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

Oct 14, 2017

Vertical asymptote at $x = - 3$, no horizontal asymptote and
slant asymptote
$y = x - 3$

#### Explanation:

$f \left(x\right) = \frac{{x}^{2} - 8}{x + 3}$ .The vertical asymptotes will occur at those

values of x for which the denominator is equal to zero.

$\therefore x + 3 = 0 \mathmr{and} x = - 3$ .Thus, the graph will have vertical

asymptote at $x = - 3$.

To find the horizontal asymptote, here the degree of the numerator

is $2$ and the degree of the denominator is $1$. Since the larger

degree occurs in the numerator, the graph will have no horizontal

asymptote.

$$If the numerator's degree is greater (by a margin of 1), then


we have a slant asymptote which can be found by doing long

division of $\frac{{x}^{2} - 8}{x + 3}$ of which quotient is $y = x - 3 \therefore$

Slant asymptote is $y = x - 3$

graph{(x^2-8)/(x+3) [-10, 10, -5, 5]} [Ans]