College Algebra
PRACTICE TEST
Chapters 2 & 3
Function Notation & Polynomial Functions
Sketch the graph of the absolute value of each relation pictured below. Also give the domain and range of the
absolute value of each relation.
1]
2]
3]
4] Sketch the graph of the function y  ( x  2)  4 by using what you know about curve translations and the
2
absolute value function.
Solve each of the following equations/inequalities; use [interval, notation) in your answers to the inequalities.
5]
3x  4  2  7
8] Graph
6]
2 x  1, x  1
h( x )  
x 1
 x,
9]
x
3  4
2
Graph
7]
 x  1, x  1
.
p( x)   2

x
,
x


1

f  2 x 2  3 and g ( x)  2  3x , find each of the following.
10]  f  g (2)
11]  f  g (2)
12]  f  g (x)
Given
14] Given
f ( x)  3x 2  4 , find the difference quotient
13]

 
 4  5   25

16]
g  f (x)
f ( x  h)  f ( x )
; simplify your answer completely.
h
Perform each of the following complex number arithmetic problems.
15] 3 
3x  2  5  8
4  3i  2  5i 
18] Write P( x)   x  6 x  7 in the form P( x)  a( x  h)
system given – be sure to also specify the parabola’s vertex.
2
2
17]
3i
2  3i
 k ; then graph the parabola on the coordinate
For each quadratic function given below, a) find the function’s vertex using the vertex formula, b) graph the
parabola, c) find the function’s x-intercepts, if any, and d) state the function’s domain & range.
19]
P( x)  x 2  2 x  3
20]
P( x)  2 x 2  7 x  4
If a ball is tossed upward from the top of a 100 foot-tall building its height above the street below after t seconds of
flight can be predicted by the equation h(t )  100  88t  16t .
21] Find the maximum height of the ball’s flight.
22] Use the Quadratic Formula to approximate the time of the ball’s flight from the moment it is launched to
the moment it hits the street below.
2
Solve each equation.
23] 5 x  3 x  0
2
26]
x 2  4 x  3x(1  x)
24]
x 2  3 x  18
25]
2x 2  4x  3  0
27]
3x 2  6 x  4
28]
x 2  3  3x
College Algebra
PRACTICE TEST
Chapters 2 & 3
Function Notation & Polynomial Functions
Solve each inequality. Graph each of your solutions on the number lines provided.
29]
x 2  4 x  3x(1  x)
30]
2x 2  x  3  0
A rectangular piece of sheet metal has a length that is 3 inches longer than its width. It is to be made into an opentop box by cutting 3-inch squares from each corner and then folding up the sides.
31] Find the function V that represents the volume of the box in terms of the width w of the original piece of
sheet metal.
32] Use this function to find w if the volume of the box is 162 cubic inches.
33] A kite is lofted by the breeze in such a way that its height is always 3 times the horizontal distance between the
kite-flyer and the point on the ground directly below the kite. If there is currently 240 feet of string let out, how
high is the kite?
Perform each division.
34]
x 4  3x 3  5 x 2  2 x  16
x3
35]
4x 3  2x  5
x4
x  2 is a zero of f ( x)  2 x 3  11x 2  10 x  8 , use synthetic division assist you in finding the rest
of the zeros of f (x ).
36] Given
x  3 is a zero of f ( x)  x 3  x 2  x  15 , use synthetic division assist you in finding the rest of
the zeros of f (x ).
38] Find a polynomial with degree 3 and with zeros –3, 0, and 1 such that P (2)  20.
39] Find a cubic polynomial having the zeros  2 and 3  i .
3
2
40] Use the Rational Roots Theorem to find all the zeros of g ( x)  x  3x  4 .
37] Given
Sketch a graph of each of the polynomials described below.
P( x)   x( x  2)( x  2) 2
42] The polynomial with zeros x  1,2 & 4 and with y-intercept (0,4).
41]
*
*
*
And now, the key….
For problems #1 – 4, reflect any part of the curve below the x-axis above it (make the y-values positive), and leave
the rest of the curve alone.
1] D   2,  , R  0, 
2]
D   3,5 , R  0,3 3]
D   3,3 , R  0,3



4] The parabola between
6]


0,4 reflects above the x-axis.
...  8  x  6  8   2,14
8] The line y  2 x  1 stops at
9] The line stops at point

7]
 
5]


 
 9  3x  4  9  x   53 , 133
 3  3x  2  3   53  x  13   , 53   13 , 
1,3 (open circle), and the line
y  x starts at 1,1 (closed circle.)
 1,0 (closed circle), the upside-down parabola starts at the point  1,1 (open.)
10] f (2)  g (2)  5  (4)  1
11]
 f g (2)  f (2  3  2)  f (4)  2(4) 2  3  29
College Algebra
PRACTICE TEST
Chapters 2 & 3
Function Notation & Polynomial Functions
12]
f ( x)  g ( x)  (2 x 2  3)(2  3x)   6 x 3  4 x 2  9 x  6
13]
 g  f ( x)  2  3 2 x 2  3  11  6 x 2
15]
 (3  2i)(5  5i)  15  5i  10i 2  25  5i
17]
3  i 2  3i 3  11i 3 11


  i
2  3i 2  3i
4  9 13 13


14]
3( x  h) 2  4  (3x 2  4)
 6 x  3h
h
16]
18]
 8  26i  15i 2  7  26i
P( x)   ( x  3) 2  2  Vertex is 3,2
19] Vertex is
1,2, no x-intercepts, domain is all real numbers, & range is 2,   .
20] Vertex is
1 34 ,10 18 , x-intercepts are  12 ,0 & 4,0 , domain is all real numbers, & range is  ,10 18 .
21] Vertex is
2 34 ,221  221 feet high
23]
x  0 &  53
25] (Q.F.)
27] (Q.F.)
29]
x
24]
22]
Set = 0 and solve   6.5 seconds
x 2  3x  18  0  x   3 & 6
4  40
10
 1
4
2
3x 2  6 x  4  0  x 
26]
4 x 2  7 x  0  x  0 & 74
6  84
21
 1
6
3
4 x 2  7 x  0   ,0  74 , 
30]
28]
(Q.F.) x

3 3
3
3
 
i
2
2 2
 1, 32 
31] V ( w)  3( w  6)( w  3  6)  V ( w)  3( w  6)( w  3)
32]
3(w  6)( w  3)  162  (w  6)( w  3)  54  w 2  9w  36  0  w  12 inches
33] (Pythagoras)
( x) 2  (3x) 2  (240) 2  10 x 2  57,600  x  75.89...  height  228 feet high
34] (Long or Synthetic)
x 3  5 x  13 

55
x2
35]
(Placeholder!)

4 x 2  16 x  62 
36]
f ( x)  ( x  2) 2 x 2  7 x  4  ( x  2)(2 x  1)( x  4)  x  2, 12 & 4
37]
f ( x)  ( x  3) x 2  2 x  5  (Q.F.)  x   3 & 1  2i


38] P( x)  a ( x  3)( x  0)( x  1)  P( x)  2 x( x  3)( x  1)
253
x4
College Algebra
PRACTICE TEST
Chapters 2 & 3
Function Notation & Polynomial Functions
39]
P( x)  ( x  2)x  (3  i)x  (3  i)  P( x)  ( x  2)( x 2  6 x  10)
40] Possible zeros are  1,2 & 4  P( x)  ( x  2)( x  2)( x  1)  x   1 & 2 (multiplicity 2)
41] X-intercepts are  2,0 & 2, ”bounces” at  2 , another point is (1,3)
42]
Goes “up-down-up”