Practice Exam 3: Be aware that the following problems are not inclusive but just examples of the
content listed above. Go over the three homework sets for further examples. The following problems
are designed to match the content above.
Quadratic Formula:
 b  b 2  4ac
2a
Vertex:
b
2a
Know the basic graphs of the following:
1.
y  x2
a.
yx
b.
d.
y x
e. y  x
c.
y  x3
f. y 
1
x
a. f x   3 x  4  5
Start as the basic
b.
y  x graph.
Move to the left 4.
Move the graph up 5.
f x   3  2( x  1)2
Start as the basic y  x 2 graph. Move to the right 1.
c.
Multiply each y-value by 3.
Multiply each y-value by -2.
Move the graph up 3.
Flip over the y-axis.
Move the graph up 4.
f x   4  3  x
Start as the basic y  x graph.
Move to the left 3.
2. Given the graph on the right find the base graph f(x) and the transformed f(x)
The original graph is the function
y x
The graph flipped over the x axis, so there should be a negative in the front: y   x
The original graph has the first point (we can easily track) after the origin at (1, 1). The new
graph has the first point doubled (from (-4, -2) to (-3, -4). So it was doubled. This puts a
2 in the front with the negative: y  2 x
The graph moved left 4: y  2 x  4
The graph moved down 2: y  2 x  4  2
3. If y  2 f ( x  2)  2 is given, describe how the graph of f has been transformed
1.
y = f(x + 2) : Move left 2.
3.
y = 2f(x + 2) – 2: Move down 2
2. y = 2f(x + 2) : Multiply y-values by 2
4. Given the graph of h(x) below, draw y  2 f ( x  1)  2
1. Move left 1.
2. Flip over the x-axis.
3. Multiply the y-values by 2.
4. Move up 2.
5. The collision impact of an automobile varies jointly as Its mass and the
square of its speed. Suppose a 2000lb car traveling at 55 miles per hour
has a collision impact of 6.1. What is the collision impact of the same
car at 65 miles per hour?
Since it is a joint variation, the right is being multiplied by some variable, k, which gives the following: C  kms
2
Now I can substitute the values I know into the equation: 6.1  k 200055
2
This allows me to solve for k: k 
6.1
2
200055
Now I can substitute k back into the original equation: C 
6.1
ms2
2
200055
Finally, I can substitute the final information to solve for the final impact: C 
6.1
2000652
2
200055
C = 8.52
6. a) Given the piecewise graph of f(x), find f(-2) and f(2)
To find f(-2), look on the graph for the y-value when the x-value is -2 which is -2.
To find f(2), look on the graph for the y-value when the x-value is 2 which is 6.
b) Find the domain of f(x)
Because the x values go to the left forever, the smallest value is   . Every x is
accounted for on the graph after that up until 6. It includes 6, so it gets a
bracket. So the domain is:  ,6 .


