e. 9 x 2  6 x  24  0
f. 3x 2  2 x  8  0
6. Which of the following intervals gives the solution to the inequality 3x 2  x  10 ?
5

5 
 5 
a. x   ,    [2, )
b. x    , 2 
c. x   , 2 
3

3 
 3 
5
5
5 


d. x  (, 2]   ,  
e. x   2,  
f. x   2, 
3
3
3 


For problems 7-10, find the coordinates of the intercepts and the vertex of the parabola
whose equation is given, then sketch a graph.
7. y  1  4 x 2
8. y  4 x 2  36
9. y    x  2 x  7 
10. y  2  x  5   18
2
In problems 11-13, write an equation for the parabola passing through the origin  0, 0 
and having its vertex at the given point. Sketch a graph for each showing the
coordinates of all intercepts.
5

5 5
13.  , 
11.  3,9
12.  , 5 
2 4
2

14. Solve the system of equations algebraically and verify your solutions with a graph.
y  32  2  x  4 
2
y  4 x  10
15. Solve the inequality: 
Chapter 4/Form C
7
 x  1 x  12   0 . Write the solution in interval notation.
11
16. Solve the inequality: 27  x  2   12  0 . Write the solution in interval notation.
2
17. The area of the shaded region shown at right is formed by
subtracting the area of a circle with radius r from the area of
a square with edge s: A  s 2   r 2 . Show that if the edge of
the square is 8 less than 3 times the radius of the circle, then
the area of the shaded region is A   9    r 2  48r  64 .
18. Find exact values for r and s in problem #17 to produce a shaded region of area 60.
Simplify the radicals in your expressions.
19. The sum of half one number and one third of another number is 4. Find two such
numbers so that their product is maximized.
20. Salt water is entering a tank full of water at the same time that the water is draining
out of the tank. The water is draining out of the tank at a faster rate than the water is
coming in so that the total amount of salt in the tank, S, (measured in grams) at time
1
1
t (measured in hours) is given by the quadratic equation, S 
t
t2 .
400 16, 000, 000
After how many hours is the amount of salt in the tank at a maximum?
21. Approximate the solutions to
7 r 2  5 r  3 , accurate to the nearest hundredth.
22. Approximate the solution interval to 7 r 2  5 r  3 where r is positive. Round to
the nearest hundredth in the solution interval.
17
41
23. Approximate the x-intercepts of the parabola described by y  19 x 2  x  .
3
3
Chapter 4/Form C
For problems 24 and 25, use the
Chesapeake Bay temperature data
tabulated at right.
24. Use linear regression to find the
best linear fit for the temperature
during June in terms of the
number of years since 1986.
25. Use quadratic regression to find
the best quadratic fit for the
temperature during 1988 in
terms of the number of months
since the beginning of the year.
Temperature data for the years 1986 - 1990 taken from
the middle of the Bay mouth (degrees Celsius)
Month 1986
1987
1988
1989
1990
Jan.
5.28
5.52
1.56
5.76
5.28
Feb.
3.72
3.60
4.68
5.28
7.20
Mar.
5.76
5.28
7.20
5.88
9.72
April 10.90
9.84
11.40
11.50
12.50
May 15.80
17.60
17.30
16.70
18.40
June 21.60
20.40
21.80
22.50
21.20
July 25.60
24.40
24.70
26.00
25.10
Aug. 25.10
25.90
22.80
25.10
26.30
Sept. 22.40
24.40
21.60
22.70
24.00
Oct. 20.00
16.90
18.00
18.20
21.10
Nov. 13.40
12.60
12.80
13.60
13.00
Dec.
9.36
6.96
9.72
7.70
8.64
Solutions For Chapter 4 Test Form C.
1. c
2. c
3. c
4. e
5. d
6. b
7. The intercepts of y  1  4 x 2 are
 1 
1 
  , 0  ,  0,1 and  ,0  .
 2 
2 
The vertex is at  0,1 .
8. The intercepts of y  4 x 2  36 are
 3,0 ,  0.  36 and 3,0.
The vertex is at  0, 36 .
9. The intercepts of y    x  2 x  7  are
 7,0 ,  0,14 and  2,0 .
 5 81 
The vertex is at   ,  .
 2 4
10. The intercepts of y  2  x  5   18
2
are  2,0 ,  0, 32 and 8,0 .
The vertex is at  5,18 .
Chapter 4/Form C
11. Plugging  x, y    0,0 into y  a  x  h   k ,
2
shows that a  
k
. If the vertex is at  3,9 ,
h2
9
then a    1 . The parabola
9
2
y    x  3  9 , has another intercept at  6, 0  .
12. a 
2
5
5 / 2
2
4
4
5
  y   x    5 gives an
5
2
5
equation for the parabola. 0ther intercept at
5,0 .
13. Again, a  
2
5/ 4
5 / 2
2
1
5 5
yields y    x   
5
2 4
as an equation for the parabola. The other
intercept is at  5,0 .
14. Substituting y  4 x  10 into y  32  2  x  4  we
2
have 4 x  10  32  2  x  4 
This is equivalent to
2x2 12x  10  0  2  x 1 x  5  0 so that
2
x  1 or x  5 . Computing the corresponding y-values,
we find solutions at 1,14  and  5,30 , as shown:
7
15.   x  1 x  12   0   x 1 x 12  0  x 1,12.
11
4
2
2
4  8 
2
2

16. 27  x  2   12  0   x  2    x  2   or x  2   x   ,    ,   .
9
3
3
3  3 

2
2
2
17. Substituting s  3r  8 into A  s   r A   9    r  48r  64 , we have
A   3r  8    r 2   9    r 2  48r  64 .
2
18.  9    r  48r  64  60  9    r  48r  4  0  r 
2
r
2
48  4 144   9   
2 9   
Chapter 4/Form C

48  482  4  9    4 
2 9   
24  2 135  
 0.08 or 8.11. But if r  0.08 , then
9 
s  3r  8  0 , which is impossible. So r 
 24  2 135  
s  3 
9 

24  2 135  
and
9 

8  6 135  
  8 
9 

19. Let a and b represent the two numbers. Then
a b
3
  4  b  12  a so that the
2 3
2
3 
3

product is ab  a 12  a    a 2  12a . This expression is maximized where
2 
2

3
12
a
 4  b  12   4   6 . Thus the numbers we seek are 4 and 6.
2
 3
2  
 2
1
1
1
400
20. S 
t
t 2 is maximized where t 
 20, 000 hours.
2
400 16, 000, 000
16, 000, 000
 5  5  12 7 2.236  36.749 2.236  6.062


 0.72 or  1.57
5.292
5.292
2 7
7 r 2  5 r  3 and r  0 if r   0,0.72 (see #21.)
21. r 
22.
17
41
x 0
3
3
 57 x 2  17 x  41  0
23. y  19 x 2 
x
17  289  4  57  41
.
114
 1.01025 or 0.71200 Above is shown how a TI85 calculator could be
programmed to find these solutions. Note that the discriminant , D, is computed.
24. The screen shots below show how the linear regression is carried out on a TI-85.
This process varies quite a bit from calculator to calculator. The third screen is
interpreted to show the regression line is T  0.13t  21.4 .
25. Screen shots for quadratic regression are shown below. The regression parabola
fitting temperature to time (in months) is T  0.57t 2  7.31t  1.86 .
Chapter 4/Form C