Question

How do you find vertical, horizontal and oblique asymptotes for `(2x^3+4x-8)/(x^3-6x)`?

Answer 1

The vertical asymptotes are `x=0`, `x=sqrt6` and `x=-sqrt6`
No oblique asymptote
the horizontal asymptote is `y=2`

Explanation:

Let `f(x)=(2x^3+4x-8)/(x^3-6x)`

The denominator is

`=x^3-6x=x(x-sqrt6)(x+sqrt6)`

The domain of `f(x)` is `D_f(x)=RR-{0, +-sqrt6}`

As you cannot divide by `0`, `x!=0`, `x!=sqrt6` and `x!=-sqrt6`

So the vertical asymptotes are `x=0`, `x=sqrt6` and `x=-sqrt6`

As the degree of the numerator `=` the degree of the denominator, there is no oblique asymptote.

To calculate the limits as `x->+-oo`, we take the terms of highest degree in the numerator and the denominator.

`lim_(x->+-oo)f(x)=lim_(x->+-oo)(2x^3)/x^3=2`

So, the horizontal asymptote is `y=2`

graph{(2x^3+48-8)/(x^3-6x) [-28.86, 28.87, -14.43, 14.45]}