# How do you graph y=-8/(x^2-4) using asymptotes, intercepts, end behavior?

Jan 11, 2017

see explanation.

#### Explanation:

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

The denominator of y cannot be zero as this would make y 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 : ${x}^{2} - 4 = 0 \Rightarrow {x}^{2} = 4 \Rightarrow x = \pm 2$

$\Rightarrow x = - 2 \text{ and " x=2" are the asymptotes}$

Horizontal asymptotes occur as

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

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

$y = - \frac{\frac{8}{x} ^ 2}{{x}^{2} / {x}^{2} - \frac{4}{x} ^ 2} = - \frac{\frac{8}{x} ^ 2}{1 - \frac{4}{x} ^ 2}$

as $x \to \pm \infty , y \to - \frac{0}{1 - 0}$

$\Rightarrow y = 0 \text{ is the asymptote}$

Oblique asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here ( numerator-degree 0,denominator-degree 2 ) Hence there are no oblique asymptotes.

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

$x = 0 \to y = - \frac{8}{- 4} = 2 \Rightarrow \left(0 , 2\right)$

$y = 0 \text{ has no solution, hence y does not cross the x-axis}$

$\textcolor{b l u e}{\text{end behaviour}}$

$\text{we already know, from above that } {\lim}_{x \to \pm \infty} f \left(x\right) = 0$

$\textcolor{b l u e}{\text{behaviour around vertical asymptotes}}$

${\lim}_{x \to - {2}^{-}} f \left(x\right) = - \infty \text{ and } {\lim}_{x \to - {2}^{+}} f \left(x\right) = + \infty$

${\lim}_{x \to {2}^{-}} = + \infty \text{ and } {\lim}_{x \to {2}^{+}} = - \infty$
graph{-8/(x^2-4) [-10, 10, -5, 5]}