c) Determine the range of f(x) in interval notation.
The lowest y-value is -4. It is’t included (not colored in), so it gets a parenthesis. All y-values are included up
until 6. Although one of the 6’s is not included, the other one is, so it will get a bracket. So, the range is (-4, 6].
d) Describe f(x) as a function.
if
x  2
 2
 x 2  2
if  2  x  2
f ( x)  
3
 x  7 if x  2
 2
The line on the left is y = -2. It only occurs
when x < -2.
The middle curve is a parabola that has moved
up 2. It is between the x-values of -2 to 2.
The third line (if you continue it at the same
slope) is y = 3/2x - 7. It occurs when x > 2
if
 3
 x  2
if
7. Given the piecewise function g ( x)  
 2 x  1 if
 3
x  2
2 x3
find f(0), f(2), f(3)
2 x6
To find f(0), find which of the three equations includes the 0 option for x. Since the second equation takes x’s from -2
to 3, it is the one we can use. So I substitute 0 into the x value and get:
0  2  2
To find f(2), I will use the second equation again because it includes the 2 (it has an equal sign with the less than). So I
substitute the 2 into the second equation and get:
2 20
This is the entire graph
Graph g(x)
This is the necessary portion
First I need to graph the first option which is y = -3.
However, this graph is only needed when x < -2. It
does not include the -2, so it gets an open circle for
that portion of the graph.
For the second graph, I graph a basic
This is the entire graph
This is the necessary portion
This is the entire graph
This is the necessary portion
y  x graph that
moves down 2. I then only need when x > -2 and x < 3.
The -2 is equal so it gets a closed circle and the 3 gets an
open circle.
For the final graph, I graph the line y 
2
x  1. I then
3
only include when x > 3 which gets a closed circle.
Finally, put the graphs together on one graph.
8. For the graphs, find the local minima and maxima, x and y-intercepts and the domain and range.
Local Min: (1, 0)
This is a low point where the graph goes from dec. to inc.
Local Max: (0, 3) (2.5, 6.8) These are high points where the graph goes from inc. to dec.
Domain:
Range:
 , 
 ,6.8
x-int: -.5, 1, 3
These are where the graph touches the x-axis.
y-int: 3
This is where the graph touches the y-axis.
This shows the x-values that work for the graph. It goes to the left and right forever.
This shows the y-values that work for the graph. It goes down forever, but only up to 6.8.
This graph is neither odd nor even. It is not symmetric to the origin (odd) or the y-axis (even).
f x   x  .5x  1  x  3
2
The three zero’s are -.5, 1, and 3. Each of these go into a parenthesis with
opposite signs. Only the 1 gets an even exponent (2) because it bounces back where it came from. The other two
zero’s get odd exponents (1) because they cross over. There is a negative in the front because the right side goes down.
Domain:
Range:
 , 
 , 
Local Min: (.5, -1.2)
This is a low point where the graph goes from dec. to inc.
Local Max: (-.5, 2.5)
This is a high point where the graph goes from inc. to dec.
x-int: -1, .2, 1
These are where the graph touches the x-axis.
y-int: 1
This is where the graph touches the y-axis.
This shows the x-values that work for the graph. It goes to the left and right forever.
This shows the y-values that work for the graph. It goes down and up forever.
This graph is neither odd nor even. It is not symmetric to the origin (odd) or the y-axis (even).
f x  x  1x  .2x  1
The three zero’s are -1, .2, and 1. Each of these go into a parenthesis with
opposite signs. They all cross over at the zero’s, so they all have odd exponents (1). There is a positive in the front
because the right side goes up.
9. Find the domain and range of the following functions :
f ( x)  2 x 3  3 x 2  5
g ( x)  2 x  1
h( x ) 
2x
x 4
2
m( x ) 
x4
2x  3
Domain is the values that can be substituted into x. The domain is normally only affected by x’s in the denominator of
a fraction or under an even root.
f ( x)  2 x 3  3 x 2  5
Domain:
Range:
g ( x)  2 x  1
 ,  This problem has no fractions or radicals.
 ,  Look at a basic cubic graph. It goes up and down forever.
Domain: You can not have a negative under an even radical. Therefore, I can set up
the inequality
2x  1  0 which gives x  .5 so the domain is .5, 
Range: Look at a basic radical graph and the y values have to be non-negative. So the
range is
2x
x 4
h( x ) 
.5, 
Domain: This problem has a fraction. You can not have a 0 in the denominator which
2
leads to x  4  0 . Solve this to get
2
x  2 which is the domain.
Range: If you graph the equation, you will see that you can get any y so it is
m( x ) 
x4
2x  3
 ,  .
Domain: The domain has two issues here. No 0 can be in the denominator so I solve
2x  3  0 so x  1.5 . You also have to deal with the even radical which leads to
the problem
x  4  0 which gives  4,  and x  1.5 as the domain.
Range: If you graph the equation, you will see that you can get any y so it is
10. Given g ( x)  1  x
h( x ) 
4
x4
 ,  .
f ( x)  x 2  2 x
a. Determine the domain of each function individually.
g(x): domain is x > -1. You can’t have a negative under the radical, so we have the inequality 1 + x > 0. Subtract
1 from both sides to get x > -1.
h(x): domain is
x  4 . You can’t have a zero in the denominator, so we have x  4  0 . Subtract 4 from
both sides to get
x  4 .
f(x): Domain is
 ,  .
There are no fractions or roots to cause problems, so all x’s work.
b. Find (f + h)(x) and the resultant domain in set notation.
f
 hx  means you need to add f(x) to h(x). This will give you x 2  2 x 
comes from the denominator.
4
. The domain issue here
x4
x  4  0 so the domain restriction is x x  4 .
c. Find (g/f)(x) and the resultant domain in set notation.
g / f x means you need to divide g(x) by f(x).
The radical gives us the restriction
This gives
1  x  0 so x  1 . You can’t have a 0 in the denominator which leads to
the restriction x  2 x  0 . I can factor that to be
2
restrictions
1 x
. This leads to two domain restrictions.
x  2x
2
xx  2  0 . I solve each part for 0 which gives the
x  0 and x  2 . My final restriction combines those to get x x  1; x  0,2 .
d. Find ( f  h)( x) and the resultant domain in set notation.
 f  hx means you need to multiply f(x) and h(x).
I get
This will give you


