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Algebra 2—2nd Semester Exam Review Problems Answers
1.

1
2.
3.

Divide using synthetic division: 10 x 4  5x 3  4 x 2  9  x  1
10
5
4
0
9
-10 5 -9
9
10 -5 9 -9
0 So answer is 10 x 3  5 x 2  9 x  9
Use a graph to estimate the real zeros of the function: f ( x)  x 3  x 2  49 x  49
Use calculator to find one zero (answer = 1). Then use synthetic division to get equation to become
quadratic (x2 + 49). Use quadratic equation to solve for last zeros. (Or in this case, set x2 + 49 = 0 and
solve. You get x = ± 7i )
Find all the zeros of the function: f ( x)  4( x  3)( 4 x  2)( 2 x  9)
Set each to 0 and solve. x = -3, x = ½, x = 9/2
7 x 8
3
 8 7 x  or  8 343x 3
3
4.
Write the following using radical notation.
5.
Find an equation for the inverse of the relation y  2 x  9 . Switch x and y, then solve for y.
6.
10.
Solve x  30  x . Check for extraneous solutions.
Square both sides, put all terms on the same side, factor, and set each piece to 0. You get x = 6 and
x = -5, but -5 is extraneous ( 25  5  5 ). Only answer is x = 6.
Write numbers in order, the one in the middle is the median: (83 + 88)/2 = 85.5
How many students got a score between 71and 80?
According to the graph, 8 students
Jane earned a 72 on her test. Describe how this compares with those of her classmates.
According to the box-and-whisker chart, the median is 72, so about ½ scored higher; about ½ scored lower
or letter “c”
Make t = 5, and solve for N. N  3000 3t  3000 35  729,000
11.
P = 250, r = 14% = 0.14, and t = 20, solve for A. A  P1  r   2501  .14  250(1.14) 20  $3435.87
12.
Simplify
7.
8.
9.
 
y
x9
2
 
t
20
54e 47
36e 26
54e 47 3e 4726 3e 21


2
2
36e 26
 $3664.21
Since you’re dividing, reduce the numbers, subtract the exponents.
13.
14.
15.
P = 3000, r = 4%= 0.04, t = 5, solve for A. A  Pe rt  3000e .045
Evaluate log 4 16
Either rewrite in exponential form and solve, or use change of base formula (use either “log” or “ln”).
log 16 ln 16
4 x  16, x  2

2
log 4
ln 4
Graph y  log 3 x
Since logs are strange, put in numbers for y, solve for x. Graph these as (x, y), connect the points. You
will need to rewrite the problem as you solve for x.
0  log 3 x
1  log 3 x
 1  log 3 x
ex:
16.
30  x
31  x
3 1  x
1 x
3 x
1
3
x
Graph looks like
(1,0)
(3,1)
( 13 ,1)
There is a vertical asymptote at x = 0. The graph gets close to the y-axis, but does not touch or cross it.
Express as a simple logarithm: log x 4  log x 27
Since both log base x, can combine them. log x 4  log x 27  log x 4  27  log x 108
17.
18.
19.
Put first row (years) in L1, and second row (time) in L2. Then do an exponential regression (STAT—
CALC—ExpReg—Enter twice). Round numbers to 2 decimals and replace a & b with numbers shown.
y  .55  1.45 x
k
Varies inversely means you’re using p  . Make p = 15, and n = 49, solve for k, and replace k in
n
equation. Then use this equation to determine p if n now equals 58.
k
p
n
k
So equation is p = 735/n. If n = 58, p equals $12.67
15 
49
k  735
x2
Graph f ( x) 
x4
To get x-intercept, set top to 0 and solve. x = - 2. To get vertical asymptote, set bottom to 0 and solve,
x = – 4. To get horizontal asymptote, look at the degrees. Since degrees are the same, divide leading
1
coefficients, y   1
1
Graph looks like:
x 2  8x  16 x 2  6 x  8 x  4x  4 x  2x  2 x  4x  2




 x  2
x  4x  2
 x  2
x2
x2  4
Since you’re diving by a fraction, flip the second over and multiply. Factor and cross-cancel matching terms.
20.
Divide:
21.
Solve
3x
6
7

 . LCD = 2x (x+1). Multiply every term by LCD, removing fractions. Solve for x.
x  1 2x x
2 x x  1 
3x
6
7
 2 x x  1 
 2 x x  1 
x 1
2x
x
2 x  3 x  6 x  1  7  2  ( x  1)
Now you can factor or use quadratic formula to find what the
x’s equal.
6 x 2  6 x  1  14 x  14
x
6 x 2  12 x  13  0
22.
23.
24.
25.
26.
12  456 6  114

