25 Advanced Algebra & Functions Practice Questions & Answers

1. 1. What is the solution (x, y) of the system 2x + 3y = 12 and x − y = 1?

   • A.(3, −2)
   • B.(3, 2)✓ Answer
   • C.(2, 3)
   • D.(5, 4)

   From x − y = 1, x = y + 1. Substituting gives 2(y + 1) + 3y = 12, so 5y = 10 and y = 2, then x = 3. The point (3, 2) satisfies both equations: 6 + 6 = 12 and 3 − 2 = 1. The pair (2, 3) swaps the coordinates, and the other pairs fail at least one equation.

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2. 2. For what value of k does the system 3x − ky = 7 and 6x + 4y = 5 have no solution?

   • A.−8
   • B.2
   • C.8
   • D.−2✓ Answer

   A linear system has no solution when the lines are parallel: the x- and y-coefficients are proportional but the constants are not. Since 6 = 2 × 3, we need 4 = 2 × (−k), so k = −2. Then the first equation becomes 3x + 2y = 7, which doubles to 6x + 4y = 14, a different constant from 5, so the lines are parallel and distinct. The value 2 comes from a sign slip, and ±8 come from inverting the ratio.

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3. 3. The line through the points (1, 4) and (3, a) is perpendicular to the line 2x + 5y = 10. What is the value of a?

   • A.9✓ Answer
   • B.16/5
   • C.−1
   • D.24/5

   Rewriting 2x + 5y = 10 as y = (−2/5)x + 2 shows its slope is −2/5. Perpendicular slopes are negative reciprocals, so the new slope is 5/2. Then (a − 4)/(3 − 1) = 5/2 gives a − 4 = 5 and a = 9. Using −5/2 (reciprocal without the sign change) gives −1, and using −2/5 or 2/5 (parallel-type slopes) gives 16/5 or 24/5.

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4. 4. The graph of a linear inequality has the boundary line y = 2x − 3 drawn as a dashed line, and the region above the line is shaded. Which inequality does the graph represent?

   • A.y ≥ 2x − 3
   • B.y > 2x − 3✓ Answer
   • C.y < 2x − 3
   • D.y ≤ 2x − 3

   A dashed boundary means points on the line are not included, so the inequality is strict (> or <). Shading above the line means y-values greater than 2x − 3, so the inequality is y > 2x − 3. The symbol ≥ would require a solid line, and < or ≤ would shade below the line.

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5. 5. Which of the following is the factored form of 8x^3 + 27?

   • A.(2x − 3)(4x^2 + 6x + 9)
   • B.(2x + 3)(4x^2 − 6x + 9)✓ Answer
   • C.(2x + 3)(4x^2 − 12x + 9)
   • D.(2x + 3)(4x^2 + 6x + 9)

   8x^3 + 27 is a sum of cubes with a = 2x and b = 3. The rule a^3 + b^3 = (a + b)(a^2 − ab + b^2) gives (2x + 3)(4x^2 − 6x + 9). The middle term of the trinomial is −ab = −6x, not +6x (that belongs to the difference of cubes) and not −12x (that is −2ab, from a perfect-square trinomial).

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6. 6. Which of the following is the complete factorization of 2x^3 − 3x^2 − 8x + 12?

   • A.(2x − 3)(x − 2)(x + 2)✓ Answer
   • B.(2x − 3)(x − 2)(x − 2)
   • C.(2x + 3)(x − 2)(x + 2)
   • D.(3 − 2x)(x − 2)(x + 2)

   Group the terms: x^2(2x − 3) − 4(2x − 3) = (2x − 3)(x^2 − 4). Then x^2 − 4 is a difference of squares, (x − 2)(x + 2). Writing (x − 2)(x − 2) treats x^2 − 4 as a perfect square, (2x + 3) comes from a sign slip when factoring −4 out of −8x + 12, and (3 − 2x) is the negative of the correct factor, so that product equals the opposite polynomial.

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7. 7. What is the vertex of the parabola y = −2x^2 + 12x − 7?

