AP Precalculus Rational Functions — Questions and Answers
Question 1: Which line is the horizontal asymptote of f(x) = (3x² − 5x + 1)/(x² + 4)?
- y = 0
- y = 3 (Correct answer)
- y = −5
- y = 1/4
Correct answer: y = 3
The numerator and denominator both have degree 2, so the horizontal asymptote is the ratio of leading coefficients, 3/1 = 3.
End behavior of a rational function depends on the degrees of its numerator and denominator. If the numerator degree is smaller, the function approaches y = 0. If the degrees are equal, it approaches the ratio of the leading coefficients. If the numerator degree is larger, there is no horizontal asymptote. Here both degrees are 2, so f(x) approaches 3/1 = 3 as x → ±∞. The lower-degree terms −5x, 1 and 4 become negligible for large |x|.
Question 2: Consider f(x) = (x − 2)/((x − 2)(x + 3)). Which statement correctly describes the graph of f?
- Vertical asymptotes at x = 2 and x = −3
- A vertical asymptote at x = −3 only, with no hole
- A vertical asymptote at x = −3 and a hole at x = 2 (Correct answer)
- A vertical asymptote at x = 2 and a hole at x = −3
Correct answer: A vertical asymptote at x = −3 and a hole at x = 2
The factor (x − 2) cancels, which leaves a hole at x = 2. The factor (x + 3) remains in the denominator, which gives a vertical asymptote at x = −3.
A zero of the denominator produces a hole if the matching factor also appears in the numerator and cancels. It produces a vertical asymptote if the factor stays in the denominator after simplifying. Here (x − 2) appears in both, so f(x) = 1/(x + 3) for x ≠ 2. The graph of f is therefore the graph of 1/(x + 3) with a single point missing at x = 2. The vertical asymptote is x = −3.
Question 3: What are the zeros of f(x) = (x² − 4)/(x² − x − 6)?
- x = −2 and x = 2
- x = −2 only
- x = 3 only
- x = 2 only (Correct answer)
Correct answer: x = 2 only
Factoring gives f(x) = (x − 2)(x + 2)/((x − 3)(x + 2)). The factor (x + 2) cancels, so x = −2 is a hole, not a zero. The only zero is x = 2.
A zero of a rational function is an input where the simplified function equals 0. Factoring gives (x − 2)(x + 2) on top and (x − 3)(x + 2) on the bottom. The shared factor (x + 2) cancels, so x = −2 is a hole in the graph and the function has no value there. At x = 2 the numerator is 0 and the denominator is (2)² − 2 − 6 = −4, which is nonzero, so f(2) = 0. The vertical asymptote is at x = 3.
Question 4: As x → ∞ and as x → −∞, the graph of f(x) = (x² + 1)/(x − 1) approaches which line?
- y = x + 1 (Correct answer)
- y = x − 1
- y = x
- y = 1
Correct answer: y = x + 1
Polynomial division gives x² + 1 = (x − 1)(x + 1) + 2, so f(x) = x + 1 + 2/(x − 1). The remainder term 2/(x − 1) approaches 0, which leaves the slant line y = x + 1.
When the numerator's degree is exactly one more than the denominator's, the function has a slant (oblique) asymptote instead of a horizontal one. Dividing x² + 1 by x − 1 gives quotient x + 1 and remainder 2, so f(x) = x + 1 + 2/(x − 1). As |x| grows, the fraction 2/(x − 1) approaches 0. The graph therefore approaches y = x + 1 in both directions.
Question 5: Which expression is equivalent to (2x² + 3x − 5)/(x − 1) for all x ≠ 1?
- 2x − 5
- 2x + 3
- 2x + 5 (Correct answer)
- 2x² + 3x − 4
Correct answer: 2x + 5
The numerator factors as (2x + 5)(x − 1), so the factor (x − 1) cancels and leaves 2x + 5 for x ≠ 1.
To rewrite a rational expression in simpler form, factor the numerator and denominator and cancel common factors. The quadratic 2x² + 3x − 5 factors as (2x + 5)(x − 1). Expanding gives 2x² − 2x + 5x − 5, which equals the original. Cancelling (x − 1) leaves 2x + 5. The restriction x ≠ 1 must stay, because the original expression is undefined at x = 1.
Question 6: For f(x) = (x + 1)/(x − 4), what happens to f(x) as x approaches 4 from the left (x → 4⁻)?
- f(x) increases without bound (→ +∞)
- f(x) decreases without bound (→ −∞) (Correct answer)
- f(x) approaches 0
- f(x) approaches 5
Correct answer: f(x) decreases without bound (→ −∞)
Near x = 4 from the left, the numerator is close to 5 (positive) and the denominator is a small negative number. The quotient is a large negative number, so f(x) → −∞.
