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NEXT-GENERATION Advanced Algebra and Functions Sample QuestionsThe College Board The College Board is a mission-driven not-for-prot organization that connects students to college success and opportunity. Founded in 1900, the College Board was created to expand access to higher education. Today, the membership association is made up of over 6,000 of the world’s leading education institutions and is dedicated to promoting excellence and equity in education. Each year, the College Board helps more than seven million students prepare for a successful transition to college through programs and services in college readiness and college success — including the SAT ® and the Advanced Placement Program ®. The organization also serves the education community through research and advocacy on behalf of students, educators, and schools. For further information, visit www.collegeboard.org. ACCUPLACER Advanced Algebra and Functions Sample Questions The next-generation ACCUPLACER ® 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 1 © 2016 The College Board. College Board, ACCUPL ACER, and the acorn logo are registered trademarks of the College Board. 00214-024
y 6 x O –6 6 –6 5 4 y = x 5 y = x + 3 4 A. B. C. D. 4 y = − x 5 4 y = − x + 3 5 A. 75 cm 2 B. 108 cm 2 C. 120 cm 2 D. 150 cm 2 A. x 3 – 3x 2 + 2 x + 14 x 3 + 4 x 2 – 19 x + 14 C. x 3 – 3x + 14 B. D. x2 – 2x + 9 Cost of Apples 8 7 6� Cost (dollars) 7 5 Cost of Pears: C = p 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 Number of pounds Sample Questions 3. Choose the best a nswer. If necessary, use the paper you were given. 3 cm 9 cm 4 cm 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 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.) 2. 4. Which of the following expressions is equivalent to ( x + 7)( x 2 – 3x + 2)? 5. 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? 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 a pproximately $0.07 less per pound than pears do. B. Apples cost a pproximately $0.04 less per pound than pears do. C. Apples cost a pproximately $0.73 less per pound than pears do. D. Apples cost a pproximately $0.62 more per pound than pears do. ACCUPLACER Sample Questions Advanced Algebra and Functions 2 © 2016 The College Board
A. 3(x + 2)( x – 4) B. 3(x – 2)( x + 4) C. ( x + 6)( x – 12) D. (x – 6)( x + 12) A . n = 50 0 (2) x B. n = 50 0 (2) 6x C. n = 5 0 0 (6) x D. n = 5 0 0 (6) 2x A . 7 B. 3 C. –2 D. –7 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) x O –4 –2 2 4 2 –2 –4 –6 –8 y A. f( x ) = x 2 – 2 x – 8 B. f( x ) = – x2 + 2x – 8 C. f( x ) = ( x – 2)( x + 4) D. f( x ) = –( x – 1) 2 – 9 A. y ≤ –2 B. y ≥ 7 C. y ≤ 7 D. All real numbers 6. Which of the following is the graph of a function where 8. y = f(x )? y x O x2 + 5x – 9 = 5 Which of the following values of x satises the equation above? 9. 10. e grap h of y = f (x ) is shown in the xy - plane below. x y O Which of the following equations could dene f ( x )? y O x 11 . Which o f the following best describes the range of y = –2 x 4 + 7? 7. Which o f the following expressions is equivalent to 3 x2 + 6 x – 24? A. x O y B. C. D. ACCUPLACER Sample Questions Advanced Algebra and Functions 3 © 2016 The College Board
A. (6x) 2 = 0 B. ( x – 6)2 = 0 C. ( x + 6)2 = 0 D. (x – 6)( x + 6) = 0 A. 2 + 3 + 3 B. ( x + 2) 2 + 3(x + 2) + 1 C. ( x + 2)( x2 + 3x + 1) D. x 2 + 3x + 9 x x 1 − 5 143 5 A. B. 7 C. D. ere is no real solution. 5 x = 2x − 3? x + 2 A. 3 and 5 3 B. 2 and − 2 3 C. –2 and 2 D. –3 and –5 A. ∠L ≅ ∠ R and JL = PR B. KL = QR and PR = JL C. JK = PQ and KL = QR D . ∠K ≅ ∠Q and ∠L ≅ ∠R 7 A. x = log 2 ( 5 ) log 2 7 B. x = 5 log 7 2 C. x = 5 log 7 5 D. x = 2 x − y ? x − y x − y A. x − yB. x − y C. x + y D. x x + y y 5 cos A = 8, 3 A. 8 5 B. 8 39 C. 8 89 D. 8 12. For which of the following equations i s x = 6 the only solution? 13. If f ( x ) = x 2 + 3 x + 1, what is f( x + 2)? 14 . What, if any, is a real solution to 5 x + 1 + 9 = 3 ? 15. 3 If x ≠ –2 a nd x ≠ − , what is the solution to 2 17. In the f unction 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 p ositive fo r 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. Which of the following equations is equivalent to 25x = 7? 18 . 19. If x > 0 and y > 0, which of the following expressions is eq uivalent to 20. In triangle ABC , a ngle C is a right angle. If 16 . wh at is the value of cos B ? J KR L Q P Triangle JKL and triangle PQR are shown above. If ∠ J is c ongruent to ∠ P, which of the following must be true in order to prove that triangles JKL and PQR are congruent? AC CUPLACER Sample Questions Advanced Algebra and Functions 4 © 2016 The College Board