4
x 2  2 x . If I multiply this out,
x4
4 x2  8x
. The domain restriction comes from the denominator, so x x  4 .
x4
e. Find ( f  h)( 3)
This problem asks for f(-3) – h(-3). I find f(-3) by substituting -3 into f(x) which gives  3  2 3  3 .
2
Likewise, h(-3) gives
4
 4 . So, f(-3) – h(-3) is 3 – 4 which is -1.
3 4
f. Find ( f  h)( x)
See part c.
g. f(g(x)) or ( f o g )( x)
This problem asks you to substitute the g(x) into the x’s in f(x). This gives
simplifying the first part, you get
 1 x  2
2
1  x . By
1  x  2 1  x . The domain restriction comes from the radical. So
1  x  0 which gives x  1 . So the domain is  1,  .
h. h(f(x)) or (h o f )( x)
This problem asks you to substitute the f(x) into the x’s in h(x). This gives
4
. The domain
x  2x  4
2
restrictions come from the denominator. This leads to x  2 x  4  0 . I solve this with the quadratic equation
2
since it can’t factor. The quad. equation gives the following: x 
x
 2  22  414
. This is
21
 2   12
. Since there is a negative under the radical, there are no domain restrictions.
2
i. g(f(2)) or ( g o f )( 2)
This problem asks you to find f(2) and then to put that answer into g(x). First, f 2   2   22  which is 8.
2
Then, I find g 8  1  8 which is 3.
j.
f x  3 * 2 f x
f(x – 3) means substitute x – 3 into the x’s of f(x). This gives
x  32  2x  3 .
I need to multiply out  x  3 . This can be rewritten as (x – 3)(x – 3). When I multiply that out, I get
2
x 2  6 x  9 . Now I add that to 2(x – 3) to get x 2  4 x  3
Now I have
x
2



2
 4 x  3 * 2 f x  . For the 2f(x), you multiply f(x) by 2. This gives 2 x  2 x . When I
distribute, I get 2 x  4 x .
2
Now I have
x
2


