# How to find the asymptotes of y =(1-x^2)/( x-1) ?

Feb 2, 2016

Notice how you can actually simplify this.

$y = \frac{- \left({x}^{2} - 1\right)}{x - 1}$

The numerator is a difference of two squares. ${x}^{2} - 1 = \left(x + 1\right) \left(x - 1\right)$.

$= \frac{- \left(x + 1\right) \cancel{\left(x - 1\right)}}{\cancel{x - 1}}$

$= \textcolor{b l u e}{- x - 1}$

Asymptotes are generally found when the denominator of the fractional equation would be $0$. Since there are no longer points where you have to divide by $0$, there are no asymptotes.

However, since we did just cancel out $x - 1$, we do have one removable discontinuity at $\left(1 , - 2\right)$, since the original denominator would be undefined when $x - 1 = 0$ (can't divide by $0$), and at $\textcolor{g r e e n}{x = 1}$, $\textcolor{g r e e n}{y} = - \left(1\right) - 1 = \textcolor{g r e e n}{- 2}$.