MATH 1040
Test 1
Fall 2014
Ch. 1, App B-D, QP 1-25
Version B
Student’s Printed Name: _______________________
CUID:___________________
Instructor: ______________________
Section # :_________
You are not permitted to use a calculator on any part of this test. You are not allowed to use any
textbook, notes, cell phone, laptop, PDA, or any technology on any part of this test. All devices
must be turned off while you are in the testing room.
During this test, any communication with any person (other than the instructor or his designated
proctor) in any form, including written, signed, verbal, or digital, is understood to be a violation
of academic integrity.
No part of this test may be removed from the testing room.
Read each question very carefully. In order to receive full credit, you must:
1.
Show legible and logical (relevant) justification which supports your final answer.
2.
Use complete and correct mathematical notation.
3.
Include proper units, if necessary.
4.
Give exact numerical values whenever possible.
You have 90 minutes to complete the entire test.
On my honor, I have neither given nor received inappropriate or unauthorized information
at any time before or during this test.
Student’s Signature: ________________________________________________
Do not write below this line.
Problem
Possible
Points
1
2
3
4
Points
Earned
Problem
Possible
Points
5
4
5
6
11
12
13
14
6
7
6
4
5
6
7
8
5
5
5
4
15
16
17
18
3
6
6
5
9
6
19
6
10
6
Test Total
100
Points
Earned
Page 1 of 8
MATH 1040
Test 1
Version B
Fall 2014
Ch. 1, App B-D, QP 1-25
Read each question carefully. In order to receive full credit you must show legible and
logical (relevant) justification which supports your final answer. Give answers as exact
answers. You are NOT permitted to use a calculator on any portion of this test.
1. (5 pts.) Find an equation of the line that passes through (−2, 3) and is perpendicular to
x − 2y −1= 0.
x − 2y −1= 0
so equation of tangent through (3,−2) is
−2 y = −x + 1
y − 3 = −2(x − −2)
1
1
x−
2
2
1
mgiven =
2
m⊥ = −2
y=
y − 3 = −2(x + 2)
2. Consider the graph of f (x) below.
a. (1 pt.) Describe in words how to obtain
f (x) from y = e x .
left π , up 2
b. (3 pts.) Write an equation for f (x) .
y = e x+π + 2
3. (5 pts.) Solve the following:
e2 x − 8e x + 7 = 0
(e
x
)(
)
− 7 ex − 1 = 0
ex = 7
ex = 1
ln e x = ln7
x = ln7
ln e x = ln1
x=0
Page 2 of 8
MATH 1040
Test 1
Fall 2014
Ch. 1, App B-D, QP 1-25
Version B
()
4. Let f (x) = sin x + 2 and g(x) = x −
a. (2 pts.) Find h(x) = f (g(x)) .
π
.
2
⎛
π⎞
h(x) = sin ⎜ x − ⎟ + 2
2⎠
⎝
()
b. (4 pts.) Graph h x on ⎡⎣ −2π ,2π ⎤⎦ .
5. (5 pts.) Prove the identity sec y − cos y = tan y sin y .
1
1
cos 2 y 1− cos 2 y sin 2 y
sec y − cos y =
− cos y =
−
=
=
cos y
cos y cos y
cos y
cos y
=
sin y
⋅sin y = tan y sin y
cos y
6. (5 pts.) Solve for x .
log10 x + log10 ( x − 3) = 1
log10 x ( x − 3) = 1
101 = x ( x − 3)
0 = x 2 − 3x − 10
0 = ( x − 5) ( x + 2 )
x = 5, x = −2
not in domain
Page 3 of 8
MATH 1040
Test 1
Fall 2014
Ch. 1, App B-D, QP 1-25
Version B
7. (5 pts.) Find the domain of the function f (x) =
State your answer as an interval.
3
−x
16 − x 2
. Show work, not just an answer.
16 − x 2 in denominator must have 16 − x 2 = 0
16 > x 2
4> x
so domain ( −4,4 )
8. a. (3 pts.) Place the following quadratic equation into the form y = a(x − h)2 + k .
f (x) =
1 2
x + 2x − 1
2
(
)
1 2
x + 4x + ___ − 1− ____
2
1
= x 2 + 4x + 2 − 1− 2
2
2
1
= ( x + 2) − 3
2
f ( x) =
(
)
b. (1 pt.) State the vertex.
( −2,−3)
9. (6 pts.) Simplify the expression
(x 2 − 2)(x + 3)
( 2 − x)
x≠ 2.
(x 2 − 2)(x + 3)
( 2 − x)
⋅
2+x
2+x
=
=
(x 2 − 2)(x + 3)
2− x
2
(
2+x
(
2+x
(x 2 − 2)(x + 3)
(
− x2 − 2
= −(x + 3)
(
2+x
)
by rationalizing the denominator. Assume
)
)
)
Page 4 of 8
MATH 1040
Test 1
Version B
Fall 2014
Ch. 1, App B-D, QP 1-25
⎧−x 2 + 4 x < 0
10. Let f (x) = ⎨
.
⎩x − 3 x ≥ 0
( )
a. (1 pt.) Evaluate f −4 = − ( −4 ) + 4 = −16 + 4 = −12
2
b. (1 pt.) Evaluate f (0) = 0 − 3 = −3
c. (4 pts.) Graph f (x) . Be sure to use the proper x-intercepts.
11. Let f (x) = 2x 2 + 1 . Answer the following questions about the function f (x) .
()
a. (1 pt.) What is the range of f x ? (Put your answer in interval notation).
