Question

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

Answer 1

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

Explanation:

`(x^3-8)/(x^2-5x+6)=(cancel(x-2)(x^2+2x+4))/((x-3)cancel(x-2))`
=`(x^2+2x+4)/(x-3)=((x-3)(x+5)+19)/(x-3)=x+5+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