 4 x  3 * 2 x 2  4 x . Now I multiply these out by distributing the 2x 2 and the 4x . This
gives 2 x  4 x  10 x  12 x .
4
3
2
11. Given f(x) is in maroon and h(x) is in blue, find the following:
a. (f - h)(2) = -1
This is asking for f(2) - h(2). The 2’s are in the x spot, to I want to use the graph to find
the y values when the x is 2. f(2) is 4 while h(2) is 5. When I subtract those, I get 1.
b. (f + h)(-6) = 3
This is asking for f(-6) + h(-6). The -6’s are in the x spot, to I want to use the graph to
find the y values when the x is -4. f(-6) is 0 while h(-6) is 3. When I subtract those,
I get 3.
c. (f * h)(-3) = -4
This is asking for f(-3) * h(-3). The -3’s are in the x spot, to I want to use the graph to find
the y values when the x is -3. f(-3) is -1 while h(-3) is 4. When I multiply those, I get -4.
d. (f + f)(-4) = -4
This is asking for f(-4) + f(-4). The -4’s are in the x spot, so I want to use the graph to find
the y-values when the x is -4. f(-4)is -2. So, -2 + -2 = -4
12. a) Given f ( x) 
To find
f
1
3x  2
find f-1(x). State the domain and range of f(x) and f-1(x).
4
x  , you switch the x and y in the equation.
first step is to multiply by 4:
y
x
This would give
3y  2
4 . You then solve for y. The
4 x  3 y  2 . Finally solve for y by adding 2 and dividing by 3 which gives
4x  2
4x  2
f 1  x  
3 or more officially,
3 . The domain and range for both is  ,  since there are no
x’s on the denominator of either one.
b) Given f ( x)  2 x  6 , find f-1(x). State the domain and range of f(x) and f-1(x).
To find
f 1 x  , you switch the x and y in the equation. This would give x  2 y  6 . Subtracting the 6 you get
x  6  2 y . To get rid of the radical, I must square both sides which gives x  62  2 y . Finally, get y by itself
y
by dividing by 2. This gives
x  62
2
f 1  x  
or more officially
x  62
2
. The domain of the original can be
found by setting the 2x > 0 since you can’t have a negative under an even root. This gives domain of f(x) to be
The range is
0,  .
6,  since it is a radical graph moved up 6. The range and domain switch when you do an inverse, so
the domain of
f 1 x  is 6,   and the range is 0,   .
1
2
13. Use the inverse function property to determine if f x   5x  2 and g x    x  are inverses.
5
5
To find an inverse of a function, switch the x and the y values. So find the inverse of f(x) and see if it matches g(x).
The inverse of f(x) would be x  5 y  2 . I need to solve for y so I will subtract 2 and divide by -5 which gives
1
2
f 1  x    x  . That is not g(x), so they are not inverse of one another.
5
5
Another way to do this is to see if f(g(x)) = x. To do this, I substitute g(x) into the x’s of f(x). f(g(x)) gives
2
 1
 5  x    2 . I distribute the -5 which is x  2  2 or x + 4. This is not just x, so they are not inverses.
5
 5
14. Find the inverse functions for the following functions. Write your answer in terms of f 1 x  .
a.
f x  
1
3x  2
2
so f 1 x  
5x
5x  3
x

3y  2
5 y . Now I need to solve for y.
x , I need to switch the y’s and the x’s. This gives
To find f
I multiply both sides by 5y which gives 5 yx  3 y  2 . Now get all y’s on one side. So I subtract the 3y which
gives 5 yx  3 y  2
y5x  3  2 .
2
2
1
To get y by itself, I divide by 5x – 3 which gives y 
. This is the inverse, so it is f  x  
.
5x  3
5x  3
Now I can factor out a y on the left which gives
b.
f x   2 x 2  4 ; x > 1 so f 1 x  
x4
; x > -2
2
f 1 x  , I need to switch the y’s and the x’s. This gives x  2 y 2  4 . Now I need to solve for y.
x4
2
 y2 .
I add 4 to both sides. This gives x  4  2 y . Now I divide by 2 which is
2
x4
x4
1
Square root both sides to get y 
which can be rewritten as f x  
.
2
2
To find
Now I deal with the original problem that says x > 1. f(1) = -2, so the y-value is -2. In the inverse, we switch the
x’s and the y’s. So for the inverse, x > -2 instead of 1.
15. Consider the quadratic function f x    x 2  x  2 .
a. Find the vertex of the parabola by completing the square and using the vertex formula.
(.5, 2.25)


px    x 2  x ___  2
To complete the square, you need to factor the beginning number
(if it is not a 1) out of the x-values. The last value stays on the
outside. There is a gap left to finish the process.


