AP Precalculus Quiz: Rational Functions And Vertical Asymptotes

What this quiz covers

This quiz focuses on Rational Functions And Vertical Asymptotes, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Precalculus.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. $f(x) = \frac{x^2 - 9}{x - 3}$
2. $f(x) = \frac{x + 2}{x - 3}$ (correct answer)
3. $f(x) = \frac{x^2 - 6x + 9}{x^2 - 4}$
4. $f(x) = \frac{x - 3}{x + 1}$

Explanation: A vertical asymptote occurs where the denominator equals zero but the numerator does not. For choice B, when $x = 3$, the denominator $x - 3 = 0$ and the numerator $x + 2 = 5 \neq 0$, creating a vertical asymptote. Choice A has a hole at $x = 3$ since both numerator and denominator equal zero. Choice C has vertical asymptotes at $x = \pm 2$, not $x = 3$. Choice D has a vertical asymptote at $x = -1$, not $x = 3$.

Question 2

1. $x = -1$ and $x = 4$ only (correct answer)
2. $x = \frac{5}{2}$ and $x = 4$ only
3. $x = -1$ and $x = \frac{5}{2}$ only
4. $x = 1$ and $x = -4$ only

Explanation: Vertical asymptotes occur where the denominator equals zero but the numerator does not. The denominator $(x + 1)(x - 4) = 0$ when $x = -1$ or $x = 4$. At $x = -1$: numerator $= 2(-1) - 5 = -7 \neq 0$. At $x = 4$: numerator $= 2(4) - 5 = 3 \neq 0$. Both create vertical asymptotes. Choice B incorrectly includes $x = \frac{5}{2}$ (where numerator equals zero). Choice C incorrectly includes $x = \frac{5}{2}$. Choice D uses incorrect signs.

Question 3

1. $k$ can be any real number except $k = -2$ and $k = -3$ (correct answer)
2. $k$ can be any real number except $k = 2$ and $k = 3$
3. $k$ must equal $2$ or $3$ for two vertical asymptotes
4. $k$ can be any real number except $k = -6$ and $k = 1$

Explanation: The denominator factors as $x^2 - 5x + 6 = (x - 2)(x - 3)$, giving potential vertical asymptotes at $x = 2$ and $x = 3$. For exactly two vertical asymptotes, the numerator must not equal zero at either $x = 2$ or $x = 3$. At $x = 2$: numerator $= 2 + k$, so $k \neq -2$. At $x = 3$: numerator $= 3 + k$, so $k \neq -3$. Therefore $k$ can be any real number except $-2$ and $-3$. Choice B uses wrong signs. Choice C would create holes, not asymptotes. Choice D uses incorrect values.

Question 4

1. $\lim_{x \to 3^-} h(x) = +\infty$ and $\lim_{x \to -3^-} h(x) = -\infty$ (correct answer)
2. $\lim_{x \to 3^-} h(x) = +\infty$ and $\lim_{x \to -3^+} h(x) = +\infty$
3. $\lim_{x \to 3^+} h(x) = -\infty$ and $\lim_{x \to -3^-} h(x) = +\infty$
4. $\lim_{x \to 3^-} h(x) = -\infty$ and $\lim_{x \to -3^+} h(x) = -\infty$

Explanation: The denominator $x^2 - 9 = (x - 3)(x + 3)$ gives vertical asymptotes at $x = 3$ and $x = -3$. As $x \to 3^-$: numerator approaches $9 > 0$ and denominator $(x-3)(x+3)$ approaches $0^- \cdot 6 = 0^-$, so the limit is $+\infty$. As $x \to -3^-$: numerator approaches $-3 < 0$ and denominator $(-6)(x+3)$ approaches $(-6)(0^-) = 0^+$, so the limit is $-\infty$. Choices B, C, and D have incorrect sign combinations for the limit behavior.

Question 5

1. There is a vertical asymptote of multiplicity 2 at $x = -3$ and multiplicity 1 at $x = 7$ (correct answer)
2. There is a vertical asymptote of multiplicity 1 at $x = -3$ and multiplicity 2 at $x = 7$
3. There are vertical asymptotes of multiplicity 1 at both $x = -3$ and $x = 7$
4. There is a vertical asymptote of multiplicity 3 at $x = -3$ only

Explanation: The denominator $(x + 3)^2(x - 7)$ has zeros at $x = -3$ (multiplicity 2) and $x = 7$ (multiplicity 1). Since the numerator $5x - 1$ doesn't equal zero at either point ($5(-3) - 1 = -16 \neq 0$ and $5(7) - 1 = 34 \neq 0$), both zeros create vertical asymptotes with their respective multiplicities. The asymptote at $x = -3$ has multiplicity 2, and at $x = 7$ has multiplicity 1. Choices B and C have incorrect multiplicities. Choice D ignores the asymptote at $x = 7$.