   • A.(3, 11)✓ Answer
   • B.(3, 47)
   • C.(−3, −61)
   • D.(6, −7)

   The vertex's x-coordinate is −b/(2a) = −12/(2 × −2) = 3. Substituting gives y = −2(9) + 36 − 7 = 11, so the vertex is (3, 11), a maximum because a < 0. Dropping the negative sign in −b/(2a) gives x = −3; evaluating −2(3)^2 as +18 gives 47; and using −b/a gives x = 6.

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8. 8. What are the solutions of x^2 − 4x + 13 = 0?

   • A.2 ± 3
   • B.4 ± 6i
   • C.2 ± 3i✓ Answer
   • D.−2 ± 3i

   The discriminant is b^2 − 4ac = 16 − 52 = −36, which is negative, so the roots are non-real. The quadratic formula gives x = (4 ± sqrt(−36))/2 = (4 ± 6i)/2 = 2 ± 3i, since sqrt(−36) = 6i. Using +b instead of −b gives −2 ± 3i, ignoring the negative under the radical gives 2 ± 3, and forgetting to divide by 2a leaves 4 ± 6i.

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9. 9. The system y = x^2 − 2x + 3 and y = 2x + k has exactly one solution. What is the value of k?

   • A.−1✓ Answer
   • B.4
   • C.1
   • D.−4

   Setting the expressions equal gives x^2 − 2x + 3 = 2x + k, or x^2 − 4x + (3 − k) = 0. Exactly one solution means the discriminant is zero: 16 − 4(3 − k) = 0, so 4 + 4k = 0 and k = −1. Then the line y = 2x − 1 touches the parabola only at (2, 3). Writing the constant as 3 + k gives 1, and treating the constant as k alone gives 4.

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10. 10. If f(x) = 2x − 1 and g(x) = x^2 + 3, what is the value of f(g(−2))?

    • A.−3
    • B.28
    • C.7
    • D.13✓ Answer

    Composition works from the inside out: first g(−2) = (−2)^2 + 3 = 7, then f(7) = 2(7) − 1 = 13. Computing g(f(−2)) instead reverses the order and gives g(−5) = 28. Stopping after g(−2) gives 7, and squaring −2 as −4 gives g(−2) = −1 and f(−1) = −3.

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11. 11. If f(x) = (3x − 4)/5, which expression is the inverse function f^(−1)(x)?

    • A.(3x + 4)/5
    • B.(5x − 4)/3
    • C.5/(3x − 4)
    • D.(5x + 4)/3✓ Answer

    Write y = (3x − 4)/5, swap x and y to get x = (3y − 4)/5, and solve: 5x = 3y − 4, so y = (5x + 4)/3. The inverse undoes the operations in reverse order, so subtracting 4 becomes adding 4. The reciprocal 5/(3x − 4) confuses an inverse function with 1/f(x), and (5x − 4)/3 keeps the original sign.

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12. 12. The point (1, 4) lies on the graph of y = f(x). Which point must lie on the graph of g(x) = −f(x − 3) + 2?

    • A.(4, 6)
    • B.(−2, −2)
    • C.(4, −6)
    • D.(4, −2)✓ Answer

    Replacing x with x − 3 shifts the graph 3 units right, so x = 1 moves to x = 4. The negative sign reflects the y-value over the x-axis (4 becomes −4), and the +2 then shifts it up 2, giving −2. So (4, −2) is on g. Shifting left gives x = −2, ignoring the reflection gives 6, and adding 2 before reflecting gives −6.

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13. 13. What is the solution set of the equation sqrt(2x + 7) = x + 2?

    • A.x = −1 or 3
    • B.x = 1 or −3
    • C.x = 1 only✓ Answer
    • D.x = −3 only

    Squaring both sides gives 2x + 7 = x^2 + 4x + 4, so x^2 + 2x − 3 = 0 and (x + 3)(x − 1) = 0, giving x = 1 or x = −3. Squaring can create extraneous roots, so each must be checked: x = 1 gives sqrt(9) = 3 = 1 + 2, which works, but x = −3 gives sqrt(1) = 1 while the right side is −1. A principal square root is never negative, so −3 is rejected. The pair −1 and 3 comes from factoring with the wrong signs.

