College Algebra Quiz: Slant Oblique Asymptotes

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.

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.

Question 1

1. $a = 3$, $b = 2$, and $d = -16$
2. $a = 3$, $b = 2$, and the numerator has $(x + 2)$ as a factor (correct answer)
3. $a = 3$, $c = 8$, and the numerator has $(x + 2)$ as a factor
4. $a = 3$, $b = 2$, and $c + d = 0$

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

Question 2

1. As $x \to \infty$, $f(x) \to \infty$ and the function approaches the line $y = x + 1$ from above
2. As $x \to \infty$, $f(x) \to \infty$ and the function approaches the line $y = x + 1$ from below (correct answer)
3. As $x \to -\infty$, $f(x) \to \infty$ and the function approaches the line $y = x + 1$ from above
4. The function has no slant asymptote because the denominator factors completely

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

Question 3

1. $y = 2x - 7$ (correct answer)
2. $y = 2x - 5$
3. $y = 2x^2 - 7x + 10$
4. $y = 2x + 3$

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

Question 4

1. $f(x) = \frac{x^4 - 2x^3 + x^2 - 5}{x^3 + 2x - 1}$
2. $g(x) = \frac{3x^3 + x^2 - 4x + 7}{x^2 - 3x + 2}$
3. $h(x) = \frac{2x^5 - x^4 + 3x^2 - 8}{x^4 + x^2 - 5}$
4. $k(x) = \frac{x^3 - 4x^2 + 6x - 9}{x^4 - 2x^3 + x^2}$ (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 $y = 0$, not a slant asymptote.