Vertical and Horizontal Asymptotes - College Algebra

1. For the rational function $f(x) = \frac{(x-1)^2(x+2)}{(x-3)(x+2)^2}$, what are the vertical asymptotes and holes?
   Answer: Vertical asymptote at $x = 3$; hole at $x = -2$ To find vertical asymptotes and holes, examine where the denominator equals zero and check for cancellation with the numerator. The denominator $(x-3)(x+2)^2$ equals zero when $x = 3$ or $x = -2$. At $x = 3$: The numerator $(x-1)^2(x+2) = (2)^2(5) = 20 \neq 0$, so there's a vertical asymptote. At $x = -2$: Both numerator and denominator have factor $(x+2)$. The numerator has $(x+2)^1$ and denominator has $(x+2)^2$. After canceling one factor of $(x+2)$, we get $f(x) = \frac{(x-1)^2}{(x-3)(x+2)}$ for $x \neq -2$. The remaining denominator still has factor $(x+2)$, so this creates a hole at $x = -2$, not a vertical asymptote. Choice B ignores the partial cancellation at $x = -2$. Choice C reverses the locations. Choice D treats both as vertical asymptotes.
2. Given $f(x) = \frac{x^2 - 9}{x^2 - x - 6}$, which statement about the asymptotes is correct?
   Answer: There is a vertical asymptote at $x = -2$ and a hole at $x = 3$, with horizontal asymptote $y = 1$ Factor the numerator and denominator: $x^2 - 9 = (x-3)(x+3)$ and $x^2 - x - 6 = (x-3)(x+2)$. So $f(x) = \frac{(x-3)(x+3)}{(x-3)(x+2)}$. The factor $(x-3)$ cancels, leaving $f(x) = \frac{x+3}{x+2}$ for $x \neq 3$. This creates a hole at $x = 3$. The simplified function has a vertical asymptote where the remaining denominator is zero: $x + 2 = 0$, so $x = -2$. For the horizontal asymptote, both numerator and denominator of the simplified form have degree 1, so $y = \frac{1}{1} = 1$. 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.
3. The function $f(x) = \frac{6x^4 - 3x^3 + x^2}{2x^4 + 5x^2 - 7}$ has a horizontal asymptote at $y = k$. What is the value of $k$?
   Answer: $k = 3$ For rational functions where the numerator and denominator have the same degree, the horizontal asymptote is $y = \frac{\text{leading coefficient of numerator}}{\text{leading coefficient of denominator}}$. Both the numerator $6x^4 - 3x^3 + x^2$ and denominator $2x^4 + 5x^2 - 7$ 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 $y = \frac{6}{2} = 3$. 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.
4. The function $h(x) = \frac{3x^3 - 12x^2}{x^2 - 16}$ can be written in a simplified form after factoring. What happens to the asymptotes when the function is simplified?
   Answer: There is one vertical asymptote at $x = -4$ and an oblique asymptote Factor the numerator: $3x^3 - 12x^2 = 3x^2(x - 4)$. Factor the denominator: $x^2 - 16 = (x-4)(x+4)$. So $h(x) = \frac{3x^2(x-4)}{(x-4)(x+4)}$. The factor $(x-4)$ cancels, giving $h(x) = \frac{3x^2}{x+4}$ for $x \neq 4$. This creates a hole at $x = 4$ and a vertical asymptote at $x = -4$ (where $x + 4 = 0$). 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.

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