Download Math141 – Practice Test # 4 Sections 3

Math141 – Practice Test # 3 Sections 3.6, 3.2, 3.3, 4.1, 4.2
Answer
1.
1/3, - 2
ANSWER KEY
Work
Use synthetic division to find the other polynomial factor.
2 3  1  12 4
3
6
10  4
5
2
0
Read the bottom row: The last zero is the remainder; the other numbers are the
coefficients of the other polynomial factor: 3x2 + 5x – 2. Set this factor = 0 
3x2 + 5x – 2 = 0  Factor or use the quadratic formula  (3x – 1)(x + 2) = 0
3x – 1 = 0 or x + 2 = 0  3x = 1  x = 1/3 or x = -2
2.
-2, 1, 2
Since the degree (highest exponent of x) is 3, there are 3 zeros. There are 2
signs changes in f(x) indicating 2 or 0 positive zeros. f(- x) = -x3 – x2 + 4x + 4
 One sign change indicates 1 negative zero. Possible rational zeros (p / q)
come from factors of 4   1, 2, 4. Using the Remainder Theorem: f(1) = 0,
factors of 1
indicating that 1 is a zero. Using synthetic division gives x2 – 4 = 0  x2 = 4 
Use square root property  x = 2, x = - 2. [Synthetic division:
1 1 1  4
1
3.
4.
4
4
1
0
0
4 0
A.
X=2
X=-2
Y=3
No oblique asymptotes
Vertical asymptotes are found by setting the denominator of the reduced
fraction = 0. x2 – 4 = 0  Using square root property x2 = 4  x 2   4 
x = 2, x = - 2. To find the horizontal or oblique asymptotes: compare the degree
of the numerator N with the degree of the denominator D. N = 2, D = 2  N =
D  Horizontal asymptote at y = a/b or lead coefficient of numerator / lead
coefficient of denominator. Y = 3 /1 = 3. There are no oblique asymptotes if
there are horizontal asymptotes.
B.
X=-3
X=3
Y=0
No oblique asymptotes.
Vertical: x2 – 9 = 0  (x + 3)(x – 3) = 0  x = -3, x = 3. Horizontal / Oblique
asymptotes: N = 1, D = 2, so N < D  Horizontal at y = 0 (the x-axis). There
are no oblique asymptotes if there are horizontal asymptotes.
C.
X=3
No horizontal asymptotes
Y = 3x + 10 (oblique)
Vertical: x – 3 = 0  x = 3. Horizontal / Oblique: N = 2, D = 1  N = D + 1
 Oblique. Using long division
3x + 10
2
X – 3 | 3x + x – 2
3x2 – 9x
Change signs to subtract
10x – 2
10x – 30
Rational function (fraction) that does not reduce. Y-intercept: Let x = 0  y =
(2*0 + 3) / (0 – 1)  y = 3 / - 1 = - 3  (0, -3)
X-intercept: Let y = 0. Fractions = 0 when the numerator = 0  2x + 3 = 0 
2x = - 3  x = -3 / or – 1.5  (- 1.5, 0)
Vertical asymptote: denominator = 0  x – 1 = 0  x = 1 (dashed line)
Horizontal / Oblique: N = 1, D = 1  N = D  Horizontal at y = a / b = 2 / 1 =
2 (No oblique asymptotes when there are horizontal ones.)
Rational function: Domain: all real numbers, x  - 2, 5. Fraction reduces to
R(x) = ( x  2)
= 1 .
5.
( x  5)( x  2)
x5
Y –intercept: Let x = 0  y = 1 / (0 – 5) = 1 / -5 or (0, -1/5)
X-intercept(s): Let y = 0 which means the numerator = 0  No x in numerator
 no x-intercepts.
Vertical asymptote: denominator = 0  x – 5 = 0  x = 5 (line)
Horizontal / Oblique: N = 0, D = 1  N < D  horizontal at y = 0 (x-axis)
GAP at x = - 2 (from the factor that divided out).
A.
