College Algebra Quiz: Domain Restrictions Holes And Discontinuities
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Domain Restrictions Holes And DiscontinuitiesQuestion 1 of 6

Consider the rational function f(x)=x2−9x2+2x−15f(x) = \frac{x^2 - 9}{x^2 + 2x - 15}. After factoring both numerator and denominator completely, what can be concluded about the domain and discontinuities of this function?

The domain is all real numbers except x=−5x = -5 and x=3x = 3, with vertical asymptotes at both values
The domain is all real numbers except x=−5x = -5 and x=3x = 3, with a hole at x=3x = 3 and vertical asymptote at x=−5x = -5
The domain is all real numbers except x=−3x = -3 and x=5x = 5, with a hole at x=−3x = -3 and vertical asymptote at x=5x = 5
The domain is all real numbers except x=−5x = -5, with a hole at x=3x = 3 and vertical asymptote at x=−5x = -5
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College Algebra Quiz

College Algebra Quiz: Domain Restrictions Holes And Discontinuities

Practice Domain Restrictions Holes And Discontinuities in College Algebra 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 Domain Restrictions Holes And Discontinuities, giving you a quick way to practice the rules, question types, and explanations that matter most for College Algebra.

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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 rational function f(x)=x2−9x2+2x−15f(x) = \frac{x^2 - 9}{x^2 + 2x - 15}. After factoring both numerator and denominator completely, what can be concluded about the domain and discontinuities of this function?

  1. The domain is all real numbers except x=−5x = -5 and x=3x = 3, with vertical asymptotes at both values
  2. The domain is all real numbers except x=−5x = -5 and x=3x = 3, with a hole at x=3x = 3 and vertical asymptote at x=−5x = -5 (correct answer)
  3. The domain is all real numbers except x=−3x = -3 and x=5x = 5, with a hole at x=−3x = -3 and vertical asymptote at x=5x = 5
  4. The domain is all real numbers except x=−5x = -5, with a hole at x=3x = 3 and vertical asymptote at x=−5x = -5
Explanation: First, factor completely: f(x)=(x−3)(x+3)(x+5)(x−3)f(x) = \frac{(x-3)(x+3)}{(x+5)(x-3)}. The factor (x−3)(x-3) appears in both numerator and denominator, creating a hole at x=3x = 3. The factor (x+5)(x+5) appears only in the denominator, creating a vertical asymptote at x=−5x = -5. The domain excludes both x=3x = 3 and x=−5x = -5. Choice A incorrectly identifies both as vertical asymptotes. Choice C uses wrong values from incorrectly factoring the denominator. Choice D incorrectly excludes x=3x = 3 from the domain restriction.

Question 2

Consider the piecewise-defined function: f(x)={x2−4x−2if x≠25if x=2f(x) = \begin{cases} \frac{x^2-4}{x-2} & \text{if } x \neq 2 \\ 5 & \text{if } x = 2 \end{cases} . What type of discontinuity, if any, occurs at x=2x = 2?

  1. No discontinuity; the function is continuous at x=2x = 2
  2. Jump discontinuity because the left and right limits are different
  3. Removable discontinuity because the rational part has a hole at x=2x = 2 (correct answer)
  4. Essential discontinuity because the function approaches infinity
Explanation: For x≠2x \neq 2, we have f(x)=x2−4x−2=(x−2)(x+2)x−2=x+2f(x) = \frac{x^2-4}{x-2} = \frac{(x-2)(x+2)}{x-2} = x+2. So lim⁡x→2f(x)=lim⁡x→2(x+2)=4\lim_{x \to 2} f(x) = \lim_{x \to 2} (x+2) = 4. However, f(2)=5f(2) = 5. Since the limit exists but doesn't equal the function value, this is a removable discontinuity. The function could be made continuous by redefining f(2)=4f(2) = 4. Choice A is wrong because f(2)≠lim⁡x→2f(x)f(2) \neq \lim_{x \to 2} f(x). Choice B is incorrect because left and right limits are both 4. Choice D is wrong because the limit exists and is finite.

Question 3

Consider the function k(x)=x2−3x−10x2−25k(x) = \frac{x^2-3x-10}{x^2-25}. If this function is graphed on a coordinate plane, which statement correctly describes its discontinuities?

