# How do you find domain and range for y=sqrt((4 - x²))?

Refer to explanation

#### Explanation:

Since its a square root we have that

$4 - {x}^{2} \ge 0 \implies \left(2 - x\right) \cdot \left(2 + x\right) \ge 0$ which holds for $\left[- 2 , 2\right]$ hence

$D \left(f\right) = \left[- 2 , 2\right]$

The range is

$y = \sqrt{4 - {x}^{2}} \implies {y}^{2} = 4 - {x}^{2} \implies {x}^{2} = 4 - {y}^{2} \ge 0 \implies \left(2 - y\right) \left(2 + y\right) \ge 0$

Because $y \ge 0$ we have that $R \left(f\right) = \left(0 , 2\right)$