px    x 2  x  .25  2
On the next line, take half of the middle number in the parenthesis,
Square that number and add that into the parenthesis on the original.
Even though this added a positive .25, it is inside the parenthesis with
a negative one being multiplied by it outside the parenthesis. This
makes the value actually a -.25. Since I put a -1 on that side of the
parenthesis, I have to do the opposite of that to keep the equation
true. Therefore, I put a +.25 on the same side. When I put that with
the +2 that is already there, it becomes a +2.25 on the outside of the
parenthesis.
p x    x  .5
2


px    x 2  x  .25  2  .25
px   x  .5  2.25
2
The vertex comes from the formula for a quadratic which is f  x   a  x  h   k where the vertex is (h, k).
Therefore, the vertex is (.5, 2.25).
2
b. Is the vertex a maximum or minimum?
Maximum
Because there is a negative in the front, this parabola opens downward. The vertex is therefore a maximum.
2
16. Given g ( x)  3x  6 x  10 , find the average rate of change between the points x = -2 and x = 3.
9
Average rate of change is another way of saying slope. To find slope, we need two points. So, I need to find the point
when x = -2 and another when x = 3.
g  2   3 2   6 2   10 which is -10. So the first point is (-2, -10)
2
g 3  33  63  10 which is 35. So the second point is (3, 35).
y  y1
35   10
The formula for slope is 2
. When I substitute in the points I get
which is 9.
x2  x1
3   2
2
17. Given the graph S(t) on the right:
a. State the domain and range or S(t) in set notation.
Domain: x x  4,3,0,1,2,4
Range: y y  3,1,3,4,5
The domain includes the x-values of the points. That includes -4, -3, 0, 1, 2, 4.
The range includes the y-values of the points. That includes -3, -1, 3, 4, 5.
Set notation puts the letter you are using in the front followed by a line. After that you
put the answers using equal or inequality signs.
b. Is S(t) a one-to-one function? Why.
No. One-to-one means each x-value is only used once and each y-value is only used once. The y-value of 4 is
used twice at (0, 4) and at (2, 4).
c. Explain in words how the domain and range of S(t) is related to the domain and range of
its inverse.
In an inverse you switch the x’s and the y’s. Therefore, the domain and range also switch. In other words, the
domain of a function is the range of the inverse and the range of the function is the domain of the inverse.
d. Draw the inverse function for S(t). Is the graph one-to-one?
I take each point from the original and switch the x and y-values. So instead of the point (-4, -1),
I have the point (-1, -4). Do this for each point. The scale is the same as the original.
Just like the original was not one-to-one, the inverse is not. This time the x-value of 4 is used
twice at (4, 0) and at (4, 2).
18. Be able to write your answer in set notation or interval notation and know the difference.
(, 2]
 {x | x  2}
(24, 15)  {x | x  R,  24  x  15}
To write an answer using interval notation, the first number is the smallest number allowed while the second number is
the largest number allowed. A parenthesis means you don’t include the number (this is always used for infinity) and a
bracket means to include it. So,  ,2 means the answers go from negative infinity to 2 and includes the 2.


Another way to write it is x < 2 Likewise,
Another way to write this is -24 < x < 15.
 24,15 means the answers go from (but don’t include) -24 to 15.
To write an answer using set notation, place an
inequalities.
x , pronounced x bar, in the front followed by the answer written using
19. 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.
a. The x-intercepts come from making the y value a 0 and solving for x. This gives 0  2 x 2  4 x  3 . I
solve this using the quadratic equation which gives
  4 
 42  4 23 .
2 2
This gives
4  4 * 10
4  40
4  2 10
which is
. I reduce the radical to get
. Finally, I divide by -2 to
4
4
4
 2  10
reduce. This is
. To graph these, I add -2 to 10 to get 1.16. I divide by 2 to get .58. I do
2
the same with -2 - 10 to get -2.58. This gives the points (.58, 0) and (-2.58, 0).
b. The y-intercept comes from substituting a 0 for the x-values.
2
This gives y  20   40   3 which is 3. This is where
the graph touches the y-axis, so I have the point (0, 3).
vertex
b
c. The x-value of the vertex comes from the formula
.
2a
  4
This gives
which is -1. I can substitute this value into
2 2
y-intercept
the equation to get the y which is 5. So the vertex is (-1, 5).
d. The axis of symmetry always comes from the x value of the
e.
f.
g.
vertex. The axis of symmetry is x = -1.
The graph has a negative in the front, so it is concave down.

