College Algebra Quiz: Slant Oblique Asymptotes
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Slant Oblique AsymptotesQuestion 1 of 4

The function g(x)=ax3+bx2+cx+dx2−4g(x) = \frac{ax^3 + bx^2 + cx + d}{x^2 - 4} has a slant asymptote with equation y=3x+2y = 3x + 2. If the function has a vertical asymptote at x=2x = 2 but not at x=−2x = -2, what must be true about the coefficients?

a=3a = 3, b=2b = 2, and d=−16d = -16
a=3a = 3, b=2b = 2, and the numerator has (x+2)(x + 2) as a factor
a=3a = 3, c=8c = 8, and the numerator has (x+2)(x + 2) as a factor
a=3a = 3, b=2b = 2, and c+d=0c + d = 0
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College Algebra Quiz

College Algebra Quiz: Slant Oblique Asymptotes

Practice Slant Oblique 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 Slant Oblique Asymptotes, 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.

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Question 1

The function g(x)=ax3+bx2+cx+dx2−4g(x) = \frac{ax^3 + bx^2 + cx + d}{x^2 - 4} has a slant asymptote with equation y=3x+2y = 3x + 2. If the function has a vertical asymptote at x=2x = 2 but not at x=−2x = -2, what must be true about the coefficients?

  1. a=3a = 3, b=2b = 2, and d=−16d = -16
  2. a=3a = 3, b=2b = 2, and the numerator has (x+2)(x + 2) as a factor (correct answer)
  3. a=3a = 3, c=8c = 8, and the numerator has (x+2)(x + 2) as a factor
  4. a=3a = 3, b=2b = 2, and c+d=0c + d = 0
Explanation: Since the slant asymptote is y=3x+2y = 3x + 2, when we divide the numerator by x2−4x^2 - 4, we get quotient 3x+23x + 2. This means ax3+bx2+cx+d=(3x+2)(x2−4)+R(x)ax^3 + bx^2 + cx + d = (3x + 2)(x^2 - 4) + R(x) where R(x)R(x) is the remainder. Expanding: (3x+2)(x2−4)=3x3+2x2−12x−8(3x + 2)(x^2 - 4) = 3x^3 + 2x^2 - 12x - 8. So a=3a = 3 and b=2b = 2. Since there's a vertical asymptote at x=2x = 2 but not at x=−2x = -2, the denominator factor (x+2)(x + 2) must cancel with a factor in the numerator, meaning (x+2)(x + 2) divides the numerator. Choice A assumes specific values for cc and dd without considering the cancellation. Choice C incorrectly determines cc. Choice D gives an insufficient condition for the cancellation.

Question 2

Consider the rational function f(x)=x3+2x2−5x+1x2+x−6f(x) = \frac{x^3 + 2x^2 - 5x + 1}{x^2 + x - 6}. After factoring the denominator as (x−2)(x+3)(x-2)(x+3), determine which statement about the function's end behavior is correct.

  1. As x→∞x \to \infty, f(x)→∞f(x) \to \infty and the function approaches the line y=x+1y = x + 1 from above
  2. As x→∞x \to \infty, f(x)→∞f(x) \to \infty and the function approaches the line y=x+1y = x + 1 from below (correct answer)
  3. As x→−∞x \to -\infty, f(x)→∞f(x) \to \infty and the function approaches the line y=x+1y = x + 1 from above
  4. The function has no slant asymptote because the denominator factors completely
Explanation: First, find the slant asymptote by polynomial division. Dividing x3+2x2−5x+1x^3 + 2x^2 - 5x + 1 by x2+x−6x^2 + x - 6: The quotient is x+1x + 1 with remainder −6x+7-6x + 7. So f(x)=(x+1)+−6x+7x2+x−6f(x) = (x + 1) + \frac{-6x + 7}{x^2 + x - 6}. The slant asymptote is y=x+1y = x + 1. To determine the approach direction as x→∞x \to \infty, we examine the sign of the remainder term −6x+7x2+x−6\frac{-6x + 7}{x^2 + x - 6}. For large positive xx: numerator −6x+7<0-6x + 7 < 0 (since −6x-6x dominates), and denominator x2+x−6>0x^2 + x - 6 > 0 (since x2x^2 dominates). So the remainder term is negative, meaning f(x)<x+1f(x) < x + 1 for large positive xx. Therefore, the function approaches the slant asymptote from below as x→∞x \to \infty. Choice A has the wrong direction. Choice C examines the wrong limit direction. Choice D incorrectly states there's no slant asymptote.

