Question

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

Answer 1

Vertical asymtotes are `x=0 , x = 4/3`. No H.A.
Oblique asymptote is `y = 1/3x+1/9`

Explanation:

`y= (x^3-x^2-10)/(3x^2-4x) = (x^3-x^2-10)/(x(3x-4))`

Vertical asymtotes are `x=0` , `3x-4=0 or x = 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 = 1/3x+1/9`

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