College Algebra for Calculus
Exam 2 Key
Instructions
1. Do NOT write your answers on these sheets. Nothing written on the test papers will be graded.
2. Please begin each section of questions on a new sheet of paper.
3. Please do not write answers side by side.
4. Please do not staple your test papers together.
5. Limited credit will be given for incomplete or incorrect justification.
6. All answers must be exact. Numeric approximations will be marked incorrect.
Questions
1. Graphing
(a) (8) Graph the rational function described in Figure 1.
20
10
-6
-4
2
-2
4
6
-10
-20
(b) (4) Write a rational function that matches the roots and vertical asymptotes in Figure 1.
(x + 2)x(x − 2)
(x + 3)(x − 4)
1
Exam 2: More Functions
2
(c) (5) Graph f (x) = 2 − 2x+3
2
1
-6
-5
-4
-3
-2
-1
-1
-2
-3
Exam 2: More Functions
3
2. Solving
(a) (5) ln(x) + ln(x − 3) = ln 10.
ln(x) + ln(x − 3)
=
ln 10.
ln(x(x − 3))
=
ln 10.
e
ln(x(x−3))
= eln 10 .
x(x − 3)
=
10.
x − 3x − 10
=
0.
2
(x − 5)(x + 2) = 0.
x = −2, 5.
(b) (5)
20
1+e−0.5t
= 15.
20
1 + e−0.5t
20
20
15
5
15
1
3
ln(1/3)
=
15.
=
15(1 + e−0.5t ).
=
1 + e−0.5t .
= e−0.5t .
= e−0.5t .
=
ln(e−0.5t ).
− ln 3 = −0.5t.
ln 3
= t.
0.5
2 ln 3 = t.
(c) (8) 3x3 − x2 − 12x + 4 = 0.
Possible roots: ±1, ±2, ±4, ±1/3, ±2/3, ±4/3
3(1)3 − (1)2 − 12(1) + 4
=
−6.
3(−1) − (−1) − 12(−1) + 4
=
12.
3(2)3 − (2)2 − 12(2) + 4
=
0.
3
x−2
2
3x2
3x3
3x3
+5x
−2
−x2 −12x +4
−6x2
5x2 −12x
5x2 −10x
−2x +4
−2x +4
0
3x2 + 5x − 2
=
(3x − 1)(x + 2).
3x − x2 − 12x + 4
=
(3x − 1)(x + 2)(x − 2).
3
x =
1/3, −2, 2.
Exam 2: More Functions
(d) (6)
√
4x + 1 −
4
√
3x − 2 = 1.
√
4x + 1 −
√
√
3x − 2
=
1.
4x + 1
=
1+
4x + 1
=
4x + 1
=
x+2
=
2
=
3x − 2.
√
(1 + 3x − 2)2 .
√
√
1 + 2 x − 1 + ( 3x − 2)2 .
√
1 + 2 3x − 2 + 3x − 2.
√
2 3x − 2.
√
(2 3x − 2)2 .
x + 4x + 4
=
4(3x − 2).
2
√
( 4x + 1)2
(x + 2)
2
p
=
x − 8x + 12
=
0.
(x − 6)(x − 2)
=
0.
x
=
2, 6.
p
3(2) − 2
√
√
9− 4
=
4(2) + 1 −
=
3−2
p
=
p
4(6) + 1 − 3(6) − 2 =
√
√
25 − 16 =
5−4
Both solutions work.
√
=
1.
1.
Exam 2: More Functions
5
3. A Few More
(a) (3)
Modify
the following to remove the negative between the radicals.
√
√
x−1− x
x+1
√
√
x−1− x
=
x+1
√
√
√
√
x−1− x
x−1+ x
=
·√
√
x+1
x−1+ x
(x − 1) − (x)
√
=
√
(x + 1)( x − 1 + x)
−1
√
√ .
(x + 1)( x − 1 + x)
(b) (4) Find the (non-vertical) asymptote for
2x3 +3x2 −14x+8
x2 +3x−5
x2 + 3x − 5
2x
−3
2x3 +3x2
2x3 +6x2
−3x2
−3x2
−14x
+8
−10x
−4x
+8
−9x +15
5x
−7
The asymptote is y = 2x − 3.
x
y
−4 −3
− VA
−3.5 −2 −1 0
+
0 − 0
1 2 3
4
+ 0 − VA
The other asymptote is y = x + 1.
Figure 1: Rational Function
5
+