Question 6

1. $f(x) = \frac{x^2 + 1}{(x - 4)(x + 1)}$
2. $f(x) = \frac{x + 1}{x - 4}$ (correct answer)
3. $f(x) = \frac{x - 4}{x^2 - 16}$
4. $f(x) = \frac{2x + 3}{x^2 - 8x + 16}$

Explanation: For exactly one vertical asymptote at $x = 4$, the denominator must have $x = 4$ as its only zero where the numerator is non-zero. Choice B has denominator $x - 4$ with only one zero at $x = 4$, and numerator $4 + 1 = 5 \neq 0$ at this point. Choice A has two asymptotes at $x = 4$ and $x = -1$. Choice C has $x^2 - 16 = (x-4)(x+4)$, giving asymptotes at both $x = 4$ and $x = -4$. Choice D has $x^2 - 8x + 16 = (x-4)^2$, giving one asymptote at $x = 4$ with multiplicity 2.

Question 7

1. $f(x) = \frac{x + 2}{(x - 1)(x + 5)}$
2. $f(x) = \frac{2x - 3}{x^2 + 4x - 5}$ (correct answer)
3. $f(x) = \frac{x^2 + 1}{(x + 1)(x - 5)}$
4. $f(x) = \frac{3x + 7}{x^2 - 6x + 5}$

Explanation: For vertical asymptotes at $x = 1$ and $x = -5$, the denominator must have these as zeros where the numerator is non-zero. Choice B has denominator $x^2 + 4x - 5 = (x - 1)(x + 5)$, giving zeros at $x = 1$ and $x = -5$. Check numerator: at $x = 1$, $2(1) - 3 = -1 \neq 0$; at $x = -5$, $2(-5) - 3 = -13 \neq 0$. Both create asymptotes. Choice A has wrong signs. Choice C has asymptotes at $x = -1$ and $x = 5$. Choice D has $x^2 - 6x + 5 = (x-1)(x-5)$, giving asymptotes at $x = 1$ and $x = 5$.

Question 8

1. $x = 0$, $x = 3$, and $x = -1$ are all vertical asymptotes (correct answer)
2. $x = 3$ and $x = -1$ are vertical asymptotes, but $x = 0$ is not
3. $x = 0$ and $x = 3$ are vertical asymptotes, but $x = -1$ is not
4. $x = 0$ is the only vertical asymptote of the function

Explanation: Factor the denominator: $x^3 - 2x^2 - 3x = x(x^2 - 2x - 3) = x(x - 3)(x + 1)$. The denominator has zeros at $x = 0, 3, -1$. Check if the numerator $2x + 1$ equals zero at any of these points: At $x = 0$: $2(0) + 1 = 1 \neq 0$. At $x = 3$: $2(3) + 1 = 7 \neq 0$. At $x = -1$: $2(-1) + 1 = -1 \neq 0$. Since the numerator is non-zero at all three zeros of the denominator, all three create vertical asymptotes. Choices B, C, and D incorrectly exclude some asymptotes.

Question 9

1. The function has exactly two vertical asymptotes
2. The function has exactly one vertical asymptote (correct answer)
3. The function has no vertical asymptotes
4. The function has exactly three vertical asymptotes

Explanation: Factor both parts: numerator $x^2 + 5x + 6 = (x + 2)(x + 3)$ and denominator $x^2 + 3x + 2 = (x + 1)(x + 2)$. So $h(x) = \frac{(x + 2)(x + 3)}{(x + 1)(x + 2)} = \frac{x + 3}{x + 1}$ for $x \neq -2$. After canceling the common factor $(x + 2)$, there's a hole at $x = -2$ and one vertical asymptote at $x = -1$ where the simplified denominator equals zero but the numerator $(-1) + 3 = 2 \neq 0$. Choice A counts the hole as an asymptote. Choices C and D are incorrect counts.

Question 10

1. $f(x) = \frac{x - 3}{(x + 2)(x - 3)}$
2. $f(x) = \frac{x + 2}{x - 3}$
3. $f(x) = \frac{2x + 1}{x + 2}$ (correct answer)
4. $f(x) = \frac{x^2 - 9}{x^2 - x - 6}$

Explanation: For a vertical asymptote at $x = -2$, the denominator must equal zero at $x = -2$ while the numerator does not. Choice C has denominator $x + 2$ which equals zero when $x = -2$, and numerator $2(-2) + 1 = -3 \neq 0$, creating a vertical asymptote. Choice A has a hole at $x = 3$ and asymptote at $x = -2$. Choice B has asymptote at $x = 3$, not $x = -2$. Choice D factors to $\frac{(x-3)(x+3)}{(x-3)(x+2)}$ with asymptote at $x = -2$ after canceling.

