Question

How do you find the asymptotes for `f(x) =(x^3-1)/(x^2-6)`?

Answer 1

The vertical asymptotes are `x=-sqrt6` and `x=sqrt6`
The slant asymptote is `y=x`
No horizontal asymptote

Explanation:

Let's factorise the denominator

`x^2-6=(x-sqrt6)(x+sqrt6)`

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

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

The vertical asymptotes are `x=-sqrt6` and `x=sqrt6`

As the degree of the numerator is `>` than the degree of the denominator, there is a slant asymptote.

We perform a long division

`color(white)(aaaa)` `x^3` `color(white)(aaaaa)` `-1` `color(white)(aaaaa)` `∣` `x^2-6`

`color(white)(aaaa)` `x^3-6x` `color(white)(aaaa)` ###color(white)(aaaa)#`∣` `x`

`color(white)(aaaaa)` `0+6x-1` `color(white)(aaaa)`

Therefore,

`f(x)=(x)+(6x-1)/(x^2-6)`

`lim_(x->-oo)(f(x)-x)=lim_(x->-oo)(6x-1)/(x^2-6)=lim_(x->-oo)(6x)/x^2=lim_(x->-oo)6/x=0^-`

`lim_(x->+oo)(f(x)-x)=lim_(x->+oo)(6x-1)/(x^2-6)=lim_(x->+oo)(6x)/x^2=lim_(x->+oo)6/x=0^+`

So, the slant asymptote is `y=x`

graph{(y-(x^3-1)/(x^2-6))(y-x)=0 [-28.86, 28.86, -14.43, 14.45]}