 .78 and 2.78
12
6
A lunch menu consists of 8 different kinds of sandwiches, 4 different kinds of soup, and 9 different
kids of drinks. How many choices are there for ordering a sandwich, a bowl of soup, and a drink?
8 * 4 * 9 = 288
You could also use 8 C1 *4 C1 *9 C1  288
Find the number of distinguishable permutations of the letters MAXWELL WALTERS.
14!
 1,816, 214, 400
Total number of letters=14. There are 2 A’s, 2 W’s, 2 E’s, 3 L’s
2!2!2!3!
You own 6 sweaters and are taking 3 on vacation. In how many ways can you choose 3 sweaters
from the 6?
Since the order doesn’t matter, use combinations 6 C3  20
A single six-sided fair die is rolled. Find the probability of obtaining the number 4.
1
P(4)   .17  17% Either leave as a reduced fraction, write as a decimal, or as a percent.
6
1
x
Identify the following function as exponential decay or growth: f ( x)  1.27 
4
This is growth since 1.27 is greater than 1. It would be decay if 1.27 was between 0 and 1
27.
28.
29.
30.
31.
32.
33.
34.
35.
True or false: x  2 is a factor of x 3  3x 2  4 x  12 ?
True. If it is a factor, you can divide x 3  3x 2  4 x  12 by x – 2. One way to check is to use synthetic
division. If there is no remainder, it is a factor.
x 2  2 x  3 x  3x  1
Using the rational function: f ( x)  2
, identify any vertical asymptotes.

x  7 x  12 x  3x  4
Vertical asymptotes are found by setting the bottom to 0 and solving. So they are x = 3, x = 4
Draw your normal distribution curve, using 1 hour and 30 minutes as the mean, and 15 minutes as your
standard deviation. Add the percentages between 45 minutes and 2 hours
P(45 min – 2 hour) = .0235 + .135+.34+.34+.135=.9735
Given a graph, can you find the equation?
Be able to see the difference between the graphs of an exponential growth (and negative of), exponential
decay (and negative of) and a log.
Look at a scatterplot to decide what type of relationship is being shown
Be able to see the difference between the graphs of an exponential growth, exponential decay, quadratic,
and positive and negative linear.
An example of an inverse variation is:
Inverse variation means as one goes up, the other goes down. The example is (b) the more time you spend
at Howie’s, the lower your grades
x 1
x 1
Determine the domain of the function: f ( x)  3

x  x x( x  1)
The domain is your choice of inputs, in this case, what x’s you can choose from. The only ones you can’t
use are the ones which make you divide by zero. Factor the bottom, set to 0, and solve. x=0, x= -1
If f(x) = 2x + 3, and g (x) = x2 – 1, find f ● g.
f  g  f ( g ( x))  2( x 2  1)  3  2 x 2  2  3  2 x 2  1
Simplify
2
3
.
1
8 4
Since the bases are the same and you are dividing, write the base, and subtract the exponents.
8
8
36.
8
Simplify
2
3
1
8
2 1
3 4
8
8 3
12 12
8
5
12
4
13
125  2 .
4
13
1
5
5 8 13
125  2   5  2   2   
4
4
4
4 4 4
Be able to match graphs given equations. Think end behavior—use leading coefficient and degree to
help you.
Graphs look like:
x


,
f
(
x
)


a)
f ( x)  5 x 3  x 2  4 x  8
x  , f ( x)  
Follow PEMDAS
37.
b)
f ( x)  2 x 3  13x 2  5x  2
c)
f ( x)  x 4  3x 2  12
d)
f ( x)  3x 2  8x  17
x  , f ( x)  
x  , f ( x)  
x  , f ( x)  
x  , f ( x)  
x  , f ( x)  
x  , f ( x)  
38.
39.
40.
Find all the zeros of the function: f(x) = (x + 5)(x – 2)(3x – 7)
Set every piece to zero and solve. x + 5 = 0, x – 2 = 0, 3x – 7 = 0
x = – 5,
x = 2,
x = 7/3
Given f(x) = 4x – 3; g(x) = x + 5, find f(g(x))
f(g(x)) = 4 (x + 5) – 3= 4x + 20 – 3
Write a rule for the nth term of the arithmetic sequence. –7, –5, –3, –1, . . .
an  a1  (n  1)(d )
an  7  (n  1)(2)
an  7  2n  2
an  9  2n
41. Find the common difference of the arithmetic sequence. –2.4, –2.9, –3.4, –3.9, . . .
– 2.4 – ( – 2.9) = – 2.4 + 2.9 = 0.5
42. Find the sum of the first 50 terms of the arithmetic sequence, if the first term is 3 and the common
difference is 3.
 a a 
Sn  n  1 n 
 2 
 3  a50 
 50 
a50  3  (50  1)(3)  150

