A2A MAT115 Practice Exam 4: Be aware that the following problems are not inclusive but just examples of the
content listed above. Go over the homework sets for further examples. Correct the problems on the hw that you got
wrong. The following problems are designed to match the content from WHS homework 10 – 13.
1.
The graph of f(x) is given in the following figure. Choose the appropriate answer from each list or fill in the blank. When writing
in roots enter from smallest to largest values and include repeated roots in your answer
a. The domain of f(x) =  , 
Because The x values go forever to the left and right.
b. The range of f(x) =  , 
Because The y values go forever down and up.
c. Estimate the local minima as an order pair (A,B). Round to the nearest 0.5
2,2.5
This is the low point in a curve
d. Estimate the local maxima as an ordered pair (A,B). Round to the nearest 0.5
 1.5,8.5
This is the high point in a curve
e. On what intervals is f(x)  0  3,1  3, 
These are the x-values that
keep the graph above or on the x-axis which is > 0
f.
On what intervals is f(x) < 0
 ,3  1,3
These are the x-values that
keep the graph below the x-axis which is < 0
2.
For the graphs G(x) and H(x) given to the right (round to the nearest integer):
G(x)
a. Find the local minima and maxima for each graph
G(x): minima (low point) at  1,3 maxima (high point) at  3,0, 2,4
H(x): minima (low point) at 2,2 maxima (high point) at  2,2
b. Find the x and y-intercepts for each graph
G(x): x-intercepts (points on x-axis) at -3, .5, and 3
y-intercepts (points on y-axis) at -1
H(x): x-intercepts (points on x-axis) at -4, 0, and 4
y-intercepts (points on y-axis) at 0
c. Find the multiplicity of the x-intercepts for each graph
G(x): -3 has multiplicity of 2 because the graph bounces
.5 and 3 have a multiplicity of 1 because the graph crosses
H(x): all 3 intercepts have a multiplicity of 1 because the graph crosses
d. Find the domain and range in interval notation.
G(x): domain
 5,  because the x-values go from -5 to the right forever
range  ,4 because the graph goes down forever and then up to 4
H(x)
 ,  because the x-values go to the left and right forever
H(x): domain
range  ,  because the graph goes down forever and up forever
e. Determine if the degree of the graphs are odd or even
G(x) is even because the left and right go in the same direction
H(x) is odd because the left and right go in opposite directions
f.
Is the graph is positive or negative?
G(x) is negative because the right side goes down
H(x) is positive because the right side goes up
g. What are the intervals of increase and decrease?
G(x): increase is  5,3   1,2 because these are the x-values that lead to the graph going up
decrease is  3,1  2,  because these are the x-values that lead to the graph going down
H(x): increase is  ,2  2,  because these are the x-values that lead to the graph going up
decrease is  2,2 because these are the x-values that lead to the graph going down
h. What is the power function for each graph
G x   a x  3 2 x  1x  3
2
Since -3, 3, and ½ are zero’s, they are the factors of the equation.
Therefore, the opposite of each one goes with x. The ½ needs to be
broken down into what would create that zero. There may be a value in
the front, so for now I will put a as a place-holder for that value. I also
included the multiplicity that attached to each one (see step c)
 1  a0  3 2(0)  10  3
2
Since the point (0, -1) is on the graph, I can substitute 0 for the x’s and -1
for the y to help me solve for a.
 1  a(9)( 1)( 3)
Simplify the right side
 1  27a
Continue simplifying
a
1
27
Gx   
Solve for a.
1 4
x
27
H x  axx  4x  4
Since x values multiply out to give x 4 , I put the a value in front of that.
Since the zero’s for H(x) are at 0, 4, and -4, The opposite of each goes
with x. Since each zero has a multiplicity of 1, each portion has an
exponent of 1.
2  a 2 2  4 2  4
Since the point (-2, 2) is on the graph, I can substitute -2 for the x’s and 2 for the y to help me solve for a.
2  a(2)( 6)( 2)
Simplify the right side
2  24a
Continue simplifying
a
1
12
H x  
i.
Solve for a.
1 3
x
12
Since x values multiply out to give x 3 , I put the a value in front of that.
Represent each graph in factored form of the lowest degree. (Don’t forget to include the sign, greatest common
factor, and include multiplicity.)
Gx   
1
x  32 2 x  1x  3
27
Once you know the a value from part h, then you can substitute it into
the equation.
H x  
1
xx  4x  4
12
Once you know the a value from part h, then you can substitute it into
the equation.
3. Consider the quadratic function f ( x)   x 2  x  2
a. Write the quadratic function in standard notation.
To go for standard notation, we have to complete the square.
f ( x)   x 2  x  2


