Question

How do you find the vertical, horizontal and slant asymptotes of: `f(x)=(2x - 2 )/( (x-1)(x^2 + x - 1))`?

Answer 1

The vertical asymptotes are `x=(-1-sqrt5)/2`
and `x=(-1+sqrt5)/2`
The horizontal asymptote is `y=0`
There is no slant symptote

Explanation:

You can start by simplifying `f(x)`
`f(x)=(2x-2)/((x-1)(x^2+x-1))=(2(cancel(x-1)))/(cancel(x-1)(x^2+x-1))`
So `f(x)=2/(x^2+x-1)`
As we cannot divide by `0`, so `x^2+x-1!=0`
The roots of this equation is `(-b+-sqrt(b^2-4ac))/(2a)`
So the roots are `(-1+-sqrt(1+5))/2=(-1+-sqrt5)/2`
So the vertical asymptotes are `x=(-1-sqrt5)/2`
and `x=(-1+sqrt5)/2`

And the limit of `f(x)` as `x->oo` is `0`
so the horizontal asymptote is `y=0`
graph{2/(x^2+x-1) [-7.474, 8.33, -4.27, 3.63]}