AP Precalculus Quiz: Rational Functions And Holes
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Rational Functions And HolesQuestion 1 of 20

Consider the function u(x)=x2−25x−5u(x)=\dfrac{x^2-25}{x-5} for a fuel-usage model; what is the location of the hole in the function u(x)u(x)?

x=−5x=-5
x=5x=5
x=1x=1
x=0x=0
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AP Precalculus Quiz

AP Precalculus Quiz: Rational Functions And Holes

Practice Rational Functions And Holes in AP Precalculus with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Rational Functions And Holes, 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.

All questions

Question 1

Consider the function u(x)=x2−25x−5u(x)=\dfrac{x^2-25}{x-5} for a fuel-usage model; what is the location of the hole in the function u(x)u(x)?

  1. x=−5x=-5
  2. x=5x=5 (correct answer)
  3. x=1x=1
  4. x=0x=0

Explanation: This question tests AP Precalculus skills: understanding of rational functions and identifying holes. A hole occurs where a rational function's numerator and denominator both equal zero, creating a removable discontinuity. In this problem, the function u(x) = (x²-25)/(x-5) can be factored as u(x) = (x+5)(x-5)/(x-5), using the difference of squares pattern, revealing a common factor of (x-5). Choice B is correct because when x = 5, both the numerator (25-25=0) and denominator (5-5=0) equal zero, and the (x-5) factor cancels out, creating a hole at x = 5. Choice A is incorrect because x = -5 only makes the numerator zero after cancellation, creating a zero of the simplified function. To help students: Remember that x²-a² = (x+a)(x-a) for difference of squares. Use algebraic cancellation to identify removable discontinuities versus non-removable ones.

Question 2

Consider the function v(x)=x2+3xxv(x)=\dfrac{x^2+3x}{x} for a unit-rate calculation; at which xx-value is vv undefined due to a removable discontinuity?

  1. x=−3x=-3
  2. x=3x=3
  3. x=0x=0 (correct answer)
  4. x=1x=1

Explanation: This question tests AP Precalculus skills: understanding of rational functions and identifying holes. A hole occurs where a rational function's numerator and denominator both equal zero, creating a removable discontinuity. In this problem, the function v(x) = (x²+3x)/x can be factored as v(x) = x(x+3)/x, showing a common factor of x. Choice C is correct because when x = 0, both the numerator (0+0=0) and denominator (0) equal zero, and the x factor cancels out, creating a hole at x = 0. Choice A is incorrect because x = -3 only makes the numerator zero after cancellation, not creating a hole. To help students: Factor out common terms from the numerator before identifying cancelable factors. Be careful with x = 0 cases, as they often create holes when x appears in both numerator and denominator.

Question 3

Consider the function f(x)=x2−1x−1f(x)=\dfrac{x^2-1}{x-1}. What is the location of the hole?

  1. x=−1x=-1
  2. x=1x=1 (correct answer)
  3. x=0x=0
  4. x=5x=5

Explanation: This question tests AP Precalculus skills: understanding of rational functions and identifying holes. A hole occurs where a rational function's numerator and denominator both equal zero, creating a removable discontinuity. In this problem, the function f(x) = (x²-1)/(x-1) can be factored as f(x) = (x+1)(x-1)/(x-1), revealing a common factor of (x-1) that cancels. Choice B is correct because when x = 1, both the original numerator (1²-1 = 0) and denominator (1-1 = 0) equal zero, confirming the hole's location. Choice A is incorrect because it identifies where the simplified function equals zero, not where the hole occurs. To help students: Emphasize factoring the numerator completely before canceling. Practice identifying the difference between zeros of the simplified function and locations of holes.

Question 4

Consider n(x)=x2+x−12x+4n(x)=\dfrac{x^2+x-12}{x+4}. What is the location of the hole in n(x)n(x)?

  1. x=3x=3
  2. x=−4x=-4 (correct answer)
  3. x=−3x=-3
  4. x=4x=4

Explanation: This question tests AP Precalculus skills: understanding of rational functions and identifying holes. A hole occurs where a rational function's numerator and denominator both equal zero, creating a removable discontinuity. In this problem, n(x) = (x²+x-12)/(x+4) requires factoring the numerator: x²+x-12 = (x+4)(x-3), revealing the common factor (x+4). Choice B is correct because when x = -4, both the numerator (16-4-12 = 0) and denominator (-4+4 = 0) equal zero, creating the hole. Choice A is incorrect as it identifies where the simplified function n(x) = x-3 equals zero, not the hole's location. To help students: Use systematic factoring to find (x+4)(x-3) from x²+x-12. Remind them to always verify by substituting the proposed hole location into both parts of the original fraction.

