Advanced
Algebra and
Functions
Sample Questions
ACCUPLACER Advanced Algebra and Functions © 2020 College Board.
1
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ACCUPLACER Advanced Algebra
and Functions Sample Questions
The Advanced Algebra and Functions placement test is a computer adaptive assessment
of test takers’ ability for selected mathematics content. Questions will focus on a range
of topics, including a variety of equations and functions, including linear, quadratic,
rational, radical, polynomial, and exponential. Questions will also delve into some
geometry and trigonometry concepts. In addition, questions may assess a student’s
math ability via computational or uency skills, conceptual understanding, or the
capacity to apply mathematics presented in a context. All questions are multiple choice
in format and appear discretely (stand alone) across the assessment. The following
knowledge and skill categories are assessed:
Linear equations
Linear applications
Factoring
Quadratics
Functions
Radical and rational equations
Polynomial equations
Exponential and logarithmic equations
Geometry concepts
Trigonometry
ACCUPLACER Advanced Algebra and Functions © 2020 College Board.
2
Sample Questions
Choose the best answer. If necessary, use the paper you
were given.
1. Function g is dened by g(x) = 3(x + 8). What is the
value of g(12)?
A. –4
B. 20
C. 44
D. 60
2.
x
y
O
6–6
–6
6
Which of the following is an equation of the line that
passes through the point (0, 0) and is perpendicular to
the line shown above?
A. y =
5
4
x
B. y =
5
4
x + 3
C. y = −
4
5
x
D. y = −
4
5
x + 3
3.
9 cm
3 cm
4 cm
e surface area of a right rectangular prism can be
found by nding the sum of the area of each of the
faces of the prism. What is the surface area of a right
rectangular prism with length 4 centimeters (cm), width
9 cm, and height 3 cm? (Area of a rectangle is equal to
length times width.)
A. 75 cm
2
B. 108 cm
2
C. 120 cm
2
D. 150 cm
2
4. Which of the following expressions is equivalent to
(x + 7)(x
2
– 3x + 2)?
A. x
3
– 3x
2
+ 2x + 14
B. x
3
+ 4x
2
– 19x + 14
C. x
3
– 3x + 14
D. x
2
– 2x + 9
5.
Number of pounds
Cost (dollars)
Cost of Pears: C = p
7
5
Cost of Apples
0
7 8 9654321
7
6
5
4
3
2
1
8
e graph above shows the cost, in dollars, of apples as
a function of the number of pounds of apples purchased
at a particular grocery store. e equation above denes
the cost C, in dollars, for p pounds of pears at the same
store. Which of the following statements accurately
compares the cost per pound of apples and the cost per
pound of pears at this store?
A. Apples cost approximately $0.07 less per pound
than pears do.
B. Apples cost approximately $0.04 less per pound
than pears do.
C. Apples cost approximately $0.73 less per pound
than pears do.
D. Apples cost approximately $0.62 more per pound
than pears do.
3ACCUPLACER Advanced Algebra and Functions © 2020 College Board.
6. Which of the following is the graph of a function where
y = f(x)?
A.
x
y
O
B.
x
y
O
C.
x
y
O
D.
x
y
O
7. Which of the following expressions is equivalent to
3x
2
+ 6x – 24?
A. 3(x + 2)(x – 4)
B. 3(x – 2)(x + 4)
C. (x + 6)(x – 12)
D. (x – 6)(x + 12)
8. A biologist puts an initial population of 500 bacteria
into a growth plate. e population is expected to
double every 4 hours. Which of the following equations
gives the expected number of bacteria, n, aer x days?
(24 hours = 1 day)
A. n = 500(2)
x
B. n = 500(2)
6x
C. n = 500(6)
x
D. n = 500(6)
2x
9. x
2
+ 5x – 9 = 5
Which of the following values of x satises the equation
above?
A. 7
B. 3
C. –2
D. –7
10. e graph of y = f(x) is shown in the xy-plane below.
x
y
O
–4 –2 2 4
2
–2
–4
–6
–8
Which of the following equations could dene f(x)?
