Question

How do you identify all asymptotes or holes for `g(x)=(x^3+3x^2-16x+12)/(x-2)`?

Answer 1

The numerator is divisible by the denominator
`g(x)=(x+6)(x-1)`
There are no asymptotes or holes

Explanation:

Let's do a long division

`x^3+3x^2-16x+12` `color(white)(aaa)` `∣` `x-2`
`x^3-2x^2` `color(white)(aaaaaaaaaaaaa)` `∣` `x^2+5x-6`
`0+5x^2-16x`
`color(white)(aaaaa)` `0-10x`
`color(white)(aaaaaaaa)` `-6x+12`
`color(white)(aaaaaaaaaaa)` `0+0`

So `g(x)=(x^3+3x^2-16x+12)/(x-2)=x^2+5x-6`

We can factorise so `g(x)=g(x)=(x+6)(x-1)`

This is a parabola
there are no asymptotes or holes

`g(x)` is defined on `RR`