# How do you find domain and range for y=sqrt(x^2+4)?

Oct 12, 2015

Domain: $\mathbb{R}$
Range: $\left[2 , + \infty\right)$

#### Explanation:

Domain
The first thing you should consider is that the number inside the radical should be nonnegative so it won't be imaginary. We can compute for the domain by solving the inequality ${x}^{2} + 4 \ge 0$.

$\textcolor{w h i t e}{X X} {x}^{2} + 4 \ge 0$
$\textcolor{w h i t e}{X X} {x}^{2} \ge - 4$

Since any real number you substitute into $x$ is squared, it will always be nonnegative and greater than $- 4$. Therefore the domain is $\mathbb{R}$.

Range
The lowest value the radical can have is $\sqrt{4}$ or $2$ (when $x = 0$). Therefore, the range is $y \ge 2$ or $\left[2 , + \infty\right)$.