x-intercept
x-intercept

The domain is  ,  since there are no radicals or fractions.
The range is  ,5 . The vertex is the highest point since the
parabola opens downward. This means the y-values go from
negative infinity to 5. The five is included because it is a used point.


20. If 100 meters of fencing will be used to fence a rectangular region, then what dimensions for the
rectangle will maximize the area of the region?
50 - x
First I want a drawing of the problem. 100 meters of fencing describes the perimeter.
Since it is a rectangle, opposite sides are equal. So, two sides can be called x. If we
have two x’s, we have 100 - 2x left. This needs to be divided in half to cover each of
the remaining two sides. So, the other sides are 50 – x.
x
Since the problem wants us to maximize area, I need to get an equation for area. Area is length times width, so for this
specific problem, area is x times 50 – x, or
Ax   x50  x  . I can distribute the x to get Ax   50 x  x 2 .
b
To minimize or maximize a quadratic, I use the vertex. To get the vertex, I use the formula 2 a . If I write the
 50
2
Ax    x  50 x so the vertex formula gives 2 1 which is 25. If x is 25, I
equation in decreasing order, I have
can get the other side by substituting 25 into the x of 50 – x. This also gives 25. So my sides need to be 25 meters by
25 meters to maximize the area. (Hint: Anytime you want to maximize a rectangle, you use sides that create a square.)
If you are asked for the area, substitute the x value (25) into the area formula and solve.
21. 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 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?
b
This is a parabola. To get the maximum (or minimum), you need the vertex. This comes from the formula 2 a . If I
Rx    x 2  49 x so the vertex formula gives  49 which is 24.5.
write the equation in decreasing order, I have
2 1
Since we are talking about people, I can’t have .5, so I can round to 25 to find the number of people necessary.
R25  25  4925 which is
2
To get revenue, I have to substitute the value back into the equation which gives
a revenue of $600.
22. For each of the graphs, find the intervals of increase and decrease.
Increase tells where a graph is going up. Specifically, it is what x-values occur when the graph is going up.
Decrease tells where a graph is going down. Specifically, it is the x-values when the graph is going down.
Increase:
 ,0 ; 1,2.5
Increase:
 ,.5 ; .6, 
Decrease:
0,1; 2.5, 
Decrease:
 .5.6
x2
5x  3
, g x  
find the following if they exist: x- and y2
x 9
10 x  2
intercepts, vertical and horizontal asymptotes, domain and range (in set notation), and graph the
function using this information.
23. Given the functions f  x  
a.
f x  
x2
x2  9
x-intercepts: We get this by substituting 0 the y (or f(x)). This gives 0 
x2
. For a fraction to be equal
x2  9
to zero, only the top needs to be 0. So, I make the equation x + 2 = 0. Solving this gives an x-intercept of -2.
02
2
which is  .
2
0 9
9
2
Vertical Asymptotes: These come from the zero’s in the denominator. I need to solve x  9  0 . I add 9
2
to both sides to get x  9 . Now square root both sides (don’t forget the plus or minus sign) to get x  3 .
y-intercept: We get this by substituting 0 for x. This gives f 0  
Horizontal Asymptotes: If the degree on top is smaller than the bottom, it is y = 0. If the degree on top is
equal to the degree on bottom, write them in descending order and the HA is y = a/b where a is the number in
the front of the top and b is the number in front on the bottom. If the degree on the bottom is larger than the
top, there is no HA. For this function, the degree on top is 1and on bottom it is 2. So the HA is y = 0.
To get the domain, I look for x’s that cause problems. In this problem, you can’t have 0’s in the denominator.
We found those when we found the vertical asymptotes. So, in set notation, the domain is
x x  3 .
Range: This is looking for y-values. The only one I won’t be able to get will come from the horizontal
asymptote. That is the value y can’t be. So the range is
yy0
Graph: Graph the x-intercepts as points on the x-axis
y-intercept as a point on the y-axis
VA as a vertical line through the x
HA as a horizontal line through the y
Don’t touch the asymptotes
Based on the given points, draw curves
that go through given points but don’t
touch the asymptotes.
To find other points, substitute in x’s
to find y’s to have points.
b.
f x  
VA
HA
VA
x-intercept
y-intercept
5x  3
10 x  2
x-intercepts: We get this by substituting 0 the y (or f(x)). This gives 0 
5x  3
. For a fraction to be
10 x  2
equal to zero, the top needs to be 0. So, make the equation 5x - 3 = 0. Solving this gives an x-intercept of .6.
3
50  3
which is .
2
100  2
Vertical Asymptotes: These come from the zero’s in the denominator. I need to solve 10 x  2  0 which is
y-intercept: We get this by substituting 0 for x. This gives
f 0 
x = .2
Horizontal Asymptotes: If the degree on top is smaller than the bottom, it is y = 0. If the degree on top is
equal to the degree on bottom, write them in descending order and the HA is y = a/b where a is the number in
the front of the top and b is the number in front on the bottom. If the degree on the bottom is larger than the
top, there is no HA. Here, the degree on top and bottom is 1. So the HA is y = 5/10 or y = 1/2.
To get the domain, I look for x’s that cause problems. In this problem, you can’t have 0’s in the denominator.
We found those when we found the vertical asymptotes. So, in set notation, the domain is
x x  .2 .
Range: This is looking for y-values. The only one I won’t be able to get will come from the horizontal
asymptote. That is the value y can’t be. So the range is
y y  .5
Graph: Graph the x-intercepts as points on the x-axis
y-intercept as a point on the y-axis
VA as a vertical line through the x
HA as a horizontal line through the y
Don’t touch the asymptotes
Based on the given points, draw curves
that go through given points but don’t
touch the asymptotes.
To find other points, substitute in x’s
to find y’s to have points.
VA
y-intercept
HA
x-intercept
24. Solve each of the following inequalities. Write your answer in interval notation.
2
1

