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

Feb 3, 2017

see explanation.

#### Explanation:

$\textcolor{b l u e}{\text{Asymptotes}}$

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 : ${x}^{2} - 9 = 0 \Rightarrow {x}^{2} = 9 \Rightarrow x = \pm 3$

$\Rightarrow x = - 3 \text{ and " x=3" are the asymptotes}$

Horizontal asymptotes occur as

${\lim}_{x \to \pm \infty} , f \left(x\right) \to c \text{ ( a constant)}$

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

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

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

$\Rightarrow y = 1 \text{ is the asymptote}$

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

$\textcolor{b l u e}{\text{Intercepts}}$

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

$\Rightarrow \text{y-intercept at } \left(0 , \frac{8}{9}\right)$

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

$\Rightarrow \text{x-intercepts at "(-2,0)" and } \left(4 , 0\right)$
graph{(x^2-2x-8)/(x^2-9) [-10, 10, -5, 5]}