ACCUPLACER Math

ACCUPLACER AAF Practice Test: 15 Advanced Algebra Questions Solved

Fifteen original ACCUPLACER AAF questions, one or more from every official area, from quadratics to logarithms and trigonometry.

Professor Chacha October 9, 2026 7 min read 3 views
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Advanced Algebra and Functions (AAF) is the hardest of the three ACCUPLACER math tests. It is the one that can place you into college algebra, precalculus, or even calculus, and it covers everything from linear equations to logarithms and trigonometry. Below are 15 original AAF-style questions, solved step by step, at least one from every area on the official list.

What AAF covers

The College Board's test specifications list these content areas for AAF. A 20-question test usually has 1 to 4 questions from each:

Area Questions here
Linear equations 1
Linear applications and graphs 2
Factoring 3
Quadratics 4, 5
Functions 6, 7
Radical and rational equations 8, 9
Polynomial equations 10
Exponential and logarithmic equations 11 to 13
Geometry concepts 14
Trigonometry 15

You only take AAF if your college sends you there, usually after a strong result on QAS or because you want a higher course. To see how the three tests fit together, start with our ACCUPLACER math practice test; if QAS is your test, try the QAS practice test first.

Linear equations, graphs, and factoring

1.Solve for \(x\): \(\dfrac{x + 2}{3} = \dfrac{x - 1}{2}\)

  1. A1
  2. B\(-7\)
  3. C5
  4. D7
Show solution

Cross-multiply, then solve.

\[\begin{gathered}2(x + 2) = 3(x - 1) \\ \Rightarrow\; 2x + 4 = 3x - 3 \\ \Rightarrow\; x = 7\end{gathered}\]

Common trap: Distributing the 3 to the \(x\) only, \(3x - 1\), which gives \(x = 5\). Multiply every term inside the parentheses.

Answer: D (7)

2.Which equation describes the line parallel to \(y = 3x - 1\) that passes through \((2, 4)\)?

  1. A\(y = 3x + 4\)
  2. B\(y = -\tfrac{1}{3}x + \tfrac{14}{3}\)
  3. C\(y = 3x - 2\)
  4. D\(y = 3x - 1\)
Show solution

Parallel lines have the same slope, 3. Use the point to find \(b\).

\[\begin{gathered}4 = 3(2) + b \\ \Rightarrow\; b = -2 \\ \Rightarrow\; y = 3x - 2\end{gathered}\]

Common trap: Using the point's \(y\)-value as the intercept, \(y = 3x + 4\). The intercept is where the line crosses the \(y\)-axis, at \(x = 0\), not at \(x = 2\). The choice with slope \(-\tfrac{1}{3}\) is perpendicular, not parallel.

Answer: C (\(y = 3x - 2\))

3.Factor completely: \(x^3 - 9x\)

  1. A\(x(x - 3)(x + 3)\)
  2. B\(x(x - 3)^2\)
  3. C\((x - 3)(x + 3)\)
  4. D\(x(x^2 - 9)\)
Show solution

Take out the common factor \(x\), then factor the difference of squares.

\[x^3 - 9x = x(x^2 - 9) = x(x - 3)(x + 3)\]

Common trap: Stopping at \(x(x^2 - 9)\). "Factor completely" means keep going while any factor can still be factored.

Answer: A (\(x(x - 3)(x + 3)\))

Quadratics and functions

4.What is the vertex of the parabola \(y = x^2 - 6x + 5\)?

  1. A\((-3, 32)\)
  2. B\((3, -4)\)
  3. C\((6, 5)\)
  4. D\((3, 4)\)
Show solution

The \(x\)-coordinate of the vertex is \(-\tfrac{b}{2a}\). Then substitute to get \(y\).

\[\begin{gathered}x = -\tfrac{-6}{2(1)} = 3 \\ y = 3^2 - 6(3) + 5 = -4\end{gathered}\]

Common trap: Dropping the minus sign in \(-\tfrac{b}{2a}\) and using \(x = -3\). With \(b = -6\), \(-\tfrac{b}{2a}\) is positive.

