How do you graph f(x)=(x^2−2x−8)/(x^2−9) using holes, vertical and horizontal asymptotes, x and y intercepts?

How do you graph $f\left(x\right)=\frac{{x}^{2}-2x-8}{{x}^{2}-9}$ using holes, vertical and horizontal asymptotes, x and y intercepts?
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Asymptotes

The denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the values that x cannot be and if the numerator is non-zero for these values then they are vertical asymptotes.

solve : ${x}^{2}-9=0⇒{x}^{2}=9⇒x=±3$

$⇒x=-3\phantom{\rule{1ex}{0ex}}\text{and}\phantom{\rule{1ex}{0ex}}x=3\phantom{\rule{1ex}{0ex}}\text{are the asymptotes}$

Horizontal asymptotes occur as

$\underset{x\to ±\infty }{lim},f\left(x\right)\to c\phantom{\rule{1ex}{0ex}}\text{( a constant)}$

divide numerator/denominator by the highest power of x, that is ${x}^{2}$

$f\left(x\right)=\frac{\frac{{x}^{2}}{{x}^{2}}-\frac{2x}{{x}^{2}}-\frac{8}{{x}^{2}}}{\frac{{x}^{2}}{{x}^{2}}-\frac{9}{{x}^{2}}}=\frac{1-\frac{2}{x}-\frac{8}{{x}^{2}}}{1-\frac{9}{{x}^{2}}}$

as $x\to ±\infty ,f\left(x\right)\to \frac{1-0-0}{1-0}$

$⇒y=1\phantom{\rule{1ex}{0ex}}\text{is the asymptote}$

Holes occur when there is a duplicate factor on the numerator/denominator. This is not the case here, hence there are no holes.

Intercepts

$x=0\to y=\frac{-8}{-9}=\frac{8}{9}$

$⇒\text{y-intercept at}\phantom{\rule{1ex}{0ex}}\left(0,\frac{8}{9}\right)$

$y=0\to {x}^{2}-2x-8=0\to \left(x-4\right)\left(x+2\right)=0$

$⇒\text{x-intercepts at}\phantom{\rule{1ex}{0ex}}\left(-2,0\right)\phantom{\rule{1ex}{0ex}}\text{and}\phantom{\rule{1ex}{0ex}}\left(4,0\right)$
graph{(x^2-2x-8)/(x^2-9) [-10, 10, -5, 5]}