College Algebra Quiz: Vertical And Horizontal Asymptotes
4 questions · exam conditions
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Vertical And Horizontal AsymptotesQuestion 1 of 4

The function f(x)=6x4−3x3+x22x4+5x2−7f(x) = \frac{6x^4 - 3x^3 + x^2}{2x^4 + 5x^2 - 7} has a horizontal asymptote at y=ky = k. What is the value of kk?

k=0k = 0
k=3k = 3
k=62=3k = \frac{6}{2} = 3
k=32k = \frac{3}{2}
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College Algebra Quiz

College Algebra Quiz: Vertical And Horizontal Asymptotes

Practice Vertical And Horizontal Asymptotes 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 Vertical And Horizontal Asymptotes, giving you a quick way to practice the rules, question types, and explanations that matter most for College Algebra.

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

The function f(x)=6x4−3x3+x22x4+5x2−7f(x) = \frac{6x^4 - 3x^3 + x^2}{2x^4 + 5x^2 - 7} has a horizontal asymptote at y=ky = k. What is the value of kk?

  1. k=0k = 0
  2. k=3k = 3 (correct answer)
  3. k=62=3k = \frac{6}{2} = 3
  4. k=32k = \frac{3}{2}
Explanation: For rational functions where the numerator and denominator have the same degree, the horizontal asymptote is y=leading coefficient of numeratorleading coefficient of denominatory = \frac{\text{leading coefficient of numerator}}{\text{leading coefficient of denominator}}. Both the numerator 6x4−3x3+x26x^4 - 3x^3 + x^2 and denominator 2x4+5x2−72x^4 + 5x^2 - 7 have degree 4. The leading coefficient of the numerator is 6, and the leading coefficient of the denominator is 2. Therefore, the horizontal asymptote is y=62=3y = \frac{6}{2} = 3. Choice A would be correct if the denominator had higher degree. Choice C shows the same calculation as the correct answer. Choice D incorrectly uses coefficients from other terms rather than the leading coefficients.

Question 2

The function h(x)=3x3−12x2x2−16h(x) = \frac{3x^3 - 12x^2}{x^2 - 16} can be written in a simplified form after factoring. What happens to the asymptotes when the function is simplified?

  1. There is one vertical asymptote at x=−4x = -4 and an oblique asymptote (correct answer)
  2. There are vertical asymptotes at x=±4x = \pm 4 and an oblique asymptote
  3. There is one vertical asymptote at x=4x = 4 and a hole at x=0x = 0
  4. There is a hole at x=4x = 4, one vertical asymptote at x=−4x = -4, and an oblique asymptote
Explanation: Factor the numerator: 3x3−12x2=3x2(x−4)3x^3 - 12x^2 = 3x^2(x - 4). Factor the denominator: x2−16=(x−4)(x+4)x^2 - 16 = (x-4)(x+4). So h(x)=3x2(x−4)(x−4)(x+4)h(x) = \frac{3x^2(x-4)}{(x-4)(x+4)}. The factor (x−4)(x-4) cancels, giving h(x)=3x2x+4h(x) = \frac{3x^2}{x+4} for x≠4x \neq 4. This creates a hole at x=4x = 4 and a vertical asymptote at x=−4x = -4 (where x+4=0x + 4 = 0). Since the degree of the simplified numerator (2) is greater than the degree of the denominator (1), there is an oblique asymptote. Choice B ignores the cancellation. Choice C incorrectly identifies the asymptote location. Choice D correctly identifies the hole but is unnecessarily detailed compared to choice A.

Question 3

For the rational function f(x)=(x−1)2(x+2)(x−3)(x+2)2f(x) = \frac{(x-1)^2(x+2)}{(x-3)(x+2)^2}, what are the vertical asymptotes and holes?

  1. Vertical asymptote at x=3x = 3; hole at x=−2x = -2 (correct answer)
  2. Vertical asymptotes at x=3x = 3 and x=−2x = -2; no holes
  3. Vertical asymptote at x=−2x = -2; hole at x=3x = 3
  4. Vertical asymptote at x=3x = 3; vertical asymptote at x=−2x = -2
Explanation: To find vertical asymptotes and holes, examine where the denominator equals zero and check for cancellation with the numerator. The denominator (x−3)(x+2)2(x-3)(x+2)^2 equals zero when x=3x = 3 or x=−2x = -2. At x=3x = 3: The numerator (x−1)2(x+2)=(2)2(5)=20≠0(x-1)^2(x+2) = (2)^2(5) = 20 \neq 0, so there's a vertical asymptote. At x=−2x = -2: Both numerator and denominator have factor (x+2)(x+2). The numerator has (x+2)1(x+2)^1 and denominator has (x+2)2(x+2)^2. After canceling one factor of (x+2)(x+2), we get f(x)=(x−1)2(x−3)(x+2)f(x) = \frac{(x-1)^2}{(x-3)(x+2)} for x≠−2x \neq -2. The remaining denominator still has factor (x+2)(x+2), so this creates a hole at x=−2x = -2, not a vertical asymptote. Choice B ignores the partial cancellation at x=−2x = -2. Choice C reverses the locations. Choice D treats both as vertical asymptotes.

Question 4

Given f(x)=x2−9x2−x−6f(x) = \frac{x^2 - 9}{x^2 - x - 6}, which statement about the asymptotes is correct?

  1. There is a vertical asymptote at x=3x = 3 and a horizontal asymptote at y=1y = 1
  2. There is a vertical asymptote at x=−2x = -2 and a hole at x=3x = 3, with horizontal asymptote y=1y = 1 (correct answer)
  3. There are vertical asymptotes at x=−2x = -2 and x=3x = 3, with horizontal asymptote y=1y = 1
  4. There is a vertical asymptote at x=3x = 3 and a hole at x=−2x = -2, with horizontal asymptote y=1y = 1
Explanation: Factor the numerator and denominator: x2−9=(x−3)(x+3)x^2 - 9 = (x-3)(x+3) and x2−x−6=(x−3)(x+2)x^2 - x - 6 = (x-3)(x+2). So f(x)=(x−3)(x+3)(x−3)(x+2)f(x) = \frac{(x-3)(x+3)}{(x-3)(x+2)}. The factor (x−3)(x-3) cancels, leaving f(x)=x+3x+2f(x) = \frac{x+3}{x+2} for x≠3x \neq 3. This creates a hole at x=3x = 3. The simplified function has a vertical asymptote where the remaining denominator is zero: x+2=0x + 2 = 0, so x=−2x = -2. For the horizontal asymptote, both numerator and denominator of the simplified form have degree 1, so y=11=1y = \frac{1}{1} = 1. Choice A incorrectly identifies a vertical asymptote at the hole location. Choice C ignores the cancellation. Choice D incorrectly identifies where the hole and asymptote occur.