 2 
 3  150 
 50 
  2550
 2 
43. Give the first four terms of the geometric sequence for which a1 = 5 and r = ½
an  a1  r 
an  5  1 2 
n 1
n 1
a1  5  1 2 
11
a3  5  1 2 
1

 
j1  4 
3
44.
Evaluate:
31
 5  1 2   5,
a2  5  1 2 
5
2
 5  12   ,
4
a4  5  1 2 
0
5
2
5

8
2 1
 5  12  
4 1
 5  12 
1
3
j
 1  r n   1   1   14 
S n  a1 
   
1
 1  r   4   1   4 
3




1 

 63 
1
1  64  1  64  1  63 3  189
 
 
   
4  3  4  3  4  64 4  1024
 4 
 4 
45.
Simplify the following: (5x3 – 2)(12x3 – 3x2 + 2x – 1)
5x 3 (12x 3 – 3x 2  2x – 1) – 2(12x 3 – 3x 2  2x – 1)
 60 x 6  15 x5  10 x3  5 x 2  24 x3  6 x 2  4 x  2
 60 x 6  15 x5  14 x3  x 2  4 x  2
Algebra 2 Semester Exam Free Response Review--5 points each
1.
6.5—Divide using long division: 5x 4  14 x 3  9 x  x 2  3x

 

5x2  x  3
x 2  3x 5 x 4  14 x3  0 x 2  9 x  0
5 x 4  15 x3
 x3  0 x 2
 x3  3 x 2
3x 2  9 x
3x 2  9 x
2.
6.7—Find all the zeros of the function f ( x)  x 3  3x 2  25x  75
f ( x)  x 3  3 x 2  25 x  75
f ( x)  ( x3  3 x 2 )( 25 x  75)
 x 2 ( x  3)  25( x  3)
 ( x  3)( x 2  25)
x3 0
3.
4.
x 2  25  0
x  3
x  5i
8.5—Completely expand log 2 45x 3 y 4
When you expand, deal with division first, then multiplication, then the exponents. Division means you
use subtraction, multiplication means you use addition, and the powers go to the front of the log.
log 2 45x 3 y 4  log 2 45  log 2 x 3  log 2 y 4  log 2 45  3 log 2 x  4 log 2 y
8.6—Solve log 2x  log 5x  15  2
log 2 x  log  5 x  15   2
log 2 x  5 x  15   2
log 10 x 2  30 x   2
10

log 10 x 2  30 x
  102
10 x 2  30 x  100
10 x 2  30 x  100  0
10  x 2  3x  10   0
10( x  5)( x  2)  0
5.
x  5, x  2
8.6—Solve 35 x  2  2  6
Since the x is in the exponent, you will need to take a log to get it to drop down. Use a log of base 3
35 x  2  2  6
35 x  2  4
log 3 35 x  2  log 3 4
5 x  2  1.26
5 x 3.26

5
5
x  .65
You use the change of base formula to figure out log 3 4  1.26
6.
6.4—Factor 8 x 3  343
It is a difference of two cubes. It follows a pattern. a  b a 2  abx  b 2
In this case, a =
7.
3
8x 3  2 x and b =
3
343  7 .

So 2 x  74 x
2

 14 x  49

7.5—Graph the following: y  2 x  1  3 . Label and list the asymptotes and vertex.
Vertex: (1, 3). Put numbers in for x, to determine y. Graph points.
y  2 2 1  3
y  2 5 1  3
 2 13
 2 4 3
 23
 2*2  3
5
7
y  2 0 1  3
 2 1  3
(2,5)
(5,7)
You can’t take a square root of negative 1, so you are done. Also, this means you can’t choose any number
less than 1. Meaning the domain is all real numbers  1 . The range is all real numbers  3
Graph looks like:
8.
8.5—Completely expand log 2 45x 3 y 4
9.
8.6—Solve 35 x  2  2  6
10.
9.3—Graph y 
See #3—it’s the same question
See #5—it’s the same question
x
. List and label x-intercepts, vertical asymptote, and horizontal asymptote.
x3
To get the x-intercepts: set top = 0. x = 0. To get vertical, set bottom to 0, x = - 3, To get horizontal, since
1
degrees are the same, y   1 . Graph looks like:
1