# What is the domain and range of (x^2+2)/(x+4)?

Apr 6, 2018

The domain is $x \in \mathbb{R} - \left\{- 4\right\}$. The range is $y \in \left(- \infty , - 16.485\right] \cup \left[0.485 , + \infty\right)$

#### Explanation:

The denominator is $\ne 0$

$x + 4 \ne 0$

$x \ne - 4$

The domain is $x \in \mathbb{R} - \left\{- 4\right\}$

To find the range, proceed as follws

Let $y = \frac{{x}^{2} + 2}{x + 4}$

$y \left(x + 4\right) = {x}^{2} + 2$

${x}^{2} - y x + 2 - 4 y = 0$

This is a quadratic equation in ${x}^{2}$ and in order to have solutions

the discriminant $\Delta \ge 0$

Therefore

$\Delta = {\left(- y\right)}^{2} - 4 \left(1\right) \left(2 - 4 y\right) \ge 0$

${y}^{2} - 16 y - 8 \ge 0$

The solutions are

$y = \frac{- 16 \pm \sqrt{{\left(- 16\right)}^{2} - 4 \left(1\right) \left(- 8\right)}}{2} = \frac{- 16 \pm 16.97}{2}$

${y}_{1} = - 16.485$

${y}_{2} = 0.485$

The range is $y \in \left(- \infty , - 16.485\right] \cup \left[0.485 , + \infty\right)$

graph{(x^2+2)/(x+4) [-63.34, 53.7, -30.65, 27.85]}