f ( x)   x 2  x  2
If there is a value in front of x 2 , factor it out of the x’s.
2
1

f ( x )   x    2
2

To complete the square, take half of the value in front of x in the
parenthesis (which is a -1). This goes after the x in this new parenthesis.
1

f ( x)   x 2  x    2
4

1
1

f ( x)   x 2  x    2 
4
4

1
and place it back in the original parenthesis.
2
1
Since I added a
to the equation, I have to “undo” that so I
4
1
don’t change the equation. The
is actually being multiplied through
4
1
the parenthesis by -1. Therefore, to get rid of a  that I added to the
4
1
equation, I have to add to the end.
4
Square the 
2
1
9
1
9

f ( x)   x   
Finally, I can add the 2 and the
to get my final value of
4
4
2
4

2
b. Find the vertex of f ( x)   x  x  2 using your answer from part (a)
1 9
The vertex comes from the opposite of the value in the parenthesis for
 , 
2 4
the x-value and the outside value for the y-value.
c. Is the vertex a maximum or minimum? Explain how you know this.
Maximum
Since there is a negative in the front, the parabola faces down which
means there must be a high point (maximum).
4. Given g ( x)  3x 2  6 x  10 , find the average rate of change between the points x = –2 and x = 3.
g (2)  3 2  6 2  10  10
Average rate of change means slope. So my first step is to find the
g (3)  33  63  10  35
points I need by substituting my x-values into the equation.
 2,10, 3,35
Now I have my points which I can substitute into the slope formula.
y 2  y1 35   10 45


9
x2  x1
3   2
5
Substitute to find slope for the final answer.
2
2
5. Given the function f ( x)  2 x 2  4 x  3 , find the x- and y-intercepts, vertex, the axis of symmetry, concavity,
domain, and range. Graph the function using what you have found.
To get the y-intercept, substitute a zero for x and solve.
f (0)  2(0) 2  4(0)  3  3
0  2 x 2  4 x  3
x
x
To get the x-intercept, substitute a zero for the y and solve
 42  4 23
2 2
4
4  40 4  2 10 2  10


4
4
2
Since this problem will not easily factor, I can use the quadratic equation.
Simplify the quadratic equation to get the x-intercepts
b
4

 1
2a 2 2
To get the x-value of the vertex, solve for
f (1)  2 1  4 1  3  5
To get the y-value of the vertex, substitute x into the equation
 1,5
This gives the vertex which is the maximum because there is a negative
x  1
in front of the x 2 value
The axis of symmetry comes from x = the x part of the vertex
Concavity: Down
This is because there is a negative in front of the x 2 value
2
 , 
 ,5
b
2a
Domain:
There are no domain restrictions (from fractions or radicals)
Range:
The highest y-value is 5 and the graph goes down forever.
6. Mike wants to enclose a rectangular play area for his rabbits next to his large barn using 30 feet of fencing. What
dimensions will maximize the fenced area if the barn is used for one side of the play area?
15 - x
x
15 - x
If we call one side x, the opposite side is also x (since it is a rectangle.) What is left for
the fence then is 30 – 2x (because there was 30 feet to begin with, and I subtract the two
x sides I have named already.) Since the two sides that are left are equal, that leaves half of
30  2 x
the 30 – 2x for each side. Therefore, the two sides are each
or 15 – x.
2
A  x15  x   15 x  x 2   x 2  15 x
Since the goal is to maximize the area, I need to find the area. Since the
formula is length times width, I can set up an equation to describe area.
 b  15