A. f(x) = x
2
– 2x – 8
B. f(x) = –x
2
+ 2x – 8
C. f(x) = (x – 2)(x + 4)
D. f(x) = –(x – 1)
2
– 9
11. Which of the following best describes the range of
y = –2x
4
+ 7?
A. y ≤ –2
B. y ≥ 7
C. y ≤ 7
D. All real numbers
ACCUPLACER Advanced Algebra and Functions © 2020 College Board. 4
12. For which of the following equations is x = 6 the
only solution?
A. (6x)
2
= 0
B. (x – 6)
2
= 0
C. (x + 6)
2
= 0
D. (x – 6)(x + 6) = 0
13. If f(x) = x
2
+ 3x + 1, what is f(x + 2)?
A. x
2
+ 3x + 3
B. (x + 2)
2
+ 3(x + 2) + 1
C. (x + 2)(x
2
+ 3x + 1)
D. x
2
+ 3x + 9
14. What, if any, is a real solution to 5x + 1 + 9 = 3?
A. −
1
5
B. 7
C.
143
5
D. ere is no real solution.
15. If x ≠ –2 and x ≠
3
2
, what is the solution to
5
x + 2
=
x
2x − 3
?
A. 3 and 5
B. 2 and −
3
2
C. –2 and
3
2
D. –3 and –5
16.
J
K
R
P
QL
Triangle JKL and triangle PQR are shown above. If ∠J
is congruent to ∠P, which of the following must be
true in order to prove that triangles JKL and PQR are
congruent?
A. ∠L ≅ ∠R and JL = PR
B. KL = QR and PR = JL
C. JK = PQ and KL = QR
D. ∠K ≅ ∠Q and ∠L ≅ ∠R
17. In the function f(x) = a(x + 2)(x – 3)
b
, a and b are both
integer constants and b is positive. If the end behavior
of the graph of y = f(x) is positive for both very large
negative values of x and very large positive values of x,
what is true about a and b?
A. a is negative, and b is even.
B. a is positive, and b is even.
C. a is negative, and b is odd.
D. a is positive, and b is odd.
18. Which of the following equations is equivalent to 2
5x
= 7?
A. x =
log
2
7
5
( )
B. x =
log
2
7
5
C. x =
log
7
2
5
D. x =
log
7
5
2
19. If x > 0 and y > 0, which of the following expressions is
equivalent to
x − y
x
− y
?
A.
x − y
x − y
B.
x − y
C.
x + y
D.
x x + y y
20. In triangle ABC, angle C is a right angle. If cos A =
5
8
,
what is the value of cos B?
A.
3
8
B.
5
8
C.
39
8
D.
89
8
5
Answer Key
1. D
2. A
3. D
4. B
5. A
6. C
7. B
8. B
9. D
10. A
11. C
12. B
13. B
14. D
15. A
16. A
17. D
18. B
19. C
20. C
ACCUPLACER Advanced Algebra and Functions
© 2020 College Board.
Rationales
1. Choice D is correct. The value of g(12) can be found by substituting 12 for x in the
equation for
g(x). This yields g(12) = 3(12 + 8) , which is equivalent to 3(20), or 60.
Choice A is incorrect. This answer represents the value of x in the equation
12 = 3(x + 8). Choice B is incorrect. This answer represents the value of the
expression in parentheses. Choice C is incorrect. This answer is a result of incorrectly
distributing the 3 through the expression in parentheses: g(12) = 3(12) + 8.
2. Choice A is correct. The slopes of perpendicular lines are negative reciprocals of
each other. The slope of the line in the graph is
.
4
−
5
The negative reciprocal of
−
5
4
is .
5
4
A line that passes through the point
(0, 0) has a y-intercept of 0. Therefore,
the equation
5 5
y
=
x
+ 0, or y =
4
x
4
, is correct. Choice B is incorrect because it is an
equation of a line that is perpendicular to the line shown, but it does not pass through
the origin. Choice C is incorrect because this equation is parallel to the line shown,
not perpendicular. Choice D is incorrect because it is the equation of the line shown in
the graph.
