# How do you find vertical, horizontal and oblique asymptotes for (x^2+6x-9)/(x-2)?

Sep 29, 2016

vertical asymptote at x = 2
oblique asymptote is y = x + 8

#### 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: $x - 2 = 0 \Rightarrow x = 2 \text{ is the asymptote}$

Horizontal asymptotes occur when the degree of the numerator is ≤ degree of the denominator. This is not the case here (numerator-degree 2 , denominator-degree 1) Hence there are no horizontal asymptotes.

Oblique asymptotes occur when the degree of the numerator > degree of the denominator which is the case here.

Use $\textcolor{b l u e}{\text{polynomial division}}$ to obtain.

$f \left(x\right) = x + 8 + \frac{7}{x - 2}$

as $x \to \pm \infty , f \left(x\right) \to x + 8 + 0$

$\Rightarrow y = x + 8 \text{ is the asymptote}$
graph{(x^2+6x-9)/(x-2) [-45, 45, -22.5, 22.5]}