 7.5
2a 2 1
Since the maximum is at the vertex, I use
15  7.5  7.5 so 7.5 ft  7.5 ft
b
to find the x-value.
2a
We know the x side, so I can find the 15 – x side to get my dimensions.
7. Mona Kalini gives a walking tour of Honolulu to one person for $49. To increase her business, she advertised at the
National Orthodontist Convention that she would lower the price by $1 per person for each additional person up to 49
people. Mona’s revenue from her revenue can be modeled as a function of the number of people on the tour:
R( x)  49 x  x 2 . What number of people on the tour would maximize her revenue? What is the maximum revenue
for her tour?
This is a quadratic, so I put the values in order of degree.
R( x)  49 x  x 2   x 2  49 x
 b  49

 24.5  25
2a 2 1
Since the maximum is at the vertex, I use
round up to 25 because we can’t have a portion of a person.
R(25)  4925  25  $600
2
8.
b
to find the x-value. I
2a
By substituting by x-value into the equation, I can find the revenue.
An object is shot upward from the floor of a canyon 160 feet deep with an initial velocity of 120 ft/sec. The height of the object
can be modeled by the function h(t )  16t 2  120t 160 .
a. What is the maximum height the object reaches?
h(3.75)  163.75  1203.75  160  65 Once you get the x-value of the vertex
2
b.
9.
(part b), substitute it back into the equation.
How long will it take the object to return to the ground?
 b  120

 3.75
2a 2 16
To maximize a quadratic, use the formula
b
.
2a
c.
How long will it take for the object rise to the height of the canyon wall?
It won’t because the canyon reaches 160 feet but the highest the object goes is 65 ft.
a.
Find the average rate of change for the tangent line on the graph to the right.
2
3
b.
c.
Find the slope of the line by doing
rise
run
What is another way of describing the average rate of change of the tangent line?
Slope
Find the equation of the tangent line.
y  mx  b
3

2
2  b
3
13
b
3
y
2
13
x
3
3
Since the slope is
2
and the line goes through the point (2, -3), I can set up an
3
equation by substituting 2 for x and -3 for y into the slope-intercept form.
Multiply the right side, then subtract to get b.
Put your slope and y-intercept into the equation for m and b.
10. Given the functions find the following if they exist: x- and y-intercepts, vertical and horizontal/slant
asymptotes, domain and range (in set notation), and graph the function using this information.
x2
a. f ( x)  2
x 9
x20
x  2
f (0) 
To find the x-intercept of a rational function, set the top = zero and solve
02
2

2
0 9
9
To find the y-intercept, substitute a zero for each x and solve.
x2  9  0
x  3x  3  0
x 3 0 x 3 0
x  3
To find vertical asymptotes, find the zero’s from the denominator that don’t cancel with
the top. For this problem I factored and set my factors = zero to solve.
x3
y0
If the degree on top is smaller than the bottom, the horizontal asymptote is y = 0
x x  , x  3
The domain cannot include the zero’s from the bottom.
y y  , y  0
Since y = 0 is an asymptote, the range cannot hit zero.
2
b. g ( x)  2 x  3 x  1 (ignore the range)
x2
2 x 2  3x  1  0
3
To find the x-intercept of a rational function, set the top = zero and solve
 32  42 1
22
3  17
4
g (0) 
Since this problem did not factor, I used the quadratic equation to find the x-int.
Reduce as far as possible.
2
20   30   1
02
x20
x2

