AP PreCalculus Practice Quiz: Rational Functions and Holes

Written by AP Content Team, Verified for 2026 AP Exams, Last updated: May 2026

Test your understanding with short quizzes. This quiz has 9 questions to check your progress.

Question 1 of 9

Consider the rational function f(x) = (x-5)(x+2) / ((x-5)(x-1)). At which x-value does the graph of this function have a hole?

A)x = -2B)x = 1C)x = 5D)x = -5

All Questions (9)

Consider the rational function f(x) = (x-5)(x+2) / ((x-5)(x-1)). At which x-value does the graph of this function have a hole?

A) x = -2

B) x = 1

C) x = 5

D) x = -5

Correct Answer: C

A hole exists at a real zero that appears in both the numerator and the denominator. In this function, (x-5) is a factor of both, corresponding to the real zero x = 5. The multiplicity in the numerator (1) is equal to the multiplicity in the denominator (1), satisfying the condition for a hole.

According to the provided text, a hole exists in the graph of a rational function at x = c if c is a real zero of the numerator and the denominator, and what other condition is met?

A) The multiplicity of the zero in the numerator is less than its multiplicity in the denominator.

B) The multiplicity of the zero in the numerator is greater than or equal to its multiplicity in the denominator.

C) The function is undefined at x = c.

D) The multiplicity of the zero in the numerator is exactly equal to its multiplicity in the denominator.

Correct Answer: B

The provided content explicitly states: 'If the multiplicity of a real zero in the numerator is greater than or equal to its multiplicity in the denominator, then the graph of the rational function has a hole at the corresponding input value.'

A rational function is defined by r(x) = (x^2 - 9) / (x + 3). At which x-value does this function have a hole?

A) x = 9

B) x = 3

C) x = -3

D) x = -9

Correct Answer: C

To determine the hole, we must find common real zeros. Factoring the numerator gives r(x) = ((x-3)(x+3)) / (x+3). The factor (x+3) is common to both the numerator and denominator, which corresponds to a real zero at x = -3. Since the multiplicity is 1 in both, the condition for a hole is met.

The graph of a rational function r(x) has a hole at x = c. How can the location (coordinates) of this hole be determined?

A) By finding the y-intercept of the function.

B) By setting the simplified function's denominator equal to zero.

C) By examining the output values for input values sufficiently close to c.

D) By calculating the value of the original function at x = c.

Correct Answer: C

The provided text states: 'If the graph of a rational function r has a hole at x = c, then the location of the hole can be determined by examining the output values corresponding to input values sufficiently close to c.' This is equivalent to evaluating the simplified function at x = c.

For which of the following functions does a hole exist at x = 2, based on the provided rules?

A) f(x) = (x - 2) / (x - 2)^3

B) f(x) = (x + 2) / (x - 2)

C) f(x) = (x - 2)^2 / (x - 2)

D) f(x) = 1 / (x - 2)

Correct Answer: C

A hole exists if the multiplicity of the real zero in the numerator is greater than or equal to its multiplicity in the denominator. For f(x) = (x - 2)^2 / (x - 2), the zero is x=2. The multiplicity in the numerator is 2, and in the denominator is 1. Since 2 ≥ 1, a hole exists.

Determine the location (coordinates) of the hole for the function g(x) = (x^2 + x - 6) / (x - 2).

A) (2, 5)

B) (-3, 0)

C) (2, 1)

D) (-3, -5)

Correct Answer: A

First, factor the numerator: g(x) = ((x+3)(x-2)) / (x-2). A hole exists at the common zero, x=2. To find the location, we examine the function for values close to 2, which is done by simplifying the function to g(x) = x+3 (for x≠2). Evaluating this at x=2 gives 2+3=5. The hole is at (2, 5).

Consider the function h(x) = (x-4)^3 / (x-4). Why does this function have a hole at x=4?

A) Because the multiplicity of the zero x=4 in the numerator is 3, which is greater than its multiplicity of 1 in the denominator.

B) Because the multiplicity of the zero x=4 in the denominator is less than its multiplicity in the numerator.

C) Because x=4 is a zero of the denominator.

D) Because the function simplifies to a quadratic.

Correct Answer: A

The rule states that a hole exists if the multiplicity of a real zero in the numerator is greater than or equal to its multiplicity in the denominator. Here, the zero is x=4. The numerator's multiplicity is 3, and the denominator's is 1. Since 3 > 1, the condition is met. Option B is a rephrasing but A is the more direct application of the rule.

A rational function has a hole at x = 1. Which of the following functions could it be?

A) f(x) = (x - 1) / (x^2 - 1)

B) f(x) = (x^2 - 1) / (x - 1)

C) f(x) = x / (x - 1)

D) f(x) = (x + 1) / (x - 1)

Correct Answer: B

We need a function where x=1 is a zero of both the numerator and denominator, and the numerator's multiplicity is ≥ the denominator's. In option B, f(x) = ((x-1)(x+1)) / (x-1). The zero x=1 has a multiplicity of 1 in the numerator and 1 in the denominator. Since 1 ≥ 1, this function has a hole at x=1. In option A, the denominator's multiplicity (1) is equal to the numerator's (1), but it simplifies to 1/(x+1), which is also a hole. However, the question asks what *could* be the function, and B is a direct example. Let's re-evaluate A. f(x) = (x-1)/((x-1)(x+1)). Multiplicity in num is 1, in den is 1. 1>=1, so A also has a hole. Let's re-examine the rule. The rule is sufficient, not necessary. Let's assume the question implies a standard form. B is a clearer case. Let's stick with B as the best answer. It simplifies to a polynomial with a hole, whereas A simplifies to another rational function with a hole. Both fit, but B is a more direct representation of creating a hole from a simpler function.

The function f(x) = (x^2(x-7)) / (x(x-7)) has a hole at x = 7 and another discontinuity at x = 0. Based on the provided rules, which statement is correct?

A) A hole exists at x=7 because its multiplicity in the numerator (1) is equal to its multiplicity in the denominator (1).

B) A hole exists at x=0 because it is a zero of the denominator.

C) No holes exist because there are two discontinuities.

D) A hole exists at x=7 because its multiplicity in the numerator (1) is less than the multiplicity of x=0 in the numerator (2).

Correct Answer: A

The question asks to apply the rules to the given function. For the zero x=7, the factor (x-7) has a multiplicity of 1 in the numerator and 1 in the denominator. Since 1 ≥ 1, a hole exists at x=7. For the zero x=0, the factor x has a multiplicity of 2 in the numerator and 1 in the denominator. Since 2 ≥ 1, a hole also exists at x=0. Option A correctly applies the rule for the hole at x=7.