Question 5

Consider m(x)=x2−2x+1x−1m(x)=\dfrac{x^2-2x+1}{x-1}. Which value of xx makes numerator and denominator zero, creating a hole?

  1. x=−1x=-1
  2. x=1x=1 (correct answer)
  3. x=0x=0
  4. x=7x=7

Explanation: This question tests AP Precalculus skills: understanding of rational functions and identifying holes. A hole occurs where a rational function's numerator and denominator both equal zero, creating a removable discontinuity. In this problem, m(x) = (x²-2x+1)/(x-1) has a numerator that factors as a perfect square: x²-2x+1 = (x-1)², revealing the common factor (x-1). Choice B is correct because when x = 1, both the numerator (1-2+1 = 0) and denominator (1-1 = 0) equal zero, creating the hole. Choice A is incorrect as it doesn't correspond to any significant feature of this function. To help students: Recognize perfect square patterns like (x-1)² = x²-2x+1. Emphasize that when a factor appears multiple times in the numerator, it still creates only one hole when canceled with the denominator.

Question 6

Consider the function u(x)=x2−1x+1u(x)=\dfrac{x^2-1}{x+1} for a simplified index; at which xx-value is it undefined due to a removable discontinuity?

  1. x=1x=1
  2. x=−1x=-1 (correct answer)
  3. x=0x=0
  4. x=2x=2

Explanation: This question tests AP Precalculus skills: understanding of rational functions and identifying holes. A hole occurs where a rational function's numerator and denominator both equal zero, creating a removable discontinuity. In this problem, the function u(x) = (x²-1)/(x+1) can be factored as u(x) = (x+1)(x-1)/(x+1), showing a common factor of (x+1). Choice B is correct because when x = -1, both the numerator (1-1=0) and denominator (-1+1=0) equal zero, and the (x+1) factor cancels, creating a hole at x = -1. Choice A is incorrect because x = 1 only makes the numerator zero after factoring, not the denominator. To help students: Recognize x²-1 as a difference of squares. Pay attention to the sign in the denominator (x+1) to identify the correct hole location.

Question 7

Consider the function g(x)=x2−4xx−4g(x)=\dfrac{x^2-4x}{x-4} for a pricing rule; which value of xx makes the numerator and denominator zero, creating a hole?

  1. x=4x=4 (correct answer)
  2. x=0x=0
  3. x=−4x=-4
  4. x=2x=2

Explanation: This question tests AP Precalculus skills: understanding of rational functions and identifying holes. A hole occurs where a rational function's numerator and denominator both equal zero, creating a removable discontinuity. In this problem, the function g(x) = (x²-4x)/(x-4) can be factored as g(x) = x(x-4)/(x-4), showing a common factor of (x-4). Choice A is correct because when x = 4, both the numerator (16-16=0) and denominator (4-4=0) equal zero, and the (x-4) factor cancels out, creating a hole at x = 4. Choice B is incorrect because x = 0 only makes the numerator zero, resulting in a zero of the function rather than a hole. To help students: Factor the numerator completely before identifying common factors. Use substitution to verify that both numerator and denominator equal zero at the hole location.

Question 8

Consider the function h(x)=x2−1x−1h(x)=\dfrac{x^2-1}{x-1} for a sensor calibration; what is the location of the hole in the function h(x)h(x)?

  1. x=−1x=-1
  2. x=1x=1 (correct answer)
  3. x=0x=0
  4. x=2x=2

Explanation: This question tests AP Precalculus skills: understanding of rational functions and identifying holes. A hole occurs where a rational function's numerator and denominator both equal zero, creating a removable discontinuity. In this problem, the function h(x) = (x²-1)/(x-1) can be factored as h(x) = (x+1)(x-1)/(x-1), revealing a common factor of (x-1). Choice B is correct because when x = 1, both the numerator (1-1=0) and denominator (1-1=0) equal zero, and the (x-1) factor cancels out, creating a hole at x = 1. Choice A is incorrect because x = -1 only makes the numerator zero, which creates a zero of the simplified function, not a hole. To help students: Remember the difference of squares factoring pattern (a²-b² = (a+b)(a-b)). Practice identifying and canceling common factors to find removable discontinuities.

