# How do you find vertical, horizontal and oblique asymptotes for (x^3-8)/(x^2-5x+6)?

Jun 13, 2018

the oblique asymptote is $y = x + 5$ and the horizontal asymptote is $x = 3$

#### Explanation:

$\frac{{x}^{3} - 8}{{x}^{2} - 5 x + 6} = \frac{\cancel{x - 2} \left({x}^{2} + 2 x + 4\right)}{\left(x - 3\right) \cancel{x - 2}}$
=$\frac{{x}^{2} + 2 x + 4}{x - 3} = \frac{\left(x - 3\right) \left(x + 5\right) + 19}{x - 3} = x + 5 + \frac{19}{x - 3}$

Therefore, the oblique asymptote is $y = x + 5$ and the horizontal asymptote is $x = 3$

The horizontal asymptote is found by letting the denominator equal to 0.

The oblique asymptote is found by long division