Question

How do you identify all asymptotes or holes and intercepts for `f(x)=(x^3-1)/((2(x^2-1))`?

Answer 1

There is a hole at `x=1`
A vertical asymptote is `x=-1`
The slant asymptote is `y=x/2`
There is no horizontal asymptote.

Explanation:

Let's factorise the denominator

`2(x^2-1)= 2(x+1)(x-1)`

Let's factorise the numerator
`x^3-1=(x-1)(x^2+x+1)`

Therefore,

`f(x)=(x^3-1)/(2(x+1)(x-1))=(cancel(x-1)(x^2+x+1))/(2(x+1)cancel(x-1))`

`=(x^2+x+1)/(2(x+1))`

There is a hole at `x=1`

The domain of `f(x)`is `D_f(x)`is `RR-{-1}`

As we cannot divide by `0`, `x!=-1`

A vertical asymptote is `x=-1`

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

Let's do a long division

`color(white)(aaaa)` `x^2+x+1` `color(white)(aaaa)` `∣` `2x+2`

`color(white)(aaaa)` `x^2+x` `color(white)(aaaaaaa)` `∣` `x/2`

`color(white)(aaaaa)` `0+0+1`

So, `(x^2+x+1)/(2(x+1))=x/2+1/(2x+2)`

The slant asymptote is `y=x/2`

To calculate the limit, we take the terms of highest degree.

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

There is no horizontal asymptote.

graph{(y-(x^2+x+1)/(2x+2))(y-x/2)=0 [-5.542, 5.553, -2.783, 2.766]}