Question 11

1. The function has vertical asymptotes at $x = 1$ and $x = -2$
2. The function has a vertical asymptote at $x = -2$ only (correct answer)
3. The function has vertical asymptotes at $x = -1$ and $x = 2$
4. The function has a vertical asymptote at $x = 1$ only

Explanation: Factor both parts: numerator $x^2 - 1 = (x - 1)(x + 1)$ and denominator $x^2 + x - 2 = (x + 2)(x - 1)$. So $f(x) = \frac{(x - 1)(x + 1)}{(x + 2)(x - 1)} = \frac{x + 1}{x + 2}$ for $x \neq 1$. After canceling the common factor $(x - 1)$, there's a hole at $x = 1$ and a vertical asymptote at $x = -2$ where the denominator equals zero but numerator does not. Choice A incorrectly includes the hole. Choices C and D use wrong values.

Question 12

1. $\lim_{x \to 2^-} g(x) = +\infty$ and $\lim_{x \to 2^+} g(x) = -\infty$ (correct answer)
2. $\lim_{x \to 2^-} g(x) = -\infty$ and $\lim_{x \to 2^+} g(x) = +\infty$
3. $\lim_{x \to 2^-} g(x) = +\infty$ and $\lim_{x \to 2^+} g(x) = +\infty$
4. $\lim_{x \to 2^-} g(x) = -\infty$ and $\lim_{x \to 2^+} g(x) = -\infty$

Explanation: As $x$ approaches 2, the numerator approaches $2 - 5 = -3 < 0$. For the denominator: as $x \to 2^-$, we have $x - 2 \to 0^-$ (negative), so $\frac{\text{negative}}{\text{negative}} = \text{positive}$, giving $+\infty$. As $x \to 2^+$, we have $x - 2 \to 0^+$ (positive), so $\frac{\text{negative}}{\text{positive}} = \text{negative}$, giving $-\infty$. Choice B has the signs reversed. Choices C and D have matching signs, which is incorrect for this function.

Question 13

1. The numerator must have factors $(x - 1)$ and $(x + 1)$
2. We must have $a = 0$, $b = 0$, and $c = 0$
3. The coefficients must satisfy $a + b + c = 0$ and $a - b + c = 0$ (correct answer)
4. The numerator must be a constant multiple of the denominator

Explanation: The denominator $x^2 - 1 = (x - 1)(x + 1)$ has zeros at $x = 1$ and $x = -1$. For no vertical asymptotes, the numerator must also equal zero at these points. At $x = 1$: $a(1)^2 + b(1) + c = 0 \Rightarrow a + b + c = 0$. At $x = -1$: $a(-1)^2 + b(-1) + c = 0 \Rightarrow a - b + c = 0$. Choice A is correct conceptually but not as precise. Choice B is too restrictive. Choice D would make the function constant, not eliminate asymptotes.

Question 14

1. Either $a = 0$ and $b = 0$, or the numerator must factor with the denominator
2. The numerator must equal zero when $x = 1$ or $x = 3$ (correct answer)
3. We must have $a = 1, b = -1$ or $a = 1, b = -3$
4. The constants must satisfy $a + b = 4$ and $ab = 3$

Explanation: The denominator $x^2 - 4x + 3 = (x - 1)(x - 3)$ has zeros at $x = 1$ and $x = 3$. For no vertical asymptotes, the numerator must also equal zero at these points (creating holes instead). At $x = 1$: $a(1) + b = 0$, so $a + b = 0$. At $x = 3$: $a(3) + b = 0$, so $3a + b = 0$. This means the numerator zeros at $x = 1$ or $x = 3$. Choice A is too restrictive. Choice C gives specific values but isn't complete. Choice D uses incorrect relationships.

Question 15

1. Vertical asymptotes at $x = 2$ and $x = -2$ only
2. A vertical asymptote at $x = -2$ only (correct answer)
3. A vertical asymptote at $x = 2$ only
4. No vertical asymptotes after complete simplification

Explanation: Factor: numerator $x^3 - 8 = (x - 2)(x^2 + 2x + 4)$ and denominator $x^2 - 4 = (x - 2)(x + 2)$. So $g(x) = \frac{(x - 2)(x^2 + 2x + 4)}{(x - 2)(x + 2)} = \frac{x^2 + 2x + 4}{x + 2}$ for $x \neq 2$. After canceling $(x - 2)$, there's a hole at $x = 2$ and a vertical asymptote at $x = -2$ where the simplified denominator equals zero. Choice A includes the hole. Choice C omits the actual asymptote. Choice D incorrectly suggests no asymptotes.