Question 9

Consider the function h(x)=x2−9x−3h(x)=\dfrac{x^2-9}{x-3} for a speed ratio; which value of xx makes numerator and denominator zero, creating a hole?

  1. x=−3x=-3
  2. x=9x=9
  3. x=0x=0
  4. x=3x=3 (correct answer)

Explanation: This question tests AP Precalculus skills: understanding of rational functions and identifying holes. A hole occurs where a rational function's numerator and denominator both equal zero, creating a removable discontinuity. In this problem, the function h(x) = (x²-9)/(x-3) can be factored as h(x) = (x+3)(x-3)/(x-3), revealing a common factor of (x-3). Choice D is correct because when x = 3, both the numerator (9-9=0) and denominator (3-3=0) equal zero, and the (x-3) factor cancels out, creating a hole at x = 3. Choice A is incorrect because x = -3 only makes the numerator zero, not the denominator. To help students: Recognize x²-9 as a difference of squares that factors to (x+3)(x-3). Always check that the suspected hole location makes BOTH parts of the fraction equal zero.

Question 10

Consider the function t(x)=x2+3xxt(x)=\dfrac{x^2+3x}{x} for a production ratio; what is the location of the hole?

  1. x=−3x=-3
  2. x=3x=3
  3. x=0x=0 (correct answer)
  4. x=1x=1

Explanation: This question tests AP Precalculus skills: understanding of rational functions and identifying holes. A hole occurs where a rational function's numerator and denominator both equal zero, creating a removable discontinuity. In this problem, the function t(x) = (x²+3x)/x can be factored as t(x) = x(x+3)/x, revealing a common factor of x. Choice C is correct because when x = 0, both the numerator (0+0=0) and denominator (0) equal zero, and the x factor cancels out, creating a hole at x = 0. Choice A is incorrect because x = -3 only makes the numerator zero after factoring out x, not the denominator. To help students: Factor out the greatest common factor from the numerator first. Remember that holes occur where common factors in numerator and denominator both equal zero.

Question 11

Consider the function f(x)=x2−9x−3f(x)=\dfrac{x^2-9}{x-3} for a shipping-rate model; at which xx-value is ff undefined due to a removable discontinuity?

  1. x=−3x=-3
  2. x=3x=3 (correct answer)
  3. x=0x=0
  4. x=5x=5

Explanation: This question tests AP Precalculus skills: understanding of rational functions and identifying holes. A hole occurs where a rational function's numerator and denominator both equal zero, creating a removable discontinuity. In this problem, the function f(x) = (x²-9)/(x-3) can be factored as f(x) = (x+3)(x-3)/(x-3), revealing a common factor of (x-3). Choice B is correct because when x = 3, both the numerator and denominator equal zero, and the (x-3) factor cancels out, creating a hole at x = 3. Choice A is incorrect because x = -3 makes only the numerator zero, resulting in a zero of the function, not a hole. To help students: Emphasize that holes require both numerator AND denominator to be zero. Practice factoring quadratics to identify common factors that create removable discontinuities.

Question 12

Consider the function p(x)=x2+5x+6x+2p(x)=\dfrac{x^2+5x+6}{x+2} for a production estimate; at which xx-value is pp undefined due to a removable discontinuity?

  1. x=−2x=-2 (correct answer)
  2. x=−3x=-3
  3. x=3x=3
  4. x=0x=0

Explanation: This question tests AP Precalculus skills: understanding of rational functions and identifying holes. A hole occurs where a rational function's numerator and denominator both equal zero, creating a removable discontinuity. In this problem, the function p(x) = (x²+5x+6)/(x+2) can be factored as p(x) = (x+2)(x+3)/(x+2), showing a common factor of (x+2). Choice A is correct because when x = -2, both the numerator (4-10+6=0) and denominator (-2+2=0) equal zero, and the (x+2) factor cancels out, creating a hole at x = -2. Choice B is incorrect because x = -3 only makes the numerator zero after cancellation, representing a zero of the simplified function. To help students: Factor trinomials systematically to find factors like (x+2)(x+3). Always verify holes by checking that both original numerator and denominator equal zero.

