MATH 1040
Test 1
Fall 2014
Ch. 1, App B-D, QP 1-25
Version A
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 11
MATH 1040
Test 1
Version A
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 (3, −2) and is perpendicular to
x − 3y + 1 = 0 .
Work on Problem
Points Awarded
x − 3y + 1 = 0
−3y = −x − 1
1
1
y= x+
3
3
1
mgiven =
3
m⊥ = −3
so equation of tangent through (3,−2) is
y − −2 = −3(x − 3)
y + 2 = −3(x − 3)
2. Consider the graph of f (x) below.
Identify slope of
1
given line
Identify slope of
1.5
perpendicular line
Perpendicular slope
0.5
properly used
Point into pt/slope
2
form
Notes:
-0.5 for any sign error
-1 go to slope/intercept incorrectly
-1 missing parentheses
-1 adding a constant after correct
point/slope
-2 no variables in final answer
a. (1 pt.) Describe in words how to obtain
f (x) from y = e x .
left π , up 2
right or wrong
b. (3 pts.) Write an equation for f (x) .
y = e x+π + 2
transformed e^x equation 0.5
left shift 1.5
vertical shift 1
notes:
-3 if e^x never used
-1 for including a reflection
-0.5 notation (in particular not naming the
function)
Page 2 of 11
MATH 1040
Test 1
Fall 2014
Ch. 1, App B-D, QP 1-25
Version A
3. (5 pts.) Solve the following:
Work on Problem
e2 x − 6e x + 5 = 0
( e − 5)( e − 1) = 0
x
ex = 5
x
ex = 1
ln e x = ln5 ln e x = ln1
x = ln5
x=0
()
4. Let f (x) = cos x + 2 and g(x) = x −
a. (2 pts.) Find h(x) = f (g(x)) .
Points Awarded
Factor
Each factor equal to 0
Correct solution (1 pt each)
Notes:
-0.5 did not carry the equals sign
-0.5 incorrect notation
-1 per incorrect number in factoring
-0.5 per incorrect sign in factoring
-5 quadratic formula with no substitution
-0.5 ln1 not simplified
-1 x=e^x substitution was used
-0.5 a substitution was used but never defined
2.5
0.5
2
π
.
2
Work on Problem
⎛
π⎞
h(x) = cos ⎜ x − ⎟ + 2
2⎠
⎝
Points
Awarded
Equation
2
Amplitude on graph
1
Period on graph
0.5
Phase Shift (direction 1)
1.5
Vertical Shift
1
Notes:
-1 cos(A-B)=cosA – cosB
-0.5 notation (not naming the
function)
-0.5 at least one period was shown but
not graphed on desired interval
-0.5 to -1.5 incorrect placement of
points, general shape
()
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
Work on Problem
Points
Awarded
In terms of cosine
Common denominator
Identity
Regroup
Final result
Notes:
-0.5 to -1 incorrect notation/bad form
-0.5 to -4 incorrect proof techniques
1
1
1
1
1
Page 3 of 11
MATH 1040
Test 1
Fall 2014
Ch. 1, App B-D, QP 1-25
Version A
6. (5 pts.) Solve for x .
log10 x + log10 ( x + 3) = 1
log10 x ( x + 3) = 1
Points
Awarded
Work on Problem
101 = x ( x + 3)
Combine logs
1
Convert to exponential
0.5
Factor = 0
2
Solutions
1
Eliminate solution not in
0.5
domain
Notes:
-0.5 for any sign error
-0.5 missing parentheses
-0.5 excessive equals
-5 combining logs with sum instead of a
product (makes rest trivial)
0 = x 2 + 3x − 10
0 = ( x + 5) ( x − 2 )
x = −5 , x = 2
not in domain
7. (5 pts.) Find the domain of the function f (x) =
State your answer as an interval.
25 − x in denominator must have 25 − x > 0
2
2
25 > x
5> x
so domain ( −5,5)
2
3
−x
25 − x
2
. Show work, not just an answer.