Question 16

1. $x=-3$ only
2. $x=3$ only (correct answer)
3. $x=-3$ and $x=3$
4. $x=-1$ and $x=3$

Explanation: This question tests understanding of rational functions and vertical asymptotes in AP Precalculus. Vertical asymptotes occur where the denominator of a rational function equals zero and the function is undefined, signaling a potential infinite discontinuity. For p(x) = (x² + 2x - 3)/(x² - 9), we factor both parts: numerator = (x + 3)(x - 1) and denominator = (x + 3)(x - 3). Choice B is correct because after canceling the common factor (x + 3), we get (x - 1)/(x - 3), which has a vertical asymptote only at x = 3. Choice C incorrectly identifies both x = -3 and x = 3 as asymptotes, missing that x = -3 creates a removable discontinuity due to the cancellation. To help students: Factor completely before identifying asymptotes, cancel common factors first, and remember that canceled factors create holes, not vertical asymptotes.

Question 17

1. $t(x)\to 0$
2. $t(x)\to 2$
3. $t(x)\to \pm\infty$ (correct answer)
4. $t(x)\to -\dfrac{7}{4}$

Explanation: This question tests understanding of rational functions and vertical asymptotes in AP Precalculus. Vertical asymptotes occur where the denominator of a rational function equals zero and the function is undefined, signaling a potential infinite discontinuity. For t(x) = (2x+1)/(x+4), setting the denominator equal to zero gives x+4=0, so x=-4 is a vertical asymptote. As x approaches -4, the denominator approaches 0 while the numerator approaches 2(-4)+1 = -7, causing the function to approach ±∞ depending on the direction of approach. Choice C is correct because t(x)→±∞ as x→-4, which is the characteristic behavior at a vertical asymptote. Choice D incorrectly suggests the function approaches a finite value. To help students: At vertical asymptotes, rational functions always approach ±∞, never finite values, and the sign depends on the direction of approach.

Question 18

1. $x=2$
2. $x=3$
3. $x=-3$ (correct answer)
4. $x=\pm 2$

Explanation: This question tests understanding of rational functions and vertical asymptotes in AP Precalculus. Vertical asymptotes occur where the denominator of a rational function equals zero and the function is undefined, signaling a potential infinite discontinuity. For s(x) = (x²-5x+6)/(x²-9), we factor the numerator as (x-2)(x-3) and the denominator as (x+3)(x-3). The factor (x-3) cancels from both, leaving s(x) = (x-2)/(x+3) for x≠3, which has a vertical asymptote only at x=-3. Choice C is correct because x=-3 is where the remaining denominator equals zero after simplification. Choices A and B are incorrect because they identify zeros of the numerator or canceled factors. To help students: Always simplify by canceling common factors first, then identify vertical asymptotes from the remaining denominator factors.

Question 19

1. $x=2$
2. $x=-2$
3. $x=0$
4. No vertical asymptotes (correct answer)

Explanation: This question tests understanding of rational functions and vertical asymptotes in AP Precalculus. Vertical asymptotes occur where the denominator of a rational function equals zero and the function is undefined, signaling a potential infinite discontinuity. For the function f(x) = (x²-4)/(x-2), we must first factor the numerator as (x+2)(x-2), giving us f(x) = (x+2)(x-2)/(x-2). Since the factor (x-2) cancels from both numerator and denominator, we get f(x) = x+2 for x≠2, which means there's a removable discontinuity (hole) at x=2, not a vertical asymptote. Choice D is correct because after simplification, no factors remain in the denominator to create vertical asymptotes. Choice A is incorrect because it identifies the location of the hole, not an asymptote. To help students: Emphasize the difference between vertical asymptotes (non-canceling factors in denominator) and holes (canceling factors), and always factor completely before identifying asymptotes.

Question 20

1. $x=-2$
2. $x=1$ and $x=-3$ (correct answer)
3. $x=1$ only
4. $x=3$ and $x=-1$

Explanation: This question tests understanding of rational functions and vertical asymptotes in AP Precalculus. Vertical asymptotes occur where the denominator of a rational function equals zero and the function is undefined, signaling a potential infinite discontinuity. For h(x) = (x+2)/[(x-1)(x+3)], the denominator is already factored, so we set each factor equal to zero: x-1=0 gives x=1, and x+3=0 gives x=-3. Choice B is correct because it identifies both vertical asymptotes at x=1 and x=-3 where the denominator equals zero. Choice A is incorrect because it confuses the numerator's zero with an asymptote location. To help students: Remember that vertical asymptotes come from denominator zeros only, not numerator zeros, and when the denominator is already factored, simply set each factor to zero.