Question

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

Answer 1

The vertical asymptote is `x=-1`
The intercepts with the x axis are `(0,0)` and `(2,0)`

Explanation:

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

As we cannot divide by `0`, the denominator `!=0`
`x^3+1!=0` `=>` `x^3!=-1` `=>` `x!=-1`
So x=-1 is a vertical asymptote

To see the intercept with the x axis,
`f(x)=0`
`(x^2-2x)/(x^3+1)=0`
`x^2-2x=0`
`x(x-2)=0`
so `x=0` and `x=2`
The points are `(0,0)` and `(2,0)`

We can find the limits at `+-oo`
limit f(x) = `x^2/x^3=1/x=0`
`xrarr-oo`

limit f(x) = `x^2/x^3=1/x=0`
`xrarr+oo`