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14. 14. Which statement correctly describes the graph of f(x) = (x^2 − 9)/(x^2 + x − 6)?

    • A.Asymptotes at x = −3 and x = 2
    • B.Hole at x = −3; asymptote x = 2✓ Answer
    • C.Hole at x = 2; asymptote x = −3
    • D.Hole at x = 3; asymptote x = −2

    Factor: f(x) = (x − 3)(x + 3)/[(x + 3)(x − 2)]. The factor x + 3 cancels, so x = −3 produces a hole (a removable discontinuity), while x − 2 remains in the denominator, so x = 2 is a vertical asymptote. Calling both zeros of the denominator asymptotes ignores the cancellation, and the other choices swap the roles or the signs.

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15. 15. What is the remainder when p(x) = x^3 − 4x^2 + 2x + 5 is divided by x − 3?

    • A.2✓ Answer
    • B.−64
    • C.5
    • D.0

    By the remainder theorem, the remainder when p(x) is divided by x − a equals p(a). Here p(3) = 27 − 36 + 6 + 5 = 2. Evaluating p(−3) = −64 uses the wrong sign of a, 5 is just the constant term, and 0 would mean x − 3 is a factor, which it is not.

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16. 16. Which statement describes the graph of f(x) = −2(x − 1)^2(x + 3)^3?

    • A.Touches at x = 1, crosses at x = −3, falls to the right✓ Answer
    • B.Crosses at x = 1, touches at x = −3, falls to the right
    • C.Touches at x = −1, crosses at x = 3, falls to the right
    • D.Touches at x = 1, crosses at x = −3, rises to the right

    The zero x = 1 has even multiplicity 2, so the graph touches the x-axis and turns there; the zero x = −3 has odd multiplicity 3, so the graph crosses. The leading term is −2x^5 (odd degree, negative coefficient), so the graph rises to the left and falls to the right. The zeros are the values that make each factor zero, so x − 1 gives 1, not −1.

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17. 17. What value of x satisfies 4^(x + 1) = 8^(x − 1)?

    • A.1
    • B.5✓ Answer
    • C.2
    • D.3

    Rewrite both sides with base 2: 4^(x + 1) = 2^(2x + 2) and 8^(x − 1) = 2^(3x − 3). Equal powers of the same base have equal exponents, so 2x + 2 = 3x − 3 and x = 5. Multiplying only the x by 2 and 3 (getting 2x + 1 = 3x − 1) gives 2, and multiplying the bases instead of using exponents (4(x + 1) = 8(x − 1)) gives 3. A sign slip when collecting terms (writing 2x − 3x = −3 + 2, so −x = −1) gives 1.

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18. 18. What is the solution set of log_2(x) + log_2(x − 2) = 3?

    • A.x = 5 only
    • B.x = 4 or −2
    • C.x = −2 only
    • D.x = 4 only✓ Answer

    The product rule combines the logs: log_2(x(x − 2)) = 3, so x(x − 2) = 2^3 = 8. Then x^2 − 2x − 8 = 0 factors as (x − 4)(x + 2) = 0, giving 4 or −2. A logarithm's argument must be positive, and x = −2 makes log_2(−2) undefined, so only x = 4 works. Adding the arguments instead of multiplying them (2x − 2 = 8) gives the wrong value 5.

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19. 19. The amount of a medication in a patient's bloodstream, in milligrams, t hours after a dose is modeled by M(t) = 300(0.82)^t. What does the number 0.82 tell you about the medication?

    • A.The amount drops 0.82 mg each hour
    • B.The amount drops by 82% each hour
    • C.18% of the amount remains each hour
    • D.The amount drops by 18% each hour✓ Answer

    In y = a(1 − r)^t, the base 1 − r is the decay factor: the fraction kept each period. Here 1 − r = 0.82, so 82% remains each hour and the decay rate is r = 0.18, an 18% drop per hour. Reading 0.82 as the percent lost confuses the factor with the rate, and a fixed drop of 0.82 mg would describe a linear model, not an exponential one.

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20. 20. The midpoint of segment AB is (2, −1). If A = (−3, 4), what are the coordinates of B?

