# How do you find the domain and range of (y^2)(x-4)=8?

Jan 23, 2018

Given: $\left({y}^{2}\right) \left(x - 4\right) = 8$

${y}^{2} = \frac{8}{x - 4}$

$y = \pm \frac{2 \sqrt{2}}{\sqrt{x - 4}}$

Separate into two equations:

$y = - \frac{2 \sqrt{2}}{\sqrt{x - 4}}$ and $y = \frac{2 \sqrt{2}}{\sqrt{x - 4}}$

In both cases, the expression under the radical cannot be negative and it must be greater than 0:

$x - 4 > 0$

$x > 4 \leftarrow$ this is the domain for both equations.

Find the range for the first equation:

${\lim}_{x \to 4} - \frac{2 \sqrt{2}}{\sqrt{x - 4}} = - \infty$

${\lim}_{x \to \infty} - \frac{2 \sqrt{2}}{\sqrt{x - 4}} = 0$

The range for the first equation is $- \infty < y < 0$

${\lim}_{x \to 4} \frac{2 \sqrt{2}}{\sqrt{x - 4}} = \infty$

${\lim}_{x \to \infty} \frac{2 \sqrt{2}}{\sqrt{x - 4}} = 0$

The range for the second equation is $0 < y < \infty$