# How do you find the vertical, horizontal and slant asymptotes of:  f(x) =(x^2+6x-9)/(x-2)?

Jul 3, 2016

V.A.: x=2, S.A.: y=x+8

#### Explanation:

The vertical asymptote depends on the domain of the function, then since it must be:

$x \ne 2$,

the vertical asymptote is the line $x = 2$

graph{(x^2+6x-9)/(x-2) [-10, 10, -5, 5]}

In fact

${\lim}_{x \rightarrow 2} \frac{{x}^{2} + 6 x - 9}{x - 2} = \frac{7}{0} = \infty$

There are no horizontal asymptote, because

${\lim}_{x \rightarrow \infty} \frac{{x}^{2} + 6 x - 9}{x - 2} = \infty$

Maybe the function has a slant asymptote, so let's calculate

$m = {\lim}_{x \rightarrow \infty} \frac{{x}^{2} + 6 x - 9}{x - 2} \cdot \frac{1}{x} = 1$

and

$n = {\lim}_{x \rightarrow 2} \frac{{x}^{2} + 6 x - 9}{x - 2} - m x =$

$= {\lim}_{x \rightarrow 2} \frac{{x}^{2} + 6 x - 9}{x - 2} - x =$

$= {\lim}_{x \rightarrow 2} \frac{{\cancel{x}}^{2} + 6 x - 9 - {\cancel{x}}^{2} + 2 x}{x - 2} =$

$= {\lim}_{x \rightarrow 2} \frac{8 x - 9}{x - 2} = 8$

So the equation of the slant asymptote is

$y = x + 8$