# How do you find vertical, horizontal and oblique asymptotes for y=(x^3-x^2-10)/(3x^2-4x)?

May 22, 2017

Vertical asymtotes are $x = 0 , x = \frac{4}{3}$. No H.A.
Oblique asymptote is $y = \frac{1}{3} x + \frac{1}{9}$

#### Explanation:

$y = \frac{{x}^{3} - {x}^{2} - 10}{3 {x}^{2} - 4 x} = \frac{{x}^{3} - {x}^{2} - 10}{x \left(3 x - 4\right)}$

Vertical asymtotes are $x = 0$ , $3 x - 4 = 0 \mathmr{and} x = \frac{4}{3}$

Since degree of numerator is greater than degree of numerator there is no horizontal asymtote.

By long division we can find oblique asymptote as $y = \frac{1}{3} x + \frac{1}{9}$

graph{(x^3-x^2-10)/(3x^2-4x) [-40, 40, -20, 20]} [Ans]