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 Sample Questions Advanced Algebra and Functions 5 © 2016 The College Board
g(12) = 3 (12 + 8) , 3(20) 12 3( 8) . = + x g(12) = 3 (12) + 8 . − 54 . − 4 5 y = 5 4 x + 0 , y = 5 4 x, x(x2) + x(− 3 x) + x(2) + 7( x2) + 7( −3 x) + 7(2) . x3 − 3 2 + 2 + 7 2 − 21 + 14 . x x x x Further simplifying resu lts in x3 + 4 x2 − 19 x + 14 . C = 57 p . 7 C is , 5 Rationales 1. Choice D is correct. The valu e of g(12) can be found by substituting 12 for x in the equation for ( ) . g x This yields which is equivalent to or 60. Choice A is incorrect. This answer represents the value of x in the equation 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: 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 The negative reciprocal of is 5 4 . A line that passes through the point (0, 0) has a y-intercept of 0. Ther efore, the equation or is correct. Choice B is incorrect becaus e i t 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 sur face area of the rectangular prism is the total area of each of the faces of the prism and can be written as 2(length × width) + 2 (height × width) + 2(length × heig ht) , 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 it is the volume of the prism. 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 Finally, adding like terms yields Choices A, C, and D are incorrect because they each result from e rrors 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 The slope of the line dened by , so the cost per pound of pears is $ 1.40. Therefore, the apples cost approximately $0.07 less per pound than pears do. Choice B is incorrect. This is the result of misreading the cost per pound of apples as $0.67 and the cost per pound of pears as $0.71 and then nding the dierence between the two values. Choice C is incorrect. This is the result of misreading the cost per pound of apples from the graph as $ 0.67 and then subtracting the cost per pound of pears, $1.4 0 . Choice D is incorrect. This is the result of misreading the cost per pound of pears as $0.71 and then subtracting this value from the cost per pound of apples, $1. 33 . 6. Ch oice C is correct. A function has one output for each input. 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. ACCUPLACER Sample Questions Advanced Algebra and Functions 6 © 2016 The College Board
3( − 2)( + 4) x x 3x2 − 6 x − 24 . x2 − 6 x − 72 x2 + 6 x − 72 . y = ab t n = 500 bt. n = 50 0 (2) t. n = 50 0 (2) 6x . x2 + 5 x − 14 = 0 . (x + 7)( x − 2) = 0 . x = − 7 and x = 2 . 72 + 5(7) − 9 32 + 5(3) − 9 ( − 2) 2 + 5(−2) − 9 y = f(x) x = −2 and x = 4 , y = 8, f(x) f(x) = x2 − 2 x − 8 . y = f(x), f(x) = − x2 + 2 x − 8 ; f(x) = − x2 + 2 x − 8 y = f(x) f(x) = − x2 + 2 x − 8 y = f(x), f(x) = ( x − 2)( x + 4) ; f(x) = ( x − 2)( x + 4) f(x) = − ( x − 1) 2 − 9 , f(x) = − ( x − 1) 2 − 9 y = f(x) f(x) = − ( x − 1) 2 − 9 y = −2 x4 + 7 y = − 2 x4 + 7 7. Choice B is correct. The expres si on can be expanded by rst multiplying (x − 2) by 3 to get (3x − 6) and then multiplying (3x − 6) by (x + 4) to get 3x2 + 6 x − 24 . Choice A is incorrect because it is equivalent to Choice C is incorrec t because it is equivalent to . Choice D is incorrect be c ause it is equivalent to 8. Choice B is co rrect. An exponential function can be written in the form 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 total number of bacteria, and the initial amount is given as 500 , which yields The growth factor i s 2 because the population is described as being expected to double, which gives the equation 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 Choices A, C, and D are inc orrect. 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 rate, and choice D uses the number of time periods per day as the growth rate and multiplies the exponent by the actual growth rate. 9. Choice D is correct. Subtracting 5 from both sides of the equation gives The left-hand side of the equation can be factored, giving Therefore, the solutions to the quadratic equation are Choice A is incorrect because is not equal to 5. Choice B is incorrect because is not equal to 5. Choice C is incorrect because is not equal to 5. 