$y=\frac{f(x)}{g(x)}=\frac{x^{2}-9}{2x^{2}-12x + 18}$

Find horizontal asymptotes by using limits as $x \rightarrow ±\infty$.

$\lim \limits_{x\rightarrow \infty }\frac{x^{2}-9}{2x^{2}-12x+18}=\frac{1}{2}$

$\lim \limits_{x\rightarrow -\infty }\frac{x^{2}-9}{2x^{2}-12x+18}=\frac{1}{2}$

$\text{Horizontal asymptote at }y=\frac{1}{2}.$

Find vertical asymptotes and holes by factoring and simplifying.

$\frac{x^{2}-9}{2x^{2}-12x+18}=\frac{(x+3)(x-3)}{2(x+3)(x-3)}$

Even though $(x − 3)$ cancels out, there remains an $(x − 3)$ in the denominator. No holes, there is a vertical asymptote at $x = 3$.

$y=\frac{g(x)}{f(x)}=\frac{2x^{2}-12x+18}{x^{2}-9}$

Find horizontal asymptotes by using limits as $x \rightarrow ±\infty$.

Find vertical asymptotes and holes by factoring and simplifying.