# How do you find the asymptotes for f(x) = (x^2 - 8)/(x+3)?

Dec 28, 2016

$\text{vertical asymptote at } x = - 3$
$\text{oblique asymptote is } y = x - 3$

#### Explanation:

The denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the value that x cannot be and if the numerator is non-zero for this value then it is a vertical asymptote.

solve : $x + 3 = 0 \Rightarrow x = - 3 \text{ is the asymptote}$

Horizontal asymptotes occur as

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

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

$f \left(x\right) = \frac{{x}^{2} / {x}^{2} - \frac{8}{x} ^ 2}{\frac{x}{x} ^ 2 + \frac{3}{x} ^ 2} = \frac{1 - \frac{8}{x} ^ 2}{\frac{1}{x} + \frac{3}{x} ^ 2}$

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

This is undefined hence there are no horizontal asymptotes.

Oblique asymptotes occur when the degree of the numerator > degree of the denominator. This is the case here hence there is an oblique asymptote.

Using $\textcolor{b l u e}{\text{polynomial division}}$

$f \left(x\right) = x - 3 + \frac{1}{x + 3}$

as $x \to \pm \infty , f \left(x\right) \to x - 3$

$\Rightarrow y = x - 3 \text{ is the asymptote}$
graph{(x^2-8)/(x+3) [-20, 20, -10, 10]}