Answer: B (\((3, -4)\))

5.How many real solutions does \(2x^2 + 3x + 5 = 0\) have?

  1. ATwo
  2. BOne
  3. CNone
  4. DInfinitely many
Show solution

Look at the discriminant, \(b^2 - 4ac\).

\[3^2 - 4(2)(5) = 9 - 40 = -31\]

It is negative, so the equation has no real solutions: the parabola never touches the \(x\)-axis.

Common trap: Assuming every quadratic has two solutions. A negative discriminant means none, zero means exactly one, and positive means two.

Answer: C (None)

6.If \(f(x) = 2x + 1\) and \(g(x) = x^2\), what is \(f(g(3))\)?

  1. A49
  2. B19
  3. C10
  4. D13
Show solution

Work from the inside out: first \(g(3)\), then put that result into \(f\).

\[\begin{gathered}g(3) = 9 \\ f(9) = 2(9) + 1 = 19\end{gathered}\]

Common trap: Doing the functions in the wrong order: \(g(f(3)) = g(7) = 49\). In \(f(g(3))\), \(g\) acts first.

Answer: B (19)

7.What is the inverse of \(f(x) = \dfrac{x - 4}{3}\)?

  1. A\(f^{-1}(x) = 3x + 4\)
  2. B\(f^{-1}(x) = \dfrac{x + 4}{3}\)
  3. C\(f^{-1}(x) = 3x - 4\)
  4. D\(f^{-1}(x) = \dfrac{3}{x - 4}\)
Show solution

Swap \(x\) and \(y\), then solve for \(y\).

\[\begin{gathered}x = \dfrac{y - 4}{3} \\ \Rightarrow\; 3x = y - 4 \\ \Rightarrow\; y = 3x + 4\end{gathered}\]

Common trap: Taking the reciprocal, \(\tfrac{3}{x - 4}\). The \(-1\) in \(f^{-1}\) means "undo," not "one over." Undo subtracting 4 and dividing by 3 by multiplying by 3 and adding 4.

Answer: A (\(f^{-1}(x) = 3x + 4\))

Radical, rational, and polynomial equations

8.Solve: \(\dfrac{x}{x - 3} - \dfrac{3}{x - 3} = 2\)

  1. A\(x = 3\)
  2. B\(x = 6\)
  3. C\(x = -3\)
  4. DNo solution
Show solution

Combine the left side: \(\dfrac{x - 3}{x - 3}\). That equals 1 for every \(x \ne 3\), and 1 is never 2.

If you multiply both sides by \(x - 3\), you get \(x - 3 = 2x - 6\), so \(x = 3\). But \(x = 3\) makes the denominator 0, so it is not allowed. There is no solution.

Common trap: Accepting \(x = 3\). Any value that makes a denominator zero must be thrown out, so always check rational-equation answers in the original.

Answer: D (No solution)

9.Solve: \(\sqrt{2x + 3} = 5\)

  1. A1
  2. B14
  3. C11
  4. D4
Show solution

Square both sides.

\[\begin{gathered}2x + 3 = 25 \\ \Rightarrow\; 2x = 22 \\ \Rightarrow\; x = 11\end{gathered}\]

Check: \(\sqrt{25} = 5\).

Common trap: Squaring only the 5 on one side but treating the left as \(2x + 3 = 5\), which gives 1. Squaring removes the root and must be applied to both sides.

Answer: C (11)

10.Multiply: \((x + 2)(x^2 - x + 3)\)

  1. A\(x^3 + x^2 + x + 6\)
  2. B\(x^3 - x^2 + 5x + 6\)
  3. C\(x^3 + x^2 + 5x + 6\)
  4. D\(x^3 + 3x + 6\)
Show solution

Multiply each term of the first factor by every term of the second, then combine like terms.

\[\begin{gathered}x^3 - x^2 + 3x + 2x^2 - 2x + 6 \\ = x^3 + x^2 + x + 6\end{gathered}\]

Common trap: Combining \(3x\) and \(2x\) into \(5x\) and forgetting the \(-2x\). Line up like terms before adding.

