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

Feb 14, 2018

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

#### Explanation:

Note that:

${x}^{3} - 2 {x}^{2} + 2 x - 1 = \left(x - 1\right) \left({x}^{2} - x + 1\right)$

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

${x}^{2} - x + 1 = {\left(x - \frac{1}{2}\right)}^{2} + \frac{3}{4}$

So we find:

$f \left(x\right) = \frac{{x}^{2} - 1}{\left(x - 1\right) \left({x}^{3} - 2 {x}^{2} + 2 x - 1\right)}$

$\textcolor{w h i t e}{f \left(x\right)} = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{\left(x - 1\right)}}} \left(x + 1\right)}{\textcolor{red}{\cancel{\textcolor{b l a c k}{\left(x - 1\right)}}} \left(x - 1\right) \left({x}^{2} - x + 1\right)}$

$\textcolor{w h i t e}{f \left(x\right)} = \frac{x + 1}{\left(x - 1\right) \left({x}^{2} - x + 1\right)}$

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

The only factor in the denominator that can be zero is $\left(x - 1\right)$, occurring when $x = 1$, when the numerator (in the simplified expression) is non-zero. We can deduce that $f \left(x\right)$ 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]}