Question

let f(x) = \\(\frac{x^2 - 9x + 18}{x^2 - 3x}\\) complete parts (a) through (c) below. \\(\lim\limits_{x \to 0^-} f(x) = \boxed{\infty}\\) (simplify your answer) \\(\lim\limits_{x \to 0^+} f(x) = \boxed{-\infty}\\) (simplify your answer) \\(\lim\limits_{x \to 3^-} f(x) = \boxed{-1}\\) (simplify your answer) \\(\lim\limits_{x \to 3^+} f(x) = \boxed{-1}\\) (simplify your answer) b. does the graph of f have any vertical asymptotes? select the correct choice below and, if necessary, fill in the answer box(es) to complete your choice. \\(\bigcirc\\) a. the graph of f has one vertical asymptote at \\(\square\\) (type an equation) \\(\bigcirc\\) b. the graph of f has two vertical asymptotes. the leftmost asymptote is \\(\square\\) and the rightmost asymptote is \\(\square\\) (type equations) \\(\bigcirc\\) c. the graph of f has no vertical asymptotes.

Explanation:

Part b: Vertical Asymptotes

To determine vertical asymptotes, we first factor the function:
$f(x) = \frac{x^2 - 9x + 18}{x^2 - 3x} = \frac{(x - 3)(x - 6)}{x(x - 3)}$.

Simplify (cancel $(x - 3)$ for $x
eq 3$):
$f(x) = \frac{x - 6}{x}$ (for $x
eq 0, 3$).

Vertical asymptotes occur where the denominator is zero (and the numerator is not zero there).

• Denominator $x = 0$: Numerator at $x = 0$ is $0 - 6 = -6

eq 0$, so $x = 0$ is a vertical asymptote.

• At $x = 3$, the function is undefined (hole, not an asymptote, since $(x - 3)$ cancels).

For vertical asymptotes, factor the function, cancel common terms, and check where the simplified denominator is zero (and numerator non - zero). Here, after simplifying $f(x)=\frac{x - 6}{x}$ (for $x
eq0,3$), the denominator is zero at $x = 0$ (numerator is non - zero here), and $x = 3$ is a hole. So there is one vertical asymptote at $x = 0$.

Answer:

A. The graph of \( f \) has one vertical asymptote at \( \boldsymbol{x = 0} \)