25
Rational function that does not factor and reduce.
y-intercept: Let x = 0  y = (02 + 4*0 + 16) / (0 – 3) 16 / - 3 (0, -5 1/3 )
x-intercept(s): Let y = 0  numerator = 0  x2 + 4x + 16 = 0 doesn’t factor 
use quadratic formula  no x-intercepts
Vertical: reduced denominator = 0  x – 3 = 0  x = 3
Horizontal / Oblique: N = 2, D = 1  N = D + 1  oblique  use long
division
x + 7  Y = x + 7 is the oblique asymptote.
x – 3 | x2 + 4x + 16
x2 – 3x
Change signs to subtract
7x + 16
7x – 21
37
F(g(3)) = f(32) = f(9) = 3(9) – 2 = 27 – 2 = 25
B.
64
G(f(-2)) = g(3*-2 – 2) = g(-6 – 2) = g(-8) = (-8)2 = 64
C.
3x2 – 2
D.
E.
(6x – 2)2 or
36x2 – 24x + 4
No
A.
209
209  function rule of f  1  function rule of f -1 (inverse)  209
B.
39
39  function rule of f -1  12  function rule of f  39
9.
Yes
Passes horizontal line test.
10.
C
Only one to pass the horizontal line test.
x2  4
5
Switch x and y and solve for y. x  5 y  4  Square both sides of the
equation.  x2 = 5y + 4  x2 – 4 = 5y  Divide by 5 on both sides to get
answer.
6.
7.
8.
11.
12.
13.
y
y3
x9
2
-1, multiplicity of 2, 3, i,
and – i
F(g(x)) = f(x2) = 3(x2) – 2 = 3x2 -2
G(f(2x)) = g(3*2x – 2) = g(6x – 2) = (6x – 2)2
To be inverses, f(g(x)) = g(f(x)) = x. f(g(x)) = 3(x2) – 2  x
Switch x and y and solve for y.  x = 2y3 – 9  x + 9 = 2y3  Divide by 2 on
both sides, then take the cube root.
Change the problem to g(x) = x5 – x4 – 4x3 – 4x2 – 5x – 3. These answers will
then be correct. Degree (highest exponent) = 5  5 solutions (roots or zeros).
Look at the sign changes in f(x) to find there is 1 positive zero. Look at f(-x) = x5 – x4 + 4x3 - 4x2 + 5x – 3 to find there are 4 or 2 or 0 negative zeros.
Possible rational zeros p 
q
factors of cons tan t term
3
  1 or 3
factors of leading coefficient 1
Since f(1) = - 16, 1 is not a zero. Since f(3) = 0, 3 is a zero. Since f(-1) = 0, - 1
is a zero. Since f(-3) = -240, -3 is not a zero. So far, we have one positive zero
(3) and one negative zero (-1). Use synthetic division for x = 3:
3 1 1  4  4  5  3
1
3
6
6
6
3
2
2
2
1
0
Then use synthetic division for x = - 1:
1 1 2 2 2 1
1 1 1 1
1 1 1
1 0
We now have a polynomial factor of x3 + x2 +x + 1 Since we have one negative
zero, we know that we have at least one more. -3 didn’t work, so it must be – 1.
Substitute – 1 into the polynomial factor (-1)3 + (-1)2 + (-1) – 1 = 0 to see that
this is true. There are now 3 zeros: 3, -1, and – 1. Use synthetic division to find
a new polynomial factor:
1 1 1 1 1
1 0 1
1 0 1 0
The new polynomial factor is x2 + 1. Since this is a quadratic, use the square
root property to solve: x2 = - 1  x 2    1  x =  i.
14.
A.
T
4800
S
k is the constant of proportionality. “Indirectly” translates as
T
k
S
. Substitute
80 for T and 60 for S, multiply by 60 on both sides of the =, to find k = 4800.
B.
15.
120 min
About 14.7 ohms
Substitute 40 for S in the formula developed in 15A.
kL
. Substitute 10 for R, 50
r2
 Multiply by 0.000036  0.0036 = 50k
Translating “directly” and “inversely” gives R 
for L and 0.006 for r  10 
50k
0.000036
 k= 0.0000072 (calculator will say 7.2 E – 6 meaning 7.2 X 10 – 6. This is
scientific notation  move the decimal point left 6 places because the exponent
is – 6.)
Substitute the second set of information into the formula  R  0.0000072  100
2
(0.007)