a.
First I need to get the inequalities on the same side, so I subtract to the left.
x 1 x 1
2
1

 0 Now I need to get a common denominator to combine the fractions.
x 1 x 1
2x  1  1x  1
x3
 0 Combine the top to get
 0.
x  1x  1
x  1x  1
Get the critical points which comes from setting the top and bottom equal to 0. This gives -3, 1, and -1.
Put the critical points in order on a number line.
-3
-1
1
Next, I need to substitute values into the equation that would fall into different sections of the chart. For
instance, in the first section, I can substitute any value less than -3. If I substitute a -4, I get
43
 4  1 4  1 . This gives a value of -.07. Since it is negative, every value in that section is negative.
So, I put a negative value for that section. I do the same process in the other sections.
23
Substitute a -2,  2  1 2  1 which is .3.
03
Substitute a 0, 0  10  1 which is -3.
-
-
+
-3
-1
+
1
23
Substitute a 2, 2  12  1 which is 1.7.
This problem is asking for the positive values (> 0), so I want the section from -3 to -1 and 1 to infinity. I
can’t include the 1 or -1 in the answer because they were zero’s from the denominator (which isn’t allowed).
The -3 is included because you are looking for > 0. Therefore, the answer is
b.
4 z 2  12 z  9
[-3, -1), 1, 
First get everything to one side by adding 9 to both sides.
4 z 2  12 z  9  0 Next, I need the zero’s, so I will factor the quadratic.
2z  32z  3  0 Solve for 0 by setting the parenthesis equal to 0.
This gives 2z – 3 = 0 so z = 1.5 (I
could have used the quadratic formula instead.)
Now I put the zero on the number line.
1.5
Now, pick numbers on either side of the zero and substitute back into the inequality.
Substitute a 0,
Substitute a 2,
2(0)  32(0)  3 which is 9.
+
2(2)  32(2)  3 which is 1.
+
1.5
Since both are positive, both sections get a positive sign.
This problem wants the positive values, ( > 0), so I need both sections. The 1.5 is not included because it is
not greater than or equal to 0. So, the answer is
 ,1.51.5, 
25. Use long division to divide P(h)  h4  3h3  h  5 by h 2  3 . Write your answer in factored form
 P ( h)  Q ( h) D ( h)  R ( h) .
First I write the problem as a division problem. I put a zero as a place-holder for missing degrees of h.
h 2  3 h 4  3h3  0h 2  h  5
2
2
2
Next, I need to decide what to multiply the front value of h by to get h . I need to multiply by h , so that goes
4
2
above the h . I now multiply that by both the h and the -4 and write it below the problem as seen.
h 2 _______________
h 2 _______________
h 2  3 h 4  3h3  0h 2  h  5 I am now going to subtract that. h 2  3 h 4  3h3  0h 2  h  5


