# How do you find the asymptotes for h(x) = (x^3-8)/(x^2-5x+6)?

Feb 15, 2017

Slant asymptote { #y=x+7.
Vertical asymptote : x = 3.

#### Explanation:

graph{((x^2+4x+4)/(x-3)-y)(y-x-7.9)(x-2+.01y)=0 [-45, 45, -20, 25]}

The graph is a hyperbola

$\left(y - x - 7\right) \left(x - 3\right) = 15$ having asymptotes given by

$\left(y - x - 7\right) \left(x - 3\right) = 0$, with a hole at (2, -16)

$h = \left(\frac{x - 2}{x - 2}\right) \left(\frac{{x}^{2} + 4 x + 4}{x - 3}\right)$.

Sans the hole at x = 2,

$h = \frac{{x}^{2} + 4 x + 4}{x - 3}$

$= x + 7 + \frac{25}{x - 3}$.

So, $y = q u o t i e n t = x + 7$ gives the slant asymptote and

$x - 3 = 0$ gives the vertical asymptote.