1
2
To find the y-intercept, substitute a zero for each x and solve.
To find vertical asymptotes, find the zero’s from the denominator that don’t cancel with
the top.
2 2  3  1
2
4
1
y  2x  1
x x  , x  2
2
1
To find the slant asymptote, the top degree has to be one higher than the bottom degree.
If this is the case, then you divide the denominator into the numerator. This time I chose
to do that using synthetic division. Ignore the remainder for the slant asymptote.
The domain cannot include the zero’s from the bottom.
11. Which of the following factors are roots or zeros of the polynomial
4x  2
1
x
2
x2
x 3
x4
x2
x 3
x4
P( x)  4 x 4  6 x 3  18 x 2  2 x  6
4x  2
x 1
1
x
x  1
2
First set every factor equal to zero to find the factor’s zero.
4
3
2
1 1
1
1
1
P   4   6   18   2   6  0
2 2
2
2
2
Substitute your x-values back into the original equation.
P2  42  62  182  22  6  54
If your answer from substitution gives a zero, then that term is
P3  43  63  183  23  6  0
a factor. If you get any other answer, it is not a factor.
4
3
4
2
3
2
P4  44  64  184  24  6  350
4
3
2
4
3
2
 1
 1
 1
 1
 1
P    4    6    18    2    6  3.5
 2
 2
 2
 2
 2
P 1  4 1  6 1  18 1  2 1  6  0
4
3
2
12. Solve the following inequalities (one is rational and one is quadratic) by finding critical points and using test points. Write you
answer in interval notation.
2
1
a.

x 1
x 1
2
x 1

1
x 1
0
Move all portions of the problem to one side of the inequality.
2 x  1 1 x  1

0
x 1
x 1
Get a common denominator by multiplying each fraction by what it is missing.
2x  x  x  1
x  1x  1  0
Distribute the values to combine fractions.
2x  1
x  1x  1  0
Combine like terms.
2x  1  0
x 1  0 x 1  0
1
x
x 1
x  1
2
Set the top and bottom equal to zero to find the critical points.
Put your critical points on a number line
-.5
-1
1
x  2 x  .75 x  0 x  2
Pick a number in each section of the number line.
2 2   1
2 .75   1
20   1
22   1
 2  1 2  1  1  .75  1 .75  1  1.14 0  10  1  1 2  12  1  1.67
Substitute each back into the equation to see what is positive and what is negative. This can be placed on the number line.
-1
-
+
-.5
 1,.5  1, 
+
1
Since the problem asked for >, take the positive sections for your answers.
4 z 2  12z   9
b.
4 z 2  12 z  9  0
Get everything to one side so we can compare to zero.
2z  32z  3  0
Solve by factoring or use quadratic equation to find the zero’s.
2z  3  0
z  1.5
+
Since I factored, set the factored portions equal to zero and solve.
+
1.5
x 1 x  2
Pick a number in each section of the number line.
41  121  9  1 42  122   9  1
2
2
Substitute each back into the equation to see what is positive and what is negative. This can be placed on the number line.
 ,1.5  1.5, 
13. Use synthetic division to divide
Since the problem asked for >0, I give the sections that are positive.
f ( x)  x3  13x  12 by x  4 . Write your answer in factored form.
 4 | 1 0  13 12
Since I am dividing out x + 4, I place the zero from that factor (-4) in the top left
|
hand corner. After that value, I place the coefficients (numbers in front of the
x’s) in order as well. Notice that a zero was put in place of the
not an
4| 1
0  13 12
4
1
x 2 since there is
x2
Drop the first number. Multiply it by the -4 and put it under the next value.
|
4| 1
0  13 12
 4 16  12
1 4 3
0
Continue this process throughout the values.
|
x2  4 x  3
The first value gets one degree less than the original degree of the problem.
Each of the following values get one less degree than the one before.
Remainders (which this one does not have because of the zero at the end) get
added as a fraction over the original divisor at the end.
14. If the degree of P ( x )  7 and the given zeros are
x  12 , x  7 , x  i 10 , x  4  6 , x  2  6i , then write
P (x ) in factored form.
Zeros: 12, -7,
 i 10 , x  4  6 , x  2  6i Conjugates of imaginary numbers always match up as zero’s.




