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

Vertical Asymptote: $x = 3$
Oblique Asymptote: $y = x + 5$
Horizontal Asymptote: None

#### Explanation:

To obtain the oblique asymptote, we divide by long division

$\frac{{x}^{3} - 8}{{x}^{2} - 5 x + 6} = x + 5 + \frac{19}{x - 3}$

the linear part (x+5) of the quotient will used for the oblique asymptote, that is

$y = x + 5$

The following includes the oblique asymptote $y = x + 5$ and the given $y = \frac{{x}^{3} - 8}{{x}^{2} - 5 x + 6}$

graph{(y-(x^3-8)/(x^2-5x+6))(y-x-5)=0[-80,80,-40,40]}

God bless....I hope the explanation is useful.