Question

find all vertical asymptotes, x = a, of the following function. for each value of a, evaluate lim f(x) as x→a⁺, lim f(x) as x→a⁻, and lim f(x) as x→a. use ∞ or -∞ when appropriate. f(x)=(x³ - 9x² + 18x)/(x² - 6x). find all vertical asymptotes. then evaluate lim f(x) as x→a⁺, lim f(x) as x→a⁻, and lim f(x) as x→a. select the correct choice below, and fill in the answer box if necessary. a. the vertical asymptote is x = . the limits at this vertical asymptote are lim f(x) as x→a⁺= , lim f(x) as x→a⁻= , and lim f(x) as x→a= . b. the vertical asymptote is x = . the limits at this vertical asymptote are lim f(x) as x→a⁺= , lim f(x) as x→a⁻= , and lim f(x) as x→a does not exist and is neither ∞ nor -∞. c. there is no vertical asymptote.

Explanation:

Step1: Factor the function

First, factor the numerator $x^{3}-9x^{2}+18x=x(x - 3)(x - 6)$ and the denominator $x^{2}-6x=x(x - 6)$. Then $f(x)=\frac{x(x - 3)(x - 6)}{x(x - 6)}=\frac{x - 3}{1}$ for $x
eq0,x
eq6$. The function is undefined at $x = 0$ and $x=6$. But after simplification, the hole is at $x = 6$ (since we can cancel out the common factors $(x - 6)$ and $x$ for non - zero values of $x$) and the vertical asymptote is at $x=0$.

Step2: Evaluate the limits

For $x
ightarrow0^{+}$:
$\lim_{x
ightarrow0^{+}}f(x)=\lim_{x
ightarrow0^{+}}\frac{x^{3}-9x^{2}+18x}{x^{2}-6x}=\lim_{x
ightarrow0^{+}}\frac{x - 3}{1}=- 3$.
For $x
ightarrow0^{-}$:
$\lim_{x
ightarrow0^{-}}f(x)=\lim_{x
ightarrow0^{-}}\frac{x^{3}-9x^{2}+18x}{x^{2}-6x}=\lim_{x
ightarrow0^{-}}\frac{x - 3}{1}=-3$.
For $x
ightarrow0$:
$\lim_{x
ightarrow0}f(x)=\lim_{x
ightarrow0}\frac{x^{3}-9x^{2}+18x}{x^{2}-6x}=\lim_{x
ightarrow0}\frac{x - 3}{1}=-3$.

Answer:

A. The vertical asymptote is $x = 0$. The limits at this vertical asymptote are $\lim_{x
ightarrow0^{+}}f(x)=-3$, $\lim_{x
ightarrow0^{-}}f(x)=-3$, and $\lim_{x
ightarrow0}f(x)=-3$