Work on Problem
Points
Awarded
Inside square root positive
Exclude 0
Solve
Proper domain
Notes:
-1 for [-5,5]
-0.5 missing absolute value notation
-1 sqrt(25-x^2)=25-x
1
1
1
2
8. a. (3 pts.) Place the following quadratic equation into the form y = a(x − h)2 + k .
f (x) =
1 2
x + x −1
2
(
)
1 2
x + 2x + ___ − 1− ____
2
1
1
= x 2 + 2x + 1 − 1−
2
2
2
1
3
= ( x + 1) −
2
2
f ( x) =
(
)
Work on Problem
Points Awarded
Factor a out of first two terms
1
Add and subtract proper constant
1
Answer in proper form
1
Notes:
-3 a not factored out but tried to complete the square
anyway
b. (1 pt.) State the vertex.
⎛
3⎞
⎜⎝ −1,− 2 ⎟⎠
right or wrong based on part a
Page 4 of 11
MATH 1040
Test 1
Fall 2014
Ch. 1, App B-D, QP 1-25
Version A
(x − 3)(x + 2)
2
9. (6 pts.) Simplify the expression
( 3 − x)
x≠ 3.
(x − 3)(x + 2)
2
( 3 − x)
⋅
3+ x
3+ x
(x − 3)(x + 2)
2
=
=
3− x 2
(
(x 2 − 3)(x + 2)
(
3+ x
(
)
by rationalizing the denominator. Assume
)
3+ x
)
− x −3
= −(x + 2)
(
2
3+ x
Work on Problem
Points Awarded
Multiply by 1 with conjugate
Simplify denominator FOIL
Factor negative (can be implied)
Cancel
Solution (award only if correct)
Notes:
1
2
1.5
0.5
1
-0.5 lack of equals (forgive one)
-0.5 incorrect notation
-0.5 lack of ( ) to show multiplication of binomial
-0.5 to -1 lack of proper “work shown”
-1 each BAD algebra instance
-0.5 wrong sign on conjugate (plus errors that followed)
-6 if the conjugate was never used
)
⎧−x 2 + 1 x < 0
10. Let f (x) = ⎨
.
x≥0
⎩x − 2
( )
a. (1 pt.) Evaluate f −4 = − ( −4 ) + 1 = −16 + 1 = −15
2
b. (1 pt.) Evaluate f (0) = 0 − 2 = −2
right or wrong,
except -0.5 for small arithmetic error
right or wrong
c. (4 pts.) Graph f (x) . Be sure to use the proper x-intercepts.
Work on Problem
Parabola
Open circle at 0 on parabola
line
Closed circle at 0 on line
Notes:
-1 x-intercept wrong (each)
-2 reversed domains
Page 5 of 11
Points
Awarded
1.5
0.5
1.5
0.5
MATH 1040
Test 1
Fall 2014
Version A
Ch. 1, App B-D, QP 1-25
2
11. Let f (x) = 2x + 3 . 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).
⎡⎣3,∞ )
right or wrong, except -0.5 for not including endpoint
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 ) + 3− 2x + 3
=
h
h
2
2 x + 2xh + h2 + 3− 2x 2 − 3
=
h
2
2x + 4xh + 2h2 + 3− 2x 2 − 3
=
h
2
4xh + 2h
=
= 4x + 2h h ≠ 0
h
2
(
)
Work on Problem
Points Awarded
Substitution
0.5
FOIL
1
Distribute
1
Simplify
1
Factor, cancel
1
Solution (award only if correct)
0.5
Notes:
-1 did not carry the equals sign
-0.5 or -1 incorrect notation or missing step(s)
-5 substitution incorrect makes the problem too simple
-1 h not carried through problem
(x − 2)(2x 2 + 5x − 12)
12. (7 pts.) State the domain of the rational function h(x) =
. Also find all
4x 2 + 8x − 32
the real root(s) and the y-intercept.
( x − 2)( 2x − 3)( x + 4) = ( x − 2)( 2x − 3)( x + 4) = 2x − 3
4
4 ( x − 2)( x + 4)
4 ( x + 2x − 8)
2
domain: den ≠ 0 means x ≠ 2, x ≠ −4
roots: num = 0, den ≠ 0 means 2x − 3 = 0
( −2)( −3)( 4) = − 3
intercept: f ( 0 ) =
( 4)( −2)( 4) 4
Domain
use interval
notation
Roots
list as xvalue(s)
y-intercept
write as a
point
( −∞,−4) ∪ ( −4,2) ∪ ( 2,∞ )
3
2
⎛
3⎞
⎜⎝ 0,− 4 ⎟⎠
x=
x=
3
2
Work on Problem
Points Awarded
Work for domain
0.5 for den=0, 0.5 for solving
for x
Correct domain
Each interval worth 0.5
Work for root
0.5 for den=0, 0.5 for solving
for x
Correct root
Work for intercept
Correct intercept
Notes:
-1 each extra root
1
1.5
1
1.5
1
1
-0.5 for brackets [] in the domain (one time deduction)
-0.5 if root was not simplified
-0.5 if incorrect sign in intercept work and/or minor error
(copy error, etc.)
