# How do you find the removable discontinuity(hole) for the graph of y=(x^2 - 9x -10)/ (2x^2 - 2)?

Jul 14, 2015

Factoring the numerator and denominator reveals a common factor of $\left(x + 1\right)$ which can be removed so the (modified) fraction is no longer discontinuous at $x = - 1$

#### Explanation:

$y = \frac{{x}^{2} - 9 x - 10}{2 {x}^{2} - 2}$

Factoring:
$\textcolor{w h i t e}{\text{XXXX}}$$y = \frac{\left(x - 10\right) \left(x + 1\right)}{2 \left({x}^{2} - 1\right)}$

Continue by factoring the difference of squares in the denominator
$\textcolor{w h i t e}{\text{XXXX}}$$y = \frac{\left(x - 10\right) \left(x + 1\right)}{2 \left(x + 1\right) \left(x - 1\right)}$

Divide the numerator and denominator by $\left(x + 1\right)$
$\textcolor{w h i t e}{\text{XXXX}}$$= y = \frac{\left(x - 10\right) \cancel{\left(x + 1\right)}}{2 \cancel{\left(x + 1\right)} \left(x - 1\right)}$

$\textcolor{w h i t e}{\text{XXXX}}$$y = \frac{x - 10}{2 \left(x - 1\right)}$

Note that the discontinuity at $x = 1$ (which would cause an attempt to divide by zero) can not be removed.