10. Choice A is correct. The graph of crosses the x-axis at crosses the y-axis at and has its ve rtex 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 dening must have a po sitive leading coecient. All of these conditions are met by the equation Choice B is incorrect. The points ( − 2 , 0), (4, 0), (0, − 8) , and (1, − 9) , which are easily identied on the graph of do not all satisfy the equation only (0, − 8) does. Therefore cannot dene the function graphed. Furthermore, because the graph opens upward, the equation dening must have a positive leading coecient, which does not. Choice C is incorrect. The points ( − 2 , 0), (4, 0), (0, − 8) , and (1, −9) , which are easily identied on the graph of do not all satisfy the equation only (0, − 8) does. Therefore, cannot dene the function graphed. Choice D is incorrect. Though the vertex (1, − 9) does satisfy the equation the points ( − 2 , 0), (4, 0), and (0, − 8) do not. Therefore, cannot dene the function graphed. Furthermore, because the graph opens upward, the equation dening must have a positive leading coecient, which 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 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 val ue of y is 7. Therefore, the range of the function dened by 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. ACCUPLACER Sample Questions Advanced Algebra and Functions 7 © 2016 The College Board
(x − 6) 2 = 0 x = 0 (6 )2 = 0 . x = − 6 ( + 6) 2 = 0 . x = 6 (x − 6)( x + 6) = 0 , x = − 6 x + 2 f(x + 2) = ( x + 2) 2 + 3( x + 2) + 1 . f(x) + 2 (x + 2) f(x). 5x + 1 = − 6 , x(x + 2) = 5(2 x − 3) . x2 + 2 x = 10 x − 15 . x2 − 8 x + 15 = 0 . ( − 3)( − 5) = 0 . x = 3 and x = 5 . f(x) = a(x + 2)( x − 3) b 12. Choice B is correct. The only va lue of x that satises the equation is 6. Choice A is incorrect because is the only solution to the equation x Choice C is incorrect because is the only solution to the equation x Choice D is incorrect because although is a solution to the equati on is another solution to the equation. 13. Choice B is correct. Substit uting for x in the original function gives Choice A is incorrect. This is . Choice C is incorrect . This is Choice D is incorrect. This is f( ) + 23 . x 14. Choice D i s correct. Subtracting 9 from both sides of the equation yields 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 e quation for x, cross multiply to yield Simplifying both sides of the new equation results in Next, subtract 10x from both sides of the equation and add 15 to both sides of the equation to yield By factoring the left-hand side, the eq u ation can be rewritten in the form x x It follows, therefore , that Choices B, C, and D are possible results from mathematical errors when solving the equation for x . 16. Choice A i s 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. 1 7. 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 to be of even degree with a positive leading coecien t , 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. Choice B 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. ACCUPLACER Sample Questions Advanced Algebra and Functions 8 © 2016 The College Board
. (b)x = y, b > 0 and b ≠ 1, x = log b y. 25x = 7 log27 = 5 x. log 2 7 = x . 5 (b)x = y, b > 0 and b ≠ 1 x − y x − y ( x 2 ) − ( y2 ) . x − y ( x + y )( x − y ) ( x − y ) x − y x − y x + y , x + y (x − y )( x + y ) , ( x − y )( x + y ) x x + x y − y x − y y x − xy + xy − y (x − y) ( x + y ) , (x − y ) x − y x − y A (cos A ) the length of the side adjacent to angle A the length of the hypotenuse 5 8 a2 + 5 2 = 8 2, a2 = 64 − 25 = 39 , so a = 39 . the length of the side adjacent to angle B 39 cos B = = . the length of the hypotenuse 8 18. Choice B is correct. By denition , if where then Therefore, the given equation can be rewritten in the form Next, solving for x by dividing both sides of the equation by 5 yields Choices A, C, and D are incorrect because they are the result of misapplying the identity, which states that if where , then x = log b y. 19. Choice C is corr ect. 2 ( x ) Since x > 0 and y > 0, x can be rewritten as and y 2 ( y ) can be rewritten as . It follows, then, that can be rewritten as . Because the numerator is a dierence of two squares, it can be factored as . Finally, dividing the common factors of ( x − y ) in the numerator and denominator yields x + y . Alternatively, if is multiplied by , which is equal to 1, and therefore does not change the value of the original expression, the result is which is equivalent to . This can be rewritten as 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 . 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 correc t. If triangle ABC is dened as a right triangle, where angle C is the right angle, then the cosine of angle is dened as the ratio . Since this ratio is dened as , 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: where a represents the length of the side opposite angle A . Solving for a yields Then, to determine the cosine of angle B, use the same ratio in relation to angle B: ACCUPLACER Sample Questions Advanced Algebra and Functions 9 © 2016 The College Board