A vertical asymptote at x = 4 means the outputs grow without bound in magnitude near 4. The direction depends on the signs. For x slightly less than 4, such as 3.99, x + 1 ≈ 4.99 is positive and x − 4 = −0.01 is negative. A positive number divided by a tiny negative number is a large negative number. So f(x) → −∞ as x → 4⁻. From the right, the denominator is positive and f(x) → +∞.
Question 7: What is the end behavior of f(x) = (x³ − 1)/(x² + 2)?
- As x → ∞, f(x) → ∞, and as x → −∞, f(x) → −∞ (Correct answer)
- As x → ∞ and as x → −∞, f(x) → ∞
- As x → ±∞, f(x) → 0
- As x → ±∞, f(x) → 1
Correct answer: As x → ∞, f(x) → ∞, and as x → −∞, f(x) → −∞
The numerator has a higher degree than the denominator. For large |x|, f(x) behaves like x³/x² = x, which goes to ∞ as x → ∞ and to −∞ as x → −∞.
For large |x|, a rational function behaves like the ratio of the leading terms of the numerator and denominator. Here that ratio is x³/x² = x. Since the numerator's degree exceeds the denominator's by 1, the graph follows a slant line, and the outputs grow without bound. The sign of x decides the direction. The outputs are positive and large for large positive x, and negative and large for large negative x.
Question 8: A rational function has zeros at x = 1 and x = −3, a vertical asymptote at x = 2, and a horizontal asymptote at y = 4. Which function fits all of these?
- f(x) = (x − 1)(x + 3)/(4(x − 2)²)
- f(x) = 4(x − 2)/((x − 1)(x + 3))
- f(x) = 4(x − 1)(x + 3)/(x − 2)
- f(x) = 4(x − 1)(x + 3)/(x − 2)² (Correct answer)
Correct answer: f(x) = 4(x − 1)(x + 3)/(x − 2)²
The numerator's zeros are 1 and −3, and the denominator's zero is 2. Both have degree 2, and the ratio of leading coefficients is 4/1 = 4, which gives the horizontal asymptote y = 4.
Zeros come from numerator factors, and vertical asymptotes come from denominator factors. So the numerator needs (x − 1)(x + 3) and the denominator needs a factor of (x − 2). For a nonzero horizontal asymptote, the degrees must be equal, so the denominator is (x − 2)². The leading-coefficient ratio then sets the asymptote at 4. The option with denominator (x − 2) has a higher numerator degree and no horizontal asymptote. One option has asymptote 1/4, and another has asymptote 0.
Question 9: Where is the hole in the graph of f(x) = (x² + x − 6)/(x² − 4)?
- (2, 5)
- (−2, 5/4)
- (2, 5/4) (Correct answer)
- (2, 0)
Correct answer: (2, 5/4)
Factoring gives f(x) = (x + 3)(x − 2)/((x − 2)(x + 2)). The factor (x − 2) cancels, so the hole is at x = 2. The simplified function (x + 3)/(x + 2) gives y = 5/4 there.
A hole occurs where a common factor cancels. Factoring gives (x + 3)(x − 2) on top and (x − 2)(x + 2) on the bottom, so the common factor is (x − 2). After cancelling, f(x) = (x + 3)/(x + 2). Substituting x = 2 into this simplified form gives 5/4, so the hole is at (2, 5/4). The remaining denominator factor (x + 2) gives a vertical asymptote at x = −2. The zero is x = −3.
Question 10: Which expression is equivalent to (x² + 5x + 7)/(x + 2) for x ≠ −2 and shows its slant asymptote?
- x + 3 + 7/(x + 2)
- x + 3 + 1/(x + 2) (Correct answer)
- x + 5 + 1/(x + 2)
- x + 2 + 3/(x + 2)
Correct answer: x + 3 + 1/(x + 2)
Dividing gives quotient x + 3 and remainder 1, because (x + 3)(x + 2) = x² + 5x + 6. So the expression equals x + 3 + 1/(x + 2).
Dividing x² + 5x + 7 by x + 2: the first term of the quotient is x, and subtracting x(x + 2) leaves 3x + 7. The next term is 3, and subtracting 3(x + 2) leaves a remainder of 1. So the quotient is x + 3 with remainder 1, and the expression equals x + 3 + 1/(x + 2). As |x| gets large, 1/(x + 2) approaches 0, so the graph approaches the slant asymptote y = x + 3.