  1. There is a hole at x=5x = 5 and a vertical asymptote at x=−5x = -5 (correct answer)
  2. There is a hole at x=−2x = -2 and a vertical asymptote at x=5x = 5
  3. There are vertical asymptotes at both x=5x = 5 and x=−5x = -5
  4. There is a hole at x=−5x = -5 and a vertical asymptote at x=5x = 5
Explanation: Factor completely: k(x)=x2−3x−10x2−25=(x−5)(x+2)(x−5)(x+5)k(x) = \frac{x^2-3x-10}{x^2-25} = \frac{(x-5)(x+2)}{(x-5)(x+5)}. The factor (x−5)(x-5) appears in both numerator and denominator, creating a hole at x=5x = 5. The factor (x+5)(x+5) appears only in the denominator, creating a vertical asymptote at x=−5x = -5. Choice B incorrectly places the hole at x=−2x = -2 (a zero of the reduced function, not a discontinuity). Choice C ignores the common factor cancellation. Choice D reverses the locations of the hole and asymptote.

Question 4

For the rational function h(x)=x3−8x2−4h(x) = \frac{x^3-8}{x^2-4}, what is the domain restriction and what type of discontinuity occurs at x=2x = 2?

  1. Domain: all real numbers except x=±2x = \pm 2; vertical asymptote at x=2x = 2
  2. Domain: all real numbers except x=±2x = \pm 2; hole at x=2x = 2 (correct answer)
  3. Domain: all real numbers except x=2x = 2; vertical asymptote at x=2x = 2
  4. Domain: all real numbers except x=±2x = \pm 2; removable discontinuity at x=2x = 2
Explanation: Factor the function: h(x)=x3−8x2−4=(x−2)(x2+2x+4)(x−2)(x+2)h(x) = \frac{x^3-8}{x^2-4} = \frac{(x-2)(x^2+2x+4)}{(x-2)(x+2)}. The factor (x−2)(x-2) appears in both numerator and denominator, creating a hole (removable discontinuity) at x=2x = 2. The factor (x+2)(x+2) appears only in the denominator, creating a vertical asymptote at x=−2x = -2. The domain excludes both x=2x = 2 and x=−2x = -2. Choice A incorrectly identifies x=2x = 2 as a vertical asymptote. Choice C incorrectly excludes x=−2x = -2 from domain restrictions. Choice D correctly identifies the removable discontinuity but uses less precise terminology.

Question 5

A student claims that the function f(x)=x2+x−6x2−9f(x) = \frac{x^2+x-6}{x^2-9} has a hole at x=3x = 3 because both the numerator and denominator equal zero when x=3x = 3. What is wrong with this reasoning?

  1. The student is correct; when both numerator and denominator are zero, there is always a hole
  2. There is no discontinuity at x=3x = 3 because the function is defined there
  3. While both expressions equal zero at x=3x = 3, the factored forms show no common factor, so there is a vertical asymptote, not a hole (correct answer)
  4. The student's logic is flawed; holes only occur when the same linear factor cancels from numerator and denominator
Explanation: Factor both expressions: numerator x2+x−6=(x+3)(x−2)x^2+x-6 = (x+3)(x-2) and denominator x2−9=(x+3)(x−3)x^2-9 = (x+3)(x-3). While both equal zero at x=3x = 3, this alone doesn't determine the type of discontinuity. The factored forms show no common factor involving (x−3)(x-3), so x=3x = 3 creates a vertical asymptote, not a hole. A hole occurs only when the same linear factor appears in both numerator and denominator and can be canceled. Choice A gives an incorrect general rule. Choice B is wrong since f(3)f(3) is undefined. Choice D correctly identifies the flaw in reasoning.

Question 6

For which value of kk will the function f(x)=x2+kx−12x2−16f(x) = \frac{x^2+kx-12}{x^2-16} have a removable discontinuity (hole) at x=4x = 4?

  1. k=−1k = -1 (correct answer)
  2. k=1k = 1
  3. k=3k = 3
  4. k=−3k = -3
Explanation: For a hole at x=4x = 4, the factor (x−4)(x-4) must divide both numerator and denominator. The denominator x2−16=(x−4)(x+4)x^2-16 = (x-4)(x+4) already contains (x−4)(x-4). For the numerator to also contain (x−4)(x-4), we need x=4x = 4 to be a root: 42+k(4)−12=04^2 + k(4) - 12 = 0, so 16+4k−12=016 + 4k - 12 = 0, giving 4k=−44k = -4 and k=−1k = -1. We can verify: when k=−1k = -1, the numerator becomes x2−x−12=(x−4)(x+3)x^2 - x - 12 = (x-4)(x+3). Choice B gives x2+x−12=(x+4)(x−3)x^2 + x - 12 = (x+4)(x-3), no (x−4)(x-4) factor. Choice C gives x2+3x−12x^2 + 3x - 12, which doesn't factor to include (x−4)(x-4). Choice D gives x2−3x−12x^2 - 3x - 12, also without (x−4)(x-4).