Question 13

Consider the function s(x)=x2+x−6x−2s(x)=\dfrac{x^2+x-6}{x-2} for a demand curve; at which xx-value is ss undefined due to a removable discontinuity?

  1. x=−3x=-3
  2. x=2x=2 (correct answer)
  3. x=3x=3
  4. x=0x=0

Explanation: This question tests AP Precalculus skills: understanding of rational functions and identifying holes. A hole occurs where a rational function's numerator and denominator both equal zero, creating a removable discontinuity. In this problem, the function s(x) = (x²+x-6)/(x-2) can be factored as s(x) = (x-2)(x+3)/(x-2), showing a common factor of (x-2). Choice B is correct because when x = 2, both the numerator (4+2-6=0) and denominator (2-2=0) equal zero, and the (x-2) factor cancels out, creating a hole at x = 2. Choice A is incorrect because x = -3 only makes the numerator zero after cancellation, not creating a hole. To help students: Factor trinomials by finding two numbers that multiply to give the constant term and add to give the middle coefficient. Always verify by substituting the x-value into both numerator and denominator.

Question 14

Consider the function t(x)=x2−6x+9x−3t(x)=\dfrac{x^2-6x+9}{x-3} for a revenue model; which value of xx makes the numerator and denominator zero, creating a hole?

  1. x=−3x=-3
  2. x=3x=3 (correct answer)
  3. x=0x=0
  4. x=6x=6

Explanation: This question tests AP Precalculus skills: understanding of rational functions and identifying holes. A hole occurs where a rational function's numerator and denominator both equal zero, creating a removable discontinuity. In this problem, the function t(x) = (x²-6x+9)/(x-3) can be recognized as t(x) = (x-3)²/(x-3), since x²-6x+9 is a perfect square trinomial, showing a common factor of (x-3). Choice B is correct because when x = 3, both the numerator (9-18+9=0) and denominator (3-3=0) equal zero, and one (x-3) factor cancels out, creating a hole at x = 3. Choice D is incorrect as x = 6 doesn't make either expression zero. To help students: Recognize perfect square trinomial patterns (a-b)² = a²-2ab+b². Note that even with repeated factors, only one cancels with the denominator to create the hole.

Question 15

Consider the function g(x)=x2−4x−2g(x)=\dfrac{x^2-4}{x-2}. At which xx-value is gg undefined due to a removable discontinuity?

  1. x=−2x=-2
  2. x=2x=2 (correct answer)
  3. x=0x=0
  4. x=4x=4

Explanation: This question tests AP Precalculus skills: understanding of rational functions and identifying holes. A hole (removable discontinuity) occurs where both numerator and denominator share a common factor that equals zero. In this problem, g(x) = (x²-4)/(x-2) can be factored as g(x) = (x+2)(x-2)/(x-2), showing the common factor (x-2). Choice B is correct because at x = 2, both the numerator (2²-4 = 0) and denominator (2-2 = 0) equal zero, creating the removable discontinuity. Choice A is incorrect as it confuses the zero of the simplified function with the hole's location. To help students: Use the difference of squares factoring pattern. Remind them that after canceling, the hole remains at the canceled factor's zero.

Question 16

Consider the function q(x)=x2−2x−3x−3q(x)=\dfrac{x^2-2x-3}{x-3} for a water-flow model; which value of xx makes the numerator and denominator zero, creating a hole?

  1. x=−1x=-1
  2. x=3x=3 (correct answer)
  3. x=1x=1
  4. x=−3x=-3

Explanation: This question tests AP Precalculus skills: understanding of rational functions and identifying holes. A hole occurs where a rational function's numerator and denominator both equal zero, creating a removable discontinuity. In this problem, the function q(x) = (x²-2x-3)/(x-3) can be factored as q(x) = (x-3)(x+1)/(x-3), revealing a common factor of (x-3). Choice B is correct because when x = 3, both the numerator (9-6-3=0) and denominator (3-3=0) equal zero, and the (x-3) factor cancels out, creating a hole at x = 3. Choice D is incorrect because x = -3 doesn't make either the numerator or denominator zero. To help students: Use factoring techniques for trinomials where the leading coefficient is 1. Check your factorization by expanding to verify it matches the original expression.