-1 if numerator and/or denominator was disregarded
completely in intercept work and/or there is a missing
factor in either
-0.5 if y-intercept was stated as (y,0)
-0.5 if y-intercept was not stated as a point
-0.5 for several missing equal signs
-0.5 if x is solved for without any "=0"
-1 each extra root for a maximum of 1.5
**It is possible to lose points for incorrect work but to get
full credit for answers (if answers are consistent)
Page 6 of 11
MATH 1040
Test 1
Fall 2014
Ch. 1, App B-D, QP 1-25
Version A
13. (6 pts.) Solve the following equation on ⎡⎣0,2π ⎤⎦ :
2cos x − sin 2x = 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
x=
1− sin x = 0
sin x = 1
π 3π
,
2 2
x=
π
2
Work on Problem
Points Awarded
Identity
Factor
Set each factor equal to zero
0.5 for each factor
3 solutions (do not have to acknowledge on is a
duplicate)
1.5
1.5
1
2
0.5 points for each solution to
cos(x)=0 and 1 point for
solution to sin(x)=1
Notes:
-0.5 incorrect notation
-0.5 extra angles or angles outside of [0,2pi] or extra solutions
-3 for dividing all terms by cosx and losing cosx = 0
-4 never factored
-5 guess and check
-0.5 for missing 2 in identity
-0.5 or -1 for incorrect factoring (-0.5 for mistake on constant/sign mistake, -1 for
missing an entire term, i.e., 2cos(x)[-sin(x)]) Work was sometimes followed through,
depending on severity of mistake
-0.5 incorrect notation (-0.5 for missing "x=solution", -0.5 for missing "=0")
-0.5 even if all solutions are written down, but only one solution is boxed, or circled as
the answer
-0.5 extra solutions or angles outside of [0,2pi]
-3 for dividing all terms by cos(x) and losing the factor cos(x)=0
-4.5 never factored
-6 guess and check (no supporting work)
**Sometimes work was followed through, depending on how the problem was started
14. (4 pts.) Give the equation of a circle that has center (2,3) and passes through the point (-1,2).
( −3) + ( −1) = 9 + 1 =
so equation of circle is ( x − 2 ) + ( y − 3) = 10
d=
( −1− 2) + ( 2 − 3)
2
2
2
=
2
2
10
2
Work on Problem
Distance formula for radius
Equation of circle
Notes:
Points
Awarded
2
2
-0.5 for several missing equal signs
-0.5 or -1 minor calculation error, dependent on
severity
-1 incorrect point as center
-2 points for equations other than circles (lines,
parabolas, etc.)
**Work was followed through with incorrect rvalue
**It was not necessary to use the distance formula,
and there was no penalty if r was found by
substituting (-1,2) into the circle equation
Page 7 of 11
MATH 1040
Test 1
Fall 2014
Version A
Ch. 1, App B-D, QP 1-25
15. (3 pts.) A certain polynomial function f has degree two, roots x = −1 and x = 7 , and a
leading coefficient of 2. Express f in the form f (x) = ax 2 + bx + c .
f ( x ) = 2 ( x + 1) ( x − 7 )
(
= 2 x 2 − 6x − 7
)
= 2x 2 − 12x − 14
Work on Problem
Proper root
Proper second root
Coefficient of 2
Notes:
Points
Awarded
1
1
1
-0.5 for several missing equal signs
-0.5 for each type of notation error
-0.5 for simplification error
-0.5 for no "f(x)=" on final answer
-0.5 for not writing in ax^2+bx+c form
-1 for incorrect distributing on coefficient (i.e.,
(2x-7) instead of (2x-14))
Page 8 of 11
MATH 1040
Test 1
Fall 2014
Ch. 1, App B-D, QP 1-25
Version A
16. (6 pts.) Suppose tan x = −
3π
3
and
< x < 2π . Find cos 2x .