Question 3

Consider the rational function f(x)=2x3−5x2+3x−1x2+x−2f(x) = \frac{2x^3 - 5x^2 + 3x - 1}{x^2 + x - 2}. After performing polynomial long division, what is the equation of the slant asymptote?

  1. y=2x−7y = 2x - 7 (correct answer)
  2. y=2x−5y = 2x - 5
  3. y=2x2−7x+10y = 2x^2 - 7x + 10
  4. y=2x+3y = 2x + 3
Explanation: To find the slant asymptote, we perform polynomial long division of the numerator by the denominator. Dividing 2x3−5x2+3x−12x^3 - 5x^2 + 3x - 1 by x2+x−2x^2 + x - 2: First term: 2x3÷x2=2x2x^3 ÷ x^2 = 2x. Multiply: 2x(x2+x−2)=2x3+2x2−4x2x(x^2 + x - 2) = 2x^3 + 2x^2 - 4x. Subtract: (2x3−5x2+3x−1)−(2x3+2x2−4x)=−7x2+7x−1(2x^3 - 5x^2 + 3x - 1) - (2x^3 + 2x^2 - 4x) = -7x^2 + 7x - 1. Second term: −7x2÷x2=−7-7x^2 ÷ x^2 = -7. Multiply: −7(x2+x−2)=−7x2−7x+14-7(x^2 + x - 2) = -7x^2 - 7x + 14. Subtract: (−7x2+7x−1)−(−7x2−7x+14)=14x−15(-7x^2 + 7x - 1) - (-7x^2 - 7x + 14) = 14x - 15. The quotient is 2x−72x - 7 with remainder 14x−1514x - 15. The slant asymptote is y=2x−7y = 2x - 7. Choice B uses the coefficient of x2x^2 incorrectly. Choice C gives the partial quotient before completing division. Choice D incorrectly adds terms.

Question 4

Which of the following rational functions does NOT have a slant asymptote?

  1. f(x)=x4−2x3+x2−5x3+2x−1f(x) = \frac{x^4 - 2x^3 + x^2 - 5}{x^3 + 2x - 1}
  2. g(x)=3x3+x2−4x+7x2−3x+2g(x) = \frac{3x^3 + x^2 - 4x + 7}{x^2 - 3x + 2}
  3. h(x)=2x5−x4+3x2−8x4+x2−5h(x) = \frac{2x^5 - x^4 + 3x^2 - 8}{x^4 + x^2 - 5}
  4. k(x)=x3−4x2+6x−9x4−2x3+x2k(x) = \frac{x^3 - 4x^2 + 6x - 9}{x^4 - 2x^3 + x^2} (correct answer)
Explanation: A rational function has a slant (oblique) asymptote when the degree of the numerator is exactly one more than the degree of the denominator. Choice A: numerator degree 4, denominator degree 3, difference = 1, so it has a slant asymptote. Choice B: numerator degree 3, denominator degree 2, difference = 1, so it has a slant asymptote. Choice C: numerator degree 5, denominator degree 4, difference = 1, so it has a slant asymptote. Choice D: numerator degree 3, denominator degree 4, difference = -1. Since the degree of the numerator is less than the degree of the denominator, this function has a horizontal asymptote at y=0y = 0, not a slant asymptote.