# How do you find the domain and range for y =sqrt(9-x^2)?

Feb 18, 2016

Domain: $x \in \left[- 3 , + 3\right]$
Range: $y \in \left[0 , 3\right]$

#### Explanation:

I assume that we are dealing only with Real numbers.

For Real numbers $\sqrt{a}$ is only defined if $a \ge 0$
$\rightarrow \left(9 - {x}^{2}\right) \ge 0$
$\rightarrow \left\mid x \right\mid \le 3$
which gives us the Domain.

The maximum value of $\sqrt{9 - {x}^{2}}$ occurs when $x = 0$ and
this maximum is $\sqrt{9} = 3$
The minimum value of $\sqrt{9 - {x}^{2}}$ occurs when $x = \pm 3$ and
this minimum is $\sqrt{9 - 9} = 0$
which together give us the Range.