3. Choice D is correct. The surface area of the right rectangular prism is the sum of the
area of each of the faces of the prism and can be written as
2(length × width) +
2(height × width) + 2(length × height), which is 2(4 cm × 9 cm) + 2(3 cm × 9 cm) +
2(4 cm × 3 cm), or 150 cm
2
. Choice A is incorrect because it is half the surface area of
the prism. Choice B is incorrect because
108 is the volume of the prism in cm
3
. Choice
C is incorrect because it is
30 units less than the surface area of the prism described.
4. Choice B is correct. Using the distribution property, the given expression can be
rewritten as
x(x
2
) + x(−3x) + x(2) + 7(x
2
) + 7(−3x) + 7(2). Further simplifying results in
x
3
− 3x
2
+ 2x + 7x
2
− 21x + 14 . Finally, adding like terms yields x
3
+ 4x
2
− 19x + 14.
Choices A, C, and D are incorrect because they each result from errors made when
performing the necessary distribution and adding like terms.
5. Choice A is correct. The cost per pound of apples can be determined by the
slope of the graph as about
$1.33 per pound. The cost per pound of pears can be
determined by the slope of the line dened by the equation C =
7
5
p. The slope of the
line dened by C is
7
5
,, so the cost per pound of pears is $1.40. Therefore, apples cost
approximately $0.07 less per pound than pears do. Choices B, C, and D are incorrect
and may result from misreading the cost per pound of pears or apples, or both.
6. Choice C is correct. A function has one output for each input in its domain. Each
x-value on this graph corresponds to only one y-value. Choices A, B, and D are
incorrect because each has
x-values that correspond to more than one y-value.
7. Choice B is correct. The expression
3(x − 2)(x + 4) can be expanded by rst
multiplying (x − 2) by 3 to get (3x − 6) and then multiplying (3x − 6) by (x + 4) to
get
3x
2
+ 6x − 24. Choice A is incorrect because it is equivalent to 3x
2
− 6x − 24.
Choice C is incorrect because it is equivalent to x
2
− 6x − 72. Choice D is incorrect
because it is equivalent to
x
2
+ 6x − 72.
ACCUPLACER Advanced Algebra and Functions © 2020 College Board. 6
7
8. Choice B is correct. An exponential function can be written in the form y = ab
t
,
where a is the initial amount, b is the growth factor, and t is the time. In the scenario
described, the variable y can be substituted with n, the expected number of bacteria,
and the initial amount is given as 500, which yields n = 500b
t
. The growth factor is
2 because the population is described as being expected to double, which gives the
equation n = 500(2)
t
. The population is expected to double every 4 hours, so for the
time to be x days, x must be multiplied by 6 (the number of 4-hour periods in 1 day).
This gives the nal equation n = 500(2)
6x
. Choices A, C, and D are incorrect. Choice A
does not account for the six 4-hour periods per day, choice C uses the number of
time periods per day as the growth factor, and choice D uses the number of time
periods per day as the growth factor and multiplies the exponent by the actual
growth factor.
9. Choice D is correct. Subtracting
5 from both sides of the equation gives
x
2
+ 5x − 14 = 0. The left-hand side of the equation can be factored, giving
(x + 7)(x − 2) = 0. Therefore, the solutions to the quadratic equation are x = −7 and
x = 2. Choice A is incorrect because 7
2
+ 5(7) − 9 is not equal to 5. Choice B is
incorrect because
3
2
+ 5(3) − 9 is not equal to 5. Choice C is incorrect because
(−2)
2
+ 5(−2) − 9 is not equal to 5.
10. Choice A is correct. The graph of
y = f(x) crosses the x-axis at x = −2 and x = 4 ,
crosses the y-axis at y = −8, and has its vertex at the point (1 −9), . Therefore, the
ordered pairs (−2 0) (4 0) (0, −8), , , , , and (1, −9) must satisfy the equation for f(x).