    • A.(−8, 9)
    • B.(7, −6)✓ Answer
    • C.(5, −5)
    • D.(−1/2, 3/2)

    The midpoint averages the endpoints, so (−3 + x)/2 = 2 and (4 + y)/2 = −1. Solving gives x = 7 and y = −6, so B = (7, −6); equivalently, B = 2M − A. Averaging A with the midpoint gives (−1/2, 3/2), computing 2A − M gives (−8, 9), and M − A gives only the step from A to M, (5, −5).

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21. 21. The point (3, −5) is rotated 90 degrees counterclockwise about the origin. What are the coordinates of its image?

    • A.(5, 3)✓ Answer
    • B.(−3, 5)
    • C.(−5, 3)
    • D.(−5, −3)

    A 90-degree counterclockwise rotation about the origin maps (x, y) to (−y, x). With x = 3 and y = −5, the image is (5, 3), which moves the point from Quadrant IV into Quadrant I as a counterclockwise turn should. The rule (y, −x) is a clockwise rotation and gives (−5, −3), (−x, −y) is a 180-degree rotation, and (y, x) is a reflection over y = x.

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22. 22. What are the center and radius of the circle x^2 + y^2 + 8x − 2y − 8 = 0?

    • A.Center (4, −1), radius 5
    • B.Center (−4, 1), radius 3
    • C.Center (−4, 1), radius 5✓ Answer
    • D.Center (−4, 1), radius 25

    Complete the square: (x^2 + 8x + 16) + (y^2 − 2y + 1) = 8 + 16 + 1, so (x + 4)^2 + (y − 1)^2 = 25. In (x − h)^2 + (y − k)^2 = r^2, the center is (h, k) = (−4, 1) and r = sqrt(25) = 5. Reading the signs directly from the equation gives (4, −1), giving r^2 = 25 as the radius gives 25, and forgetting to add 16 to the right side gives r^2 = 9 and radius 3.

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23. 23. A circle has a radius of 12 centimeters. What is the length, in centimeters, of the arc cut off by a central angle of 150 degrees?

    • A.5pi
    • B.1800
    • C.10pi✓ Answer
    • D.60pi

    Arc length s = r × theta requires theta in radians. Converting, 150 × pi/180 = 5pi/6, so s = 12 × 5pi/6 = 10pi centimeters. Multiplying 12 by 150 degrees gives 1800, a degrees-radians mix-up; 60pi is the sector area (1/2)r^2 theta; and 5pi comes from using the radius 6 instead of 12.

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24. 24. In a right triangle, one acute angle measures 60 degrees and the hypotenuse is 14 units long. What is the length of the leg opposite the 60-degree angle?

    • A.14sqrt(2)
    • B.14sqrt(3)
    • C.7sqrt(3)✓ Answer
    • D.7sqrt(2)

    A 30-60-90 triangle has sides in the ratio 1 : sqrt(3) : 2, with the short leg opposite 30 degrees and the long leg opposite 60 degrees. With hypotenuse 14, the short leg is 7 and the long leg is 7sqrt(3); this matches sin(60°) = sqrt(3)/2, since 14 × sqrt(3)/2 = 7sqrt(3). Multiplying the hypotenuse by sqrt(3) gives 14sqrt(3), and the sqrt(2) answers apply the 45-45-90 ratio to the wrong triangle.

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25. 25. What are all solutions of 2sin(x) + sqrt(3) = 0 on the interval 0 ≤ x < 2pi?

    • A.pi/3 and 2pi/3
    • B.5pi/6 and 7pi/6
    • C.4pi/3 and 5pi/3✓ Answer
    • D.7pi/6 and 11pi/6

    Isolating the sine gives sin(x) = −sqrt(3)/2. The reference angle with sine sqrt(3)/2 is pi/3, and sine is negative in Quadrants III and IV, so x = pi + pi/3 = 4pi/3 and x = 2pi − pi/3 = 5pi/3. The angles pi/3 and 2pi/3 ignore the negative sign, 7pi/6 and 11pi/6 solve sin(x) = −1/2, and 5pi/6 and 7pi/6 solve cos(x) = −sqrt(3)/2.

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