Question

How do you find the asymptote(s) or holes of `f(x) = (x^2-1) / ((x-1)(x^3-2x^2+2x-1)`?

Answer 1

`f(x)` has a horizontal asymptote `y=0` and a vertical asymptote `x=1`. It has no slant asymptote or any holes.

Explanation:

Note that:

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

The quadratic factor `x^2-x+1` is always positive, as you can see by checking its discriminant, or by completing the square:

`x^2-x+1 = (x-1/2)^2+3/4`

So we find:

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

`color(white)(f(x)) = (color(red)(cancel(color(black)((x-1))))(x+1))/(color(red)(cancel(color(black)((x-1))))(x-1)(x^2-x+1))`

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

Note that the denominator is of higher degree than the numerator. Hence we can deduce that `f(x)` has a horizontal asymptote `y=0`.

The only factor in the denominator that can be zero is `(x-1)`, occurring when `x=1`, when the numerator (in the simplified expression) is non-zero. We can deduce that `f(x)` has a vertical asymptote `x=1`.

The factors that we cancelled were also at `x=1`, so would have resulted in a hole instead of an asymptote if the extra factor in the denominator were not present. As it is, we have a vertical asymptote there.

There are no slant (a.k.a. oblique) asymptotes. Those only happen when the degree of the numerator exceeds the denominator by exactly `1`.

graph{(x^2-1)/((x-1)(x^3-2x^2+2x-1)) [-12.66, 12.65, -6.33, 6.33]}