Slant (Oblique) Asymptotes - College Algebra

1. The function $g(x) = \frac{ax^3 + bx^2 + cx + d}{x^2 - 4}$ has a slant asymptote with equation $y = 3x + 2$. If the function has a vertical asymptote at $x = 2$ but not at $x = -2$, what must be true about the coefficients?
   Answer: $a = 3$, $b = 2$, and the numerator has $(x + 2)$ as a factor 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.
2. Consider the rational function $f(x) = \frac{2x^3 - 5x^2 + 3x - 1}{x^2 + x - 2}$. After performing polynomial long division, what is the equation of the slant asymptote?
   Answer: $y = 2x - 7$ 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.
3. Which of the following rational functions does NOT have a slant asymptote?
   Answer: $k(x) = \frac{x^3 - 4x^2 + 6x - 9}{x^4 - 2x^3 + x^2}$ 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.
4. Consider the rational function $f(x) = \frac{x^3 + 2x^2 - 5x + 1}{x^2 + x - 6}$. After factoring the denominator as $(x-2)(x+3)$, determine which statement about the function's end behavior is correct.
   Answer: As $x \to \infty$, $f(x) \to \infty$ and the function approaches the line $y = x + 1$ from below 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.

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