# What are the asymptote(s) and hole(s), if any, of  f(x) =(1-x)^2/(x^2-1)+(1-x)^2/(x^2-2x+1)?

##### 1 Answer
Dec 29, 2015

Asymptotes are $y = 2 , x = 1 \mathmr{and} x = - 1$

#### Explanation:

The solutions are easier to find if the expression is first simplified.
$f \left(x\right) = {\left(1 - x\right)}^{2} / \left(\left(x - 1\right) \left(x + 1\right)\right) + {\left(1 - x\right)}^{2} / {\left(x - 1\right)}^{2}$
$f \left(x\right) = {\left(1 - x\right)}^{2} / \left(- \left(1 - x\right) \left(x + 1\right)\right) + {\left(1 - x\right)}^{2} / {\left(- \left(1 - x\right)\right)}^{2}$
$f \left(x\right) = - \frac{1 - x}{x + 1} + 1$
$f \left(x\right) = \frac{- \left(1 - x\right) + \left(x + 1\right)}{x + 1}$
$f \left(x\right) = \frac{2 x}{x + 1}$
$L i m f \left(x\right) x \to \infty = 2$ giving the horizontal asymptoteas $y = 2$

Going back to the original form of the equation, the denominator approaches zero as x approaches $1$ or $- 1$ so therefore there are vertical asymptotes at $x = - 1$ and $x = 1$