Answer: A (\(x^3 + x^2 + x + 6\))

Exponents and logarithms

11.Solve for \(x\): \(4^x = 2^{x + 3}\)

  1. A3
  2. B1
  3. C\(-3\)
  4. D6
Show solution

Rewrite 4 as \(2^2\) so both sides have base 2, then set the exponents equal.

\[\begin{gathered}(2^2)^x = 2^{x + 3} \\ \Rightarrow\; 2x = x + 3 \\ \Rightarrow\; x = 3\end{gathered}\]

Common trap: Setting \(x = x + 3\) without changing the base, which has no solution. Exponents can only be compared once the bases match.

Answer: A (3)

12.Solve: \(\log_2 x = 5\)

  1. A10
  2. B25
  3. C\(\tfrac{5}{2}\)
  4. D32
Show solution

A logarithm asks "what power?" So \(\log_2 x = 5\) means \(2^5 = x\).

\[x = 2^5 = 32\]

Common trap: Multiplying, \(2 \times 5 = 10\), or squaring, \(5^2 = 25\). Rewrite the log as an exponent: base to the power gives the number.

Answer: D (32)

13.Evaluate: \(\log 4 + \log 25\) (logarithms base 10)

  1. A\(\log 29\)
  2. B100
  3. C2
  4. D1
Show solution

The sum of logs is the log of the product.

\[\begin{gathered}\log 4 + \log 25 = \log(4 \times 25) \\ = \log 100 = 2\end{gathered}\]

Common trap: Adding inside the log, \(\log 29\). The rule is \(\log a + \log b = \log(ab)\): a sum of logs becomes a product.

Answer: C (2)

Geometry and trigonometry

14.What are the center and radius of the circle \((x - 2)^2 + (y + 1)^2 = 16\)?

  1. ACenter \((-2, 1)\), radius 4
  2. BCenter \((2, -1)\), radius 4
  3. CCenter \((2, -1)\), radius 16
  4. DCenter \((-2, 1)\), radius 16
Show solution

The standard form is \((x - h)^2 + (y - k)^2 = r^2\). Here \(h = 2\), \(k = -1\) (because \(y + 1 = y - (-1)\)), and \(r^2 = 16\).

\[\begin{gathered}\text{center } (2, -1), \\ r = \sqrt{16} = 4\end{gathered}\]

Common trap: Taking the signs as written, \((-2, 1)\), or reading 16 as the radius. The equation shows \(r^2\), so the radius is its square root.

Answer: B (Center \((2, -1)\), radius 4)

15.In a right triangle, the hypotenuse is 10 cm and one angle is \(30^\circ\). How long is the side opposite the \(30^\circ\) angle?

  1. A8.66 cm
  2. B20 cm
  3. C5 cm
  4. D10 cm
Show solution

Opposite and hypotenuse go with sine.

\[\begin{gathered}\sin 30^\circ = \frac{\text{opposite}}{10} \\ \Rightarrow\; \text{opposite} = 10 \times \tfrac{1}{2} = 5\end{gathered}\]

Common trap: Using cosine instead of sine, which gives \(10 \cos 30^\circ \approx 8.66\), the side next to the angle. Remember SOH-CAH-TOA: sine is opposite over hypotenuse.

Answer: C (5 cm)

Answer key

  1. 1 D
  2. 2 C
  3. 3 A
  4. 4 B
  5. 5 C
  6. 6 B
  7. 7 A
  8. 8 D
  9. 9 C
  10. 10 A
  11. 11 A
  12. 12 D
  13. 13 C
  14. 14 B
  15. 15 C

Where to go from here

Teacher's note: AAF has a family of questions where the algebra gives you an answer that isn't really an answer. Question 8 is the clearest example, and the same risk comes with every radical or rational equation. My rule for students is that the last line of every solution is a check in the original equation, not in one of your later steps. On this test, that check is where the point is won.

Tags: AAF ACCUPLACER Advanced algebra Functions Logarithms
Professor Chacha
Professor Chacha Math teacher and educational psychologist

Math teacher and educational psychologist with more than 20 years of classroom experience. He writes every practice question on this site from scratch and solves it step by step, the way he explains it to his own students.

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