2
4
Work on Problem
Triangle third side
Sine and cosine
Double angle formula
Correct answer
Notes:
Points Awarded
1
2
2
1
the hypotenuse of 5 can be calculated with the Pythagorean Thm
in the fourth quadrant, tangent and sine are negative
cosine positive
3
sin x = −
5
4
cos x =
5
cos 2x = cos 2 x − sin 2 x
2
2
⎛ 4 ⎞ ⎛ 3⎞
16 9
7
= ⎜ ⎟ −⎜− ⎟ =
−
=
25 25 25
⎝ 5 ⎠ ⎝ 5⎠
Page 9 of 11
MATH 1040
Test 1
Fall 2014
Version A
Ch. 1, App B-D, QP 1-25
3
2
17. (6 pts.) Completely factor the polynomial p(x) = x − 7x + 7x + 15 by using the fact that
exactly one of x = 0 , x = 1 , and x = 3 is a root of the polynomial. Show all of your work in a
clear, logical manner.
p(0) = 15 ≠ 0
x 2 − 4x − 5
p(1) = 1− 7 + 7 + 15 = 16 ≠ 0
x − 3 x 3 − 7x 2 + 7x + 15
p(3) = 27 − 63+ 21+ 15 = 0
x 3 − 3x 2
so 3 is the root
− 4x 2 + 7x
− 4x 2 + 12x
− 5x + 15
−5x + 15
0
(
)(
)(
)
Final Factored Form: x − 3 x − 5 x + 1
Work on Problem
Points Awarded
Use proper root
1
Correctly perform long division of polynomial
3
Correctly factor remaining quadratic
1.5
Write completely factored form
0.5
Notes:
Full credit for correct synthetic division
-0.5 to – 1.5 for each error in polynomial long division depending on the
severity and cascading effect
-0.5 to -2 for incomplete work in polynomial long division
award 0.5 points of the 1.5 for factoring the quadratic with one sign error
-1.5 incorrect division results in irreducible quadratic
-0.5 =0 in final answer
-1 lack of work
-4 no work at all
Page 10 of 11
MATH 1040
Fall 2014
Version A
Ch. 1, App B-D, QP 1-25
2
18. (5 pts.) Find the x-intercept(s) of y = 5x + 3x − 1 . State your answer(s) as point(s).
does not factor, so use quadratic formula
x=
Test 1
−3 ± 9 − 4 (5) ( −1)
2(5)
=
−3 ± 9 + 20 −3 ± 29
=
10
10
Work on Problem
⎛ −3+ 29 ⎞
⎛ −3− 29 ⎞
so x-intercepts as points are ⎜
,0⎟ and ⎜
,0⎟
⎝ 10
⎠
⎝ 10
⎠
Constants into quadratic
formula properly
Simplified
Answers presented as points
Notes:
19. (6 pts.) Evaluate the following. Simplify completely.
⎛ 11π ⎞
3
3
3+2
cos ⎜
− cos (53π )
− −1
+1
⎝ 6 ⎟⎠
3+2 2
3+2
= 2
= 2
= 2 =
⋅ =
2
1
5
2
5
5
⎛ π⎞
⎛ 3π ⎞
2+
sec ⎜ − ⎟ + sin 2 ⎜ ⎟ 2 + ⎛ 2 ⎞
2
2
⎝ 3⎠
⎝ 4⎠
⎜ 2 ⎟
⎝
⎠
⎛ 11π ⎞
3
cos ⎜
=
⎟
2
⎝ 6 ⎠
Work on Problem
fourth quad, cosine positive
cos (53π ) = cos (π ) = −1
⎛ π⎞
sec ⎜ − ⎟ =
⎝ 3⎠
Each trig evaluation
simplification
Notes:
Points
Awarded
1
2
1
1
= =2
⎛ π⎞ 1
cos ⎜ − ⎟
⎝ 3⎠ 2
⎛ 3π ⎞
2
sin ⎜ ⎟ =
2
⎝ 4⎠
second quad, sine positive
Page 11 of 11
Points
Awarded
2.5
1.5
1