Furthermore, because the graph opens upward, the equation defining f(x) must
have a positive leading coefficient. All of these conditions are met by the equation
f(x) = x
2
− 2x − 8. Choice B is incorrect. The points (−2, 0), (4, 0), (0, −8), and (1, −9),
which are easily identified on the graph of y = f(x), do not all satisfy the equation
f(x) = − x
2
+ 2x − 8; only (0, −8) does. Therefore, f(x) = −x
2
+ 2x − 8 cannot define the
function graphed. Furthermore, because the graph opens upward, the equation
defining y = f(x) must have a positive leading coefficient, which f(x) = −x
2
+ 2x − 8
does not. Choice C is incorrect. The points (−2 0) (4 0) (0 −8), , , , , , and (1, −9), which
are easily identified on the graph of y = f(x), do not all satisfy the equation
f(x) = (x − 2)(x + 4); only (0, −8) does. Therefore, f(x) = (x − 2)(x + 4) cannot define the
function graphed. Choice D is incorrect. Though the vertex (1, −9) does satisfy the
equation
2
f(x) = −(x − 1) − 9, the points (−2, 0), (4, 0), and (0, −8) do not. Therefore,
f(x) = −(x − 1)
2
− 9 cannot define the function graphed. Furthermore, because the
graph opens upward, the equation defining y = f(x) must have a positive leading
coefficient, which f(x) = −(x − 1)
2
− 9 does not.
11. Choice C is correct. The range of a function describes the set of all outputs,
y, that
satisfy the equation dening the function. In the xy-plane, the graph of y = −2x
4
+ 7 is
a U-shaped graph that opens downward with its vertex at (0, 7). Because the graph
opens downward, the vertex indicates that the maximum value of y is 7. Therefore, the
range of the function dened by y = −2x
4
+ 7 is the set of y-values less than or equal
to 7. Choices A, B, and D are incorrect in that choice A doesn’t cover the entire range,
while choices B and D include values that aren’t part of the range.
12. Choice B is correct. The only value of
x that satises the equation (x − 6)
2
= 0 is 6.
Choice A is incorrect because x = 0 is the only solution to the equation
2
(6x) = 0 .
Choice C is incorrect because x = −6 is the only solution to the equation (x + 6)
2
= 0.
Choice D is incorrect because although x = 6 is a solution to the equation
(x − 6)(x + 6) = 0, x = −6 is another solution to the equation.
13. Choice B is correct. Substituting
x + 2 for x in the original function gives
2
f(x + 2) = (x + 2) + 3(x + 2) + 1. Choice A is incorrect. This is f(x) + 2. Choice C is
incorrect. This is (x + 2)f(x). Choice D is incorrect. This is f(x) + 2
3
.
ACCUPLACER Advanced Algebra and Functions
© 2020 College Board.
14. Choice D is correct. Subtracting 9 from both sides of the equation yields
5x + 1 = − 6, and there are no real values of x that result in the square root of a
number being negative, so the equation has no real solution. Choices A and C are
incorrect due to computational errors in solving for
x and not checking the solution
in the original equation. Choice B is incorrect because it is the extraneous solution to
the equation.
15. Choice A is correct. To solve the equation for
x, cross multiply to yield x(x + 2) = 5(2x − 3).
Simplifying both sides of the new equation results in x
2
+ 2x = 10x − 15. Next, subtract 10x
from both sides of the equation and add 15 to both sides of the equation to yield
x
2
− 8x + 15 = 0. By factoring the left-hand side, the equation can be rewritten in the form
(x − 3)(x − 5) = 0. It follows, therefore, that x = 3 and x = 5. Choices B, C, and D are incorrect
and are possible results from mathematical errors when solving the equation for x.
16. Choice A is correct. If two angles and the included side of one triangle are
congruent to corresponding parts of another triangle, the triangles are congruent.
Since angles
J and L are congruent to angles P and R, respectively, and the side
lengths between each pair of angles, JL and PR, are also equal, then it can be proven
that triangles JKL and PQR are congruent. Choices B and C are incorrect because
only when two sides and the included angle of one triangle are congruent to
corresponding parts of another triangle can the triangles be proven to be congruent,
and angles J and P are not included within the corresponding pairs of sides given.