h 4  3h 2
 h 4  3h 2
The h 4 and the  3h 2 cancelled, so now I need to see if h 2 can divide into anything else. The next thing that is left is
the h which has a smaller degree, so you can’t divide any further. The h – 5 is the remainder.
(h 2 ) times the divisor (h2  3) plus the remainder (h  5) which is
P(h)  h 2  3(h 2 ) h  5
26. Use synthetic division to divide f ( x)  x3  13x  12 by x  4 . Write your answer in factored form.
Now write it as the quotient
First, I take the zero from x + 4 = 0, so I use -4 in the synthetic division.
Next, I can rewrite the function as
placeholder.
f ( x)  x3  0 x 2  13x  12 . I added 0x 2 to make sure each degree had a
Now I write the problem with the zero (-4) on the outside and the coefficients on the inside.
-4 1
0 -13 12
Next, bring down the 1. Multiply the 1 by the -4 and write it below the 0. Add those, and write it. Keep up the
process until you finish through the 12.
-4 1
0 -13 12
-4 16 -12
1 -4 -3 0
The last value in the line is the remainder. Each value up from there adds a degree of x, with the first value being a
number by itself. So the answer is: x  4 x  3 .
2
27. Given R( x)  x3  3x 2  4 x  12
a. Find all the possible rational roots.
To find possible rational roots, write the function in decreasing order of degree.
The factors of the front number become the q’s and the factors of the last number become the p’s. Since the
first number is a 1, the q’s are 1 and -1. Since the last number is a 12, the factors are 1, -1, 2, -2, 3, -3, 4, -4, 6,
-6, 12, and -12.
Now find the
 1,2,3,4,6,12
p' s
. This will be
which is  1,2,3,4,6,12 .
1
q' s
b. Find all of the actual roots of R(x) over the complex numbers.
First I need to find which of the rational roots work. To do this, I can substitute them into the original
equation to see which ones give me a zero as an answer. 3 works because
R(3)  3  33  43  12 which is 0.
3
2
Now factor the 3 out using synthetic division (see problem 26 for help).
3
1
1
-3
3
0
4 -12
0 -12
4 0
This leaves x  4 . To get the last zero’s, I need to solve x  4  0 . I subtract the 4 to get x  4 .
2
2
2
Square root both sides to get x    4 . The square root of a negative gives i and the square root of 4 is 2,
so the zero’s from this part is  2i . So all my zero’s are 3 and  2i .
c. Write R(x) in factored form.
To write it in factored form, each zero gets a parenthesis with a changed sign. So the 3 becomes
x  2i  and the -2i becomes x  2i  .
Rx  x  3x  2i x  2i  .
2i becomes
So, when I write it in factored form, I get
x  3, the