Question 17

Consider h(x)=x2+5x+6x+2h(x)=\dfrac{x^2+5x+6}{x+2}. Which value of xx makes numerator and denominator zero, creating a hole?

  1. x=−2x=-2 (correct answer)
  2. x=−3x=-3
  3. x=2x=2
  4. x=5x=5

Explanation: This question tests AP Precalculus skills: understanding of rational functions and identifying holes. A hole occurs where a rational function's numerator and denominator both equal zero, creating a removable discontinuity. In this problem, h(x) = (x²+5x+6)/(x+2) requires factoring the numerator: x²+5x+6 = (x+2)(x+3), revealing the common factor (x+2). Choice A is correct because when x = -2, both the numerator (4-10+6 = 0) and denominator (-2+2 = 0) equal zero, confirming the hole's location. Choice B is incorrect as it identifies where the simplified function h(x) = x+3 equals zero. To help students: Practice factoring quadratic expressions. Emphasize checking that both original numerator and denominator equal zero at the proposed hole location.

Question 18

Consider the function s(x)=x2−25x−5s(x)=\dfrac{x^2-25}{x-5} for a conversion factor; how would you describe the hole in its graph?

  1. Hole at x=5x=5 (correct answer)
  2. Hole at x=−5x=-5
  3. Hole at x=0x=0
  4. Hole at x=25x=25

Explanation: This question tests AP Precalculus skills: understanding of rational functions and identifying holes. A hole occurs where a rational function's numerator and denominator both equal zero, creating a removable discontinuity. In this problem, the function s(x) = (x²-25)/(x-5) can be factored as s(x) = (x+5)(x-5)/(x-5), showing a common factor of (x-5). Choice A is correct because when x = 5, both the numerator (25-25=0) and denominator (5-5=0) equal zero, and the (x-5) factor cancels, creating a hole at x = 5. Choice B is incorrect because x = -5 only zeros the numerator, creating a zero of the simplified function, not a hole. To help students: Use difference of squares factoring for x²-25 = (x+5)(x-5). Distinguish between zeros of the function and holes in the graph.

Question 19

Consider the function p(x)=x2+xxp(x)=\dfrac{x^2+x}{x} for a per-item cost model; at which xx-value is it undefined due to a removable discontinuity?

  1. x=1x=1
  2. x=−1x=-1
  3. x=0x=0 (correct answer)
  4. x=2x=2

Explanation: This question tests AP Precalculus skills: understanding of rational functions and identifying holes. A hole occurs where a rational function's numerator and denominator both equal zero, creating a removable discontinuity. In this problem, the function p(x) = (x²+x)/x can be factored as p(x) = x(x+1)/x, showing a common factor of x. Choice C is correct because when x = 0, both the numerator (0+0=0) and denominator (0) equal zero, and the x factor cancels, creating a hole at x = 0. Choice A is incorrect because x = 1 doesn't make the denominator zero. To help students: Factor out common terms from the numerator before simplifying. Remind students that division by zero creates discontinuities, but when both parts are zero, it's removable.

Question 20

Consider the function q(x)=x2−6x+9x−3q(x)=\dfrac{x^2-6x+9}{x-3} for a scaling rule; what is the location of the hole?

  1. x=−3x=-3
  2. x=3x=3 (correct answer)
  3. x=9x=9
  4. x=6x=6

Explanation: This question tests AP Precalculus skills: understanding of rational functions and identifying holes. A hole occurs where a rational function's numerator and denominator both equal zero, creating a removable discontinuity. In this problem, the function q(x) = (x²-6x+9)/(x-3) has a numerator that factors as (x-3)², so q(x) = (x-3)²/(x-3), with a common factor of (x-3). Choice B is correct because when x = 3, both the numerator (9-18+9=0) and denominator (3-3=0) equal zero, and one (x-3) factor cancels, creating a hole at x = 3. Choice A is incorrect because x = -3 doesn't make either part zero. To help students: Recognize perfect square trinomials like x²-6x+9 = (x-3)². Practice factoring quadratics to identify common factors with the denominator.