AP Precalculus: Rational Functions: Asymptotes & Holes (Drill 1)

Question 1. Which of the following rational functions has a vertical asymptote at x = −3 and a hole at x = 4?

• A) \( f(x) = \dfrac{x - 4}{(x + 3)(x - 4)} \) ✓
• B) \( f(x) = \dfrac{x + 3}{(x - 4)(x + 3)} \)
• C) \( f(x) = \dfrac{x - 4}{x + 3} \)
• D) \( f(x) = \dfrac{1}{(x + 3)(x - 4)} \)

Explanation: Choice A is correct. The factor (x − 4) appears in both the numerator and denominator, so it cancels, creating a hole at x = 4. The factor (x + 3) appears only in the denominator and does not cancel, creating a vertical asymptote at x = −3. The simplified form is f(x) = 1/(x + 3), valid for x ≠ 4. Choice B is incorrect because in that function, (x + 3) is the canceling factor, producing a hole at x = −3 and a zero at x = 4, the opposite of what is required. Choice C is incorrect because (x − 4) appears only in the numerator, making x = 4 a zero of the function (where f = 0), not a hole (where f is undefined due to cancellation). Choice D is incorrect because neither factor cancels; both (x + 3) and (x − 4) remain in the denominator, producing vertical asymptotes at both x = −3 and x = 4 with no holes.

Question 2. Let \( h(x) = \dfrac{x^2 - 9}{x^2 + 2x - 15} \). Which of the following correctly identifies all holes and vertical asymptotes of h?

• A) Vertical asymptote at x = 3 and x = −5; no holes
• B) Vertical asymptote at x = −5 and x = 5; no holes
• C) Vertical asymptote at x = −5; hole at x = 3 ✓
• D) Hole at x = −5; vertical asymptote at x = 3

Explanation: Choice C is correct. Factor the numerator as (x − 3)(x + 3) and the denominator as (x − 3)(x + 5). The common factor (x − 3) cancels, producing a hole at x = 3. The remaining denominator factor (x + 5) does not cancel, giving a vertical asymptote at x = −5. The simplified form is h(x) = (x + 3)/(x + 5), valid for x ≠ 3. Choice A is incorrect because it treats x = 3 as a vertical asymptote; since (x − 3) cancels, x = 3 is a hole, not an asymptote. Choice B is incorrect because it factors the denominator incorrectly, x² + 2x − 15 = (x − 3)(x + 5), not (x − 5)(x + 3). Choice D is incorrect because it reverses which factor cancels; (x + 5) does not appear in the numerator, so it cannot cancel to produce a hole.

Question 3. The table below shows values of a rational function r(x) for inputs near x = 2 and x = 5.

x	1.9	1.99	2.01	2.1	4.9	4.99	5.01	5.1
r(x)	0.31	0.330	0.336	0.36	−98	−998	1002	102

Which statement is best supported by the table?

• A) r has a vertical asymptote at x = 2 and a hole at x = 5
• B) r has vertical asymptotes at both x = 2 and x = 5
• C) r has a zero at x = 2 and a vertical asymptote at x = 5
• D) r has a hole at x = 2 and a vertical asymptote at x = 5 ✓

Explanation: Choice D is correct. Near x = 2, the values of r(x) approach a finite value (approximately 1/3) from both sides, which is characteristic of a hole, a removable discontinuity where the function is undefined but the limit exists. Near x = 5, r(x) decreases without bound from the left (−998 at x = 4.99) and increases without bound from the right (+1002 at x = 5.01), which is characteristic of a vertical asymptote. Choice A is incorrect because it reverses the two behaviors: the blow-up occurs at x = 5, not x = 2. Choice B is incorrect because it misidentifies the finite approach near x = 2 as asymptotic behavior, if there were a vertical asymptote at x = 2, those values would also grow without bound. Choice C is incorrect because r(x) approaching a finite nonzero value (≈ 1/3) near x = 2 does not indicate a zero; a zero would require r(x) → 0.

Question 4. Let \( p(x) = \dfrac{x^3 - 4x}{x^2 - x - 2} \). Which of the following correctly identifies the x-intercepts, holes, and vertical asymptotes of p?

• A) x-intercepts at x = 0, 2, and −2; no holes; vertical asymptote at x = −1
• B) x-intercepts at x = 0 and −2; vertical asymptotes at x = 2 and x = −1; no holes
• C) x-intercepts at x = 0 and −2; hole at x = 2; vertical asymptote at x = −1 ✓
• D) x-intercepts at x = 0, 2, and −2; hole at x = 2; vertical asymptote at x = −1

Explanation: Choice C is correct. Factor the numerator as x(x² − 4) = x(x − 2)(x + 2) and the denominator as (x − 2)(x + 1). The common factor (x − 2) cancels, creating a hole at x = 2. The simplified function is x(x + 2)/(x + 1), which has zeros at x = 0 and x = −2, and a vertical asymptote at x = −1. Choice A is incorrect because x = 2 is a hole, the function is undefined there due to cancellation, so it cannot be an x-intercept. Choice B is incorrect because it treats x = 2 as a vertical asymptote; since (x − 2) cancels, x = 2 is a removable discontinuity, not an asymptote. Choice D commits both errors simultaneously: it lists x = 2 as both an x-intercept and a hole, which is a contradiction, a hole means the function is undefined at that point, so it cannot equal zero there.

Question 5. A student analyzes \( r(x) = \dfrac{x^2 - 5x + 6}{x^2 - 4} \) and claims: "Since the denominator equals zero at x = 2, the function has a vertical asymptote at x = 2." Which of the following best evaluates this claim?

• A) The claim is correct because any value that makes the denominator zero produces a vertical asymptote.
• B) The claim is incorrect because x = 2 is not in the domain of r, so the behavior there is irrelevant.
• C) The claim is correct because the degree of the numerator equals the degree of the denominator.
• D) The claim is incorrect because (x − 2) is also a factor of the numerator, so it cancels, producing a hole at x = 2 rather than a vertical asymptote. ✓

Explanation: Choice D is correct. A vertical asymptote occurs at a value where the denominator is zero AND the numerator is nonzero after full simplification. Here, the numerator x² − 5x + 6 = (x − 2)(x − 3) and the denominator x² − 4 = (x − 2)(x + 2). The factor (x − 2) appears in both and cancels, leaving a removable discontinuity (hole) at x = 2, not a vertical asymptote. The actual vertical asymptote is at x = −2, where (x + 2) remains after simplification. Choice A is incorrect because it overstates the rule: a zero of the denominator produces a vertical asymptote only when the numerator is nonzero at that point; cancellation changes the classification. Choice B is incorrect because being outside the domain does not make behavior irrelevant, distinguishing holes from asymptotes is precisely the point of this analysis. Choice C is incorrect because equal degrees of numerator and denominator determine the horizontal asymptote (here y = 1), not whether a discontinuity is a hole or a vertical asymptote.