Further, side-side-angle congruence works only for right triangles, and it is not
given that triangles JKL and PQR are right triangles. Choice D is incorrect because
the triangles can only be proven to be similar (not congruent) if all three sets of
corresponding angles are congruent.
17. Choice D is correct. A polynomial function of even degree with a positive leading
coecient will have positive end behavior for both very large negative values of
x
and very large positive values of x. For a polynomial function in the form
f(x) = a(x + 2)(x − 3)
b
to be of even degree with a positive leading coecient,
a must be positive and b must be odd. Choice A is incorrect. If a is negative and
b is even, the polynomial function will be of odd degree, with a negative leading
coecient. This results in positive end behavior for very large negative values of x
and negative end behavior for very large positive values of x. ChoiceB is incorrect.
If a is positive and b is even, the polynomial function will be of odd degree with a
positive leading coecient. This results in negative end behavior for very large
negative values of x and positive end behavior for very large positive values of x.
Choice C is incorrect. If a is negative and b is odd, the polynomial function will be of
even degree with a negative leading coecient. This results in negative end behavior
on both sides of the function.
18. Choice B is correct. By denition, if
(b)
x
= y, where b > 0 and b ≠ 1, then x = log
b
y.
Therefore, the given equation
2
5x
= 7 can be rewritten in the form log
2
7 = 5x. Next,
solving for
x by dividing both sides of the equation by 5 yields
log
2
7
5
= x. Choices A,
C, and D are incorrect because they are the result of misapplying the identity, which
states that if
(b)
x
= y, where b > 0 and b ≠ 1, then x = log
b
y.
ACCUPLACER Advanced Algebra and Functions © 2020 College Board. 8
9
19. Choice C is correct. Since x > 0 and y > 0, x can be rewritten as x
( )
2
and y can be
rewritten as
y
( )
2
.
. It follows, then, that
x − y
x
− y
can be rewritten as
x
( )
2
− y
( )
2
x − y
.
.
Because the numerator is a dierence of two squares, it can be factored as
x + y
( )
x − y
( )
x − y
( )
. Finally, dividing the common factors of
x − y
( )
in the
numerator and denominator yields
x + y . Alternatively, if
x − y
x
− y
is multiplied by
x + y
x + y
,
, which is equal to 1, and therefore does not change the value of the
original expression, the result is
x − y
( )
x + y
( )
x − y
( )
x + y
( )
, which is equivalent to
x x + x y − y x − y y
x − xy + xy − y
. This can be rewritten as
(x − y) x + y
( )
x − y
( )
, which can be
simplied to
x + y . Choice A is incorrect and may be the result of incorrectly
combining
x − y . Choice B is incorrect because it is equivalent to
x − y
x − y
.
Choice D is incorrect and may be the result of misusing the conjugate strategy.
Instead of multiplying the numerator and denominator by the quantity
x + y
( )
,
they may have been multiplied by
x − y
( )
and then improperly distributed.
20. Choice C is correct. If triangle
ABC is dened as a right triangle, where angleC
is the right angle, then the cosine of angle
A (cos A) is dened as the ratio
the length of the side adjacent to angle A
the length of the hypotenuse
. Since this ratio is dened as
5
8
, one can take
the length of the side adjacent to angle
A to be 5 and the length of the hypotenuse to
be
8. Then the length of the side opposite angle A, which is also the side adjacent to
angle
B, can be derived from the Pythagorean theorem: a
2
+ 5
2
= 8
2
, where a represents
the length of the side opposite angle
A. Solving for a yields a
2
= 64 − 25 = 39,
so
a = 39. Then, to determine the cosine of angle B, use the same ratio in relation to
angle
B:
cosB =
the length of the side adjacent to angle B
the length of the hypotenuse
=
39
8
.
Choices A and D are
incorrect and likely result from an error in nding the length of side
CB. Choice B is
incorrect because it is the value of
cos A and sin B.
ACCUPLACER Advanced Algebra and Functions
© 2020 College Board.