Px   x  12x  7  x  i 10 x  i 10 x  4  6 x  2  6i x  2  6i 
The opposite of each zero (change the signs) goes with an x in each parenthesis to write the polynomial in factored form.
15. Find each of the following for the rational graph given to the right.
a.
b.
c.
d.
e.
x  4 Vertical line we don’t touch
y  3 Horizontal line we don’t touch
x-intercept
x  2 Where the graph touches the x-axis
y  1.5 Where the graph touches the y-axis
y-intercept
Domain x x  , x  4 The vertical asymptote is not an x
Vertical asymptote
Horizontal asymptote
that can be used for the graph.
16. Graph the following functions using transformations. State the domain and range. (Remember you must reflect across the y-axis
before you shift horizontally. Don’t forget to include the asymptote as part of your graph.)
a.
g ( x)  e  x  2  1
g ( x)  e x
Domain:
Range:
b.
 , 
 1, 
F: Flip y-axis
Domain:
Range:
 , 
 ,4
S: No Stretch
g ( x)  e  x  2
g ( x)  e x  2
Range:
 3, 
 , 
g ( x)  e  x  2  1
The range of this graph starts at the asymptote and goes up forever.
F: Flip x-axis
f ( x)  e x  3
S: No Stretch
f ( x)  e x 3
U: Up 4
f ( x)  e x 3  4
This is always the domain for an exponential graph.
The range of this graph goes down forever and up to the asymptote.
L: Left 3
F: No Flip
p( x)  log 2 ( x  3)
p( x)  log 2 ( x)
p( x)  log 2 ( x  3)
Domain:
U: Down 1
This is always the domain for an exponential graph.
f ( x)  e x 3  4 L: Right 3
f ( x)  e x
c.
L: Left 2
S: No Stretch
The domain goes from the asymptote to the right forever.
This is always the range for a log graph.
U: No Up/Down
d.
L: Right 3
F: Flip x-axis
S: No Stretch
U: Up 1
k ( x)  1  log 2 ( x  3)
k ( x)  log 2 ( x)
k ( x)  log 2 ( x  3)
k ( x)   log 2 ( x  3)
k ( x)  1  log 2 ( x  3)
Domain:
Range:
e.
3, 
 , 
The domain goes from the asymptote to the right forever.
This is always the range for a log graph.
F: Flip y-axis
S: No Stretch
U: Up 1
k ( x)  log 2 ( x  2)  1 L: Left 2
k ( x)  log 2 ( x)
k ( x)  log 2 ( x  2)
k ( x)  log 2 ( x  2)
k ( x)  log 2 ( x  2)  1
Domain:
Range:
 ,2
 , 
The domain goes from left forever to the asymptote of 2.
This is always the range for a log graph.
17. Write each logarithmic equation in exponential form each exponential equation in logarithmic form.
a b  c can be rewritten as log a c  b . Follow this pattern for each answer.
a.
3  log 2 8
23  8
b.
1
 log 81 9
2
1
2
81  9
c. 3
2

3
1
9
 2  log 3
d. e 4  2.117
1
9
3
 ln 2.117
4
18. Determine the value of each logarithm without using a calculator.
a. log 11 121
b.
log 11 121  x
Since I don’t know the answer right now, I can set it equal to x to solve.
11x  121
Now I can follow the pattern (see #17) to get an exponential equation.
112  121
x2
Since 11 squared is 121, x must be a 2.
log 81 9
log 81 9  x
Since I don’t know the answer right now, I can set it equal to x to solve.
81x  9
Now I can follow the pattern (see #17) to get an exponential equation.
81  9
1
x
2
c.
Since the square root of 81 is 9, x must be
1
2
log 7
1
49
log 7
1
x
49
Since I don’t know the answer right now, I can set it equal to x to solve.
7x 
1
49
Now I can follow the pattern (see #17) to get an exponential equation.
7 2  49
1
72 
49
x  2
Since
7 2 is 49 and to get the value on the denominator we make the exponent negative,
then x must be -2.