# How do you find the domain and range of y= x^2 / (x^2-16)?

Nov 17, 2017

The domain is $x \in \left(- \infty , - 4\right) \cup \left(- 4 , 4\right) \cup \left(4 , + \infty\right)$. The range is $y \in \left(- \infty , 1\right) \cup \left(1 , + \infty\right)$

#### Explanation:

As we cannot divide by $0$, the denominator is

${x}^{2} - 16 \ne 0$

${x}^{2} \ne 16$

$x \ne 4$ and $x \ne - 4$

The domain is $x \in \left(- \infty , - 4\right) \cup \left(- 4 , 4\right) \cup \left(4 , + \infty\right)$

Calculation of the range :

$y = {x}^{2} / \left({x}^{2} - 16\right)$

$y \left({x}^{2} - 16\right) = {x}^{2}$

$y {x}^{2} - 16 y = {x}^{2}$

$y {x}^{2} - {x}^{2} = 16 y$

${x}^{2} \left(y - 1\right) = 16 y$

${x}^{2} = \frac{16 y}{y - 1}$

$x = \pm \sqrt{\frac{16 y}{y - 1}}$

Therefore,

$y \ne 1$

The range is $y \in \left(- \infty , 1\right) \cup \left(1 , + \infty\right)$

graph{x^2/(x^2-16) [-10, 10, -5, 5]}