AP Precalculus: Rational Functions: End Behavior & Zeros (Drill 6)

Question 1. What is the horizontal asymptote of \( f(x) = \dfrac{4x^3 - 2x + 1}{2x^3 + 5x^2 - 3} \)?

• A) y = 0
• B) y = 4
• C) y = 2 ✓
• D) There is no horizontal asymptote.

Explanation: Choice C is correct. When the degree of the numerator equals the degree of the denominator (both degree 3 here), the horizontal asymptote is the ratio of the leading coefficients: 4/2 = 2. So y = 2. Choice A is incorrect because y = 0 is the horizontal asymptote only when the degree of the numerator is strictly less than the degree of the denominator. Here the degrees are equal. Choice B is incorrect because a student who reads only the numerator's leading coefficient (4) without dividing by the denominator's leading coefficient (2) would incorrectly get y = 4. Choice D is incorrect because there is no horizontal asymptote when the degree of the numerator is greater than the degree of the denominator. If the numerator's degree is exactly one more than the denominator's, the function has a slant (oblique) asymptote; if the degree difference is larger, the end behavior is governed by a higher-degree polynomial expression, neither case produces a horizontal asymptote.

Question 2. Which of the following gives all real zeros of \( g(x) = \dfrac{(x-3)(x+1)(x-5)}{(x+4)(x-3)} \)?

• A) x = 3, x = −1, and x = 5
• B) x = −4 and x = 3
• C) x = −1 and x = 5 ✓
• D) x = −4, x = −1, and x = 5

Explanation: Choice C is correct. The zeros of a rational function come from values of x that make the numerator zero, but only when those values do not simultaneously make the denominator zero. Here, (x − 3) appears in both numerator and denominator, so x = 3 produces a hole (the function is undefined there), not a zero. The remaining numerator factors (x + 1) and (x − 5) give potential zeros at x = −1 and x = 5. Neither of these makes the denominator zero (denominator factors are x + 4 and x − 3), so both are genuine zeros. The complete list of real zeros is x = −1 and x = 5. Choice A is incorrect because it includes x = 3, which is a hole, the function is undefined at x = 3, so it cannot be a zero. Choice B is incorrect because x = −4 makes the denominator zero (vertical asymptote) and x = 3 is a hole; neither is a zero of the function. Choice D is incorrect because x = −4 is a vertical asymptote, not a zero, including it confuses denominator roots with numerator roots.

Question 3. The concentration C (in milligrams per liter) of a medication in the bloodstream t hours after administration is modeled by \( C(t) = \dfrac{8t}{t^2 + 4} \) for t ≥ 0. What does the end behavior of this model indicate about the concentration as t → ∞?

• A) The concentration approaches 0 mg/L, meaning the medication is eventually eliminated from the bloodstream. ✓
• B) The concentration approaches 8 mg/L, the initial dose.
• C) The concentration increases without bound, indicating the medication accumulates indefinitely according to the model.
• D) The concentration approaches 2 mg/L, a long-term stable level.

Explanation: Choice A is correct. The numerator has degree 1 and the denominator has degree 2. When the degree of the denominator is strictly greater than the degree of the numerator, the horizontal asymptote is y = 0, meaning \( \lim_{t \to \infty} C(t) = 0 \). In context, this indicates that the medication concentration decreases toward zero over time, the body eventually eliminates the drug. Choice B is incorrect because 8 is the numerator's leading coefficient, not the asymptote, the asymptote rule requires dividing leading coefficients only when the degrees are equal, which is not the case here. Choice C is incorrect because the concentration does not grow without bound; the denominator t² + 4 grows faster than the numerator 8t, so C(t) → 0. Choice D is incorrect because y = 2 would require equal degrees with leading coefficient ratio 8/4 = 2, but the degrees differ (numerator degree 1, denominator degree 2), so the horizontal asymptote is y = 0, not y = 2.

Question 4. Let \( r(x) = \dfrac{x^2 - 4x - 5}{x^2 - x - 20} \). On which of the following intervals is r(x) < 0?

• A) (−∞, −4)
• B) (−4, −1) ✓
• C) (−1, 5)
• D) (5, ∞)

Explanation: Choice B is correct. Factor both numerator and denominator: numerator x² − 4x − 5 = (x − 5)(x + 1); denominator x² − x − 20 = (x − 5)(x + 4). The factor (x − 5) appears in both, so x = 5 is a hole, the function is undefined there. Cancel (x − 5) to get the simplified form r(x) = (x + 1)/(x + 4) for x ≠ 5. The zero of r(x) is x = −1 (numerator = 0) and the vertical asymptote is x = −4 (denominator = 0). Now perform sign analysis on (x + 1)/(x + 4) using critical values x = −4 and x = −1: on (−∞, −4), test x = −5: (−4)/(−1) = 4 > 0; on (−4, −1), test x = −2: (−1)/(2) = −1/2 0. Therefore r(x) < 0 exactly on (−4, −1). Choice A is incorrect because on (−∞, −4) both (x + 1) and (x + 4) are negative, giving a positive quotient. Choice C is incorrect because on (−1, 5) both factors are positive, giving a positive quotient. The hole at x = 5 does not create a sign change, approaching x = 5 from either side, the simplified expression gives the same positive sign. Choice D is incorrect because on (5, ∞) the simplified expression also gives a positive value.

Question 5. A student claims: "For any rational function \( r(x) = \dfrac{p(x)}{q(x)} \), if the degree of p(x) is greater than the degree of q(x), then r(x) has no horizontal asymptote and therefore \( \lim_{x \to \infty} r(x) \) does not exist." Which of the following best evaluates this claim?

• A) The claim is partially correct but imprecise: r(x) has no finite horizontal asymptote, but \( \lim_{x \to \infty} r(x) \) does exist in the sense that r(x) → +∞ or r(x) → −∞ depending on the sign of the leading coefficients. ✓
• B) The claim is fully correct: no horizontal asymptote means the limit does not exist.
• C) The claim is incorrect: any rational function where degree of p(x) > degree of q(x) must have a horizontal asymptote at y = 1.
• D) The claim is incorrect: if the degree of p(x) is exactly one more than the degree of q(x), the function has a horizontal asymptote determined by polynomial long division.

Explanation: Choice A is correct. The student is right that there is no horizontal asymptote when the degree of the numerator exceeds the degree of the denominator, but wrong to conclude the behavior is indeterminate. The end behavior can still be determined: as \( x \to \infty \), \( r(x) \to +\infty \) or \( r(x) \to -\infty \) depending on the sign of the leading coefficients, the function grows or decays without bound, which is a specific and meaningful conclusion about its end behavior. For example, r(x) = x²/x = x → +∞ as x → ∞. "Does not exist" is typically reserved for oscillating or otherwise indeterminate behavior, not for infinite growth. Choice B is incorrect because "no horizontal asymptote" does not mean the behavior is undefined; it means the output is unbounded, not finite, which is a determinate conclusion. Choice C is incorrect because there is no rule that gives y = 1 in this case; the degree comparison tells us the function increases or decreases without bound, rather than approaching a constant value. Choice D is incorrect because when the degree of p exceeds the degree of q by exactly 1, there is a slant (oblique) asymptote, not a horizontal one, and a slant asymptote is a line y = mx + b with m ≠ 0. If the degree difference is 2 or more, the end behavior follows a higher-degree polynomial expression, not a slant line.