⎡⎣1,∞ )
b. (5 pts.) Evaluate and simplify
(
f (x + h) − f (x)
. Assume ℎ ≠ 0. Show step by step work.
h
)
2
f (x + h) − f (x) 2 ( x + h ) + 1− 2x + 1
=
h
h
2
2 x + 2xh + h2 + 1− 2x 2 − 1
=
h
2
2x + 4xh + 2h2 + 1− 2x 2 − 1
=
h
2
4xh + 2h
=
= 4x + 2h h ≠ 0
h
2
(
)
Page 5 of 8
MATH 1040
Test 1
Fall 2014
Version B
Ch. 1, App B-D, QP 1-25
(x − 3)(3x 2 + 10x − 8)
12. (7 pts.) State the domain of the rational function h(x) =
. Also find all
4x 2 + 4x − 48
the real root(s) and the y-intercept.
( x − 3)(3x − 2)( x + 4) = ( x − 3)(3x − 2)( x + 4) = 3x − 2
4
4 ( x − 3) ( x + 4 )
4 x 2 + x − 12
(
)
domain: den ≠ 0 means x ≠ 3, x ≠ −4
roots: num = 0, den ≠ 0 means 3x − 2 = 0
intercept: f ( 0 ) =
x=
( −3)( −2)( 4) = − 2 = − 1
( 4)( 4)( −3) 4 2
2
3
Domain
use interval notation
Roots
list as x-value(s)
y-intercept
write as a point
( −∞,−4) ∪ ( −4,3) ∪ (3,∞ )
2
3
⎛
1⎞
⎜⎝ 0,− 2 ⎟⎠
x=
13. (6 pts.) Solve the following equation on ⎡⎣0,2π ⎤⎦ :
2 cos x + sin 2 x = 0
2cos x − sin 2x = 0
2cos x − 2sin x cos x = 0
2cos x(1+ sin x) = 0
2cos x = 0
cos x = 0
π 3π
x= ,
2 2
1+ sin x = 0
sin x = −1
3π
x=
2
14. (4 pts.) Give the equation of a circle that has center (2,4) and passes through the point (-1,3).
( −3) + ( −1)
so equation of circle is ( x − 2 ) + ( y − 4 )
d=
( −1− 2) + (3− 4)
2
2
2
=
2
2
= 9 + 1 = 10
2
= 10
Page 6 of 8
MATH 1040
Test 1
Fall 2014
Ch. 1, App B-D, QP 1-25
Version B
15. (3 pts.) A certain polynomial function f has degree two, roots x = 1 and x = −5 , and a
leading coefficient of 3. Express f in the form f (x) = ax 2 + bx + c .
f ( x ) = 3( x − 1) ( x + 5)
(
)
= 3 x 2 + 4x − 5
= 3x 2 + 12x − 15
16. (6 pts.) Suppose tan x = −
4
3π
and
< x < 2π . Find cos 2x .
3
2
the hypotenuse of 5 can be calculated with the Pythagorean Thm
in the fourth quadrant, tangent and sine are negative
cosine positive
4
sin x = −
5
3
cos x =
5
cos 2x = cos 2 x − sin 2 x
2
2
⎛ 3⎞ ⎛ 4 ⎞
9 16
7
= ⎜ ⎟ −⎜− ⎟ =
−
=−
25 25
25
⎝ 5⎠ ⎝ 5 ⎠
17. (6 pts.) Completely factor the polynomial p(x) = x 3 − 6x 2 + 5x + 12 by using the fact that
exactly one of x = 0 , x = 1 , and x = 4 is a root of the polynomial. Show all of your work in a
clear, logical manner.
p(0) = 12 ≠ 0
x 2 − 2x − 3
p(1) = 1− 6 + 5 + 12 = 12 ≠ 0
x − 4 x 3 − 6x 2 + 5x + 12
p(4) = 64 − 96 + 20 + 12 = 0
x 3 − 4x 2
so 4 is the root
− 2x 2 + 5x
− 2x 2 + 8x
− 3x + 12
−3x + 12
0
(
)(
)(
)
Final Factored Form: x − 4 x − 3 x + 1
Page 7 of 8
MATH 1040
Test 1
Version B
Fall 2014
Ch. 1, App B-D, QP 1-25
18. (5 pts.) Find the x-intercept(s) of y = 4x 2 + 7x − 1 . State your answer(s) as point(s).
does not factor, so use quadratic formula
x=
−7 ± 49 − 4 ( 4 ) ( −1)
2(4)
=
−7 ± 49 + 16 −7 ± 65
=
8
8
⎛ −7 + 65 ⎞
⎛ −7 − 65 ⎞
so x-intercepts as points are ⎜
,0⎟ and ⎜
,0⎟
8
8
⎝
⎠
⎝
⎠
19. (6 pts.) Evaluate the following. Simplify completely.
⎛ 3π ⎞
2
2
2+2
sin ⎜ ⎟ − cos (57π )
− −1
+1
2 2+2
⎝ 4⎠
2
2 = 2+2⋅ 4 =
= 2
=
=
2
3
11
2
11
11
⎛ π⎞
2 ⎛ 11π ⎞
⎛
⎞
3
2+
sec ⎜ − ⎟ + cos ⎜
⎟
2
+
4
4
⎝ 3⎠
⎝ 6 ⎠
⎜ 2 ⎟
⎝ ⎠
(
⎛ 3π ⎞
2
sin ⎜ ⎟ =
2
⎝ 4⎠
)
second quad, sine positive
cos (57π ) = cos (π ) = −1
⎛ π⎞
sec ⎜ − ⎟ =
⎝ 3⎠
1
1
= =2
⎛ π⎞ 1
cos ⎜ − ⎟
⎝ 3⎠ 2
⎛ 11π ⎞
3
cos ⎜
=
⎟
2
⎝ 6 ⎠
fourth quad, cosine positive
Page 8 of 8