COLLEGE ALGEBRA – Final Exam Review – KEY
1. b 2  4ac  52  x 
2. DQ =
3.
2  52
8
 x
2  2 13
8
 x
6( x  h)2  9  (6 x 2  9) 6 x 2  12 xh  6h 2  9  6 x 2  9 12 xh  6h 2


 12 x  6h
h
h
h
a) original:
b) original:
x
x
Up 4, Right 1
Vertical Stretch of 5, Down 2
c) original:
x
Left 3, Reflected over the x-axis
4.
a) Left 1, Down 3
b) Reflected over the y-axis, up 2
c) Reflected over the x-axis, vertical shrink
5.
a) 13  5 x  0   5 x  13  x  135 
b) 4 x  9  0  4 x  9  x 
c)
x
9
4

x
x
x
x  135  or (, 135 ]
9
4

or ( 49 , )
x  3,1 or (, 3)  (3,1)  (1, )
b 617.4

 63; C (63)  $151.90
2a 2(4.9)
6.
63 clubs; $151.90
7.
End Behavior: both ends point up
Zero
1
–1
–3
8.
1  13
4
Multiplicity
2
1
1
x
a) Domain:
x  0
VA: x  0 HA: y = 0
x
b) Domain:
VA: x 
c) Domain:
5
2
x
5
2

HA: None
x
VA: x   52
x   52 
HA: y  54
Cross/Bounce
B
C
C
Degree: 4
p  1,2,3,6
1
3
  1,  ,  2,  3,  ,  6
q  1,2
2
2
Use –3 and 2 in synthetic division, and you’ll get: f ( x)  ( x  3)( x  2)(2 x 2  1)
9. a) Possibilities of rational zeros are:
Set each factor equal to zero and solve and you’ll get 4 solutions: x  3, 2,  i

As a product of linear factors: f ( x)   x  3 x  2  x  i
1
2
 x  i 
1
2
1
2
p  1,2,4,5,10,20
  1,  2,  4,  5,  10,  20
q 1
Use 2 twice in synthetic division, and you’ll get: f ( x)  ( x  2)2 ( x 2  5)
b) Possibilities of rational zeros are:
Set each factor equal to zero and solve and you’ll get 3 solutions: x  2,  i 5


As a product of linear factors: f ( x)   x  2  x  i 5 x  i 5

10.
(f◦g)(x) = 2  (3x 2  1)  1  3x 2
11.
x  53 y  7  x  7  53 y  y  53  x  7  
12.
617  x
13.
log 4 1024  5
14.
2 log 2 a 
15.
log 2 a 3  log 2 54  log 2 26  log 2  a 3  54   6  log 2  625a 3   6
16.
1
2
f 1 ( x)  53 x  215
log 2 b
a) log 3 ( x  2)  5  35  x  2  243  x  2  x  241
ln(6)  1
b) 2e5 x 1  12  e5 x 1  6  5 x  1  ln 6  x 
5
2 x 1
5
c) 3  3  2 x  1  5  x  2
17. Let year 2003 represent t = 0. Then, A(t) = 6,302,486,693e.0116t
Solve 10,000,000,000 = 6,302,486,693e.0116t  t = 39.79 ≈ 40 years ≈ year 2043
 12 , 43 
b)
 3,
18.
a)
19.
a) Arithmetic
b) a20  3  19(4)  79
1
2
, 1
c) S20  202  3  79  820
21.
3
6
(1  12 )
b) Cannot sum since | r | > 1
a) Sum =
20.
a) Geometric
19
b) a20  1 2   524, 288
c) S20 
1(1  220 )
 1,048,575
(1  2)