# How do I find the range of these functions without using a graph y=-2+sqrt(16-x^2)?

Jun 9, 2018

$- 2 \le - 2 + \sqrt{16 - {x}^{2}} \le 2$

#### Explanation:

Since $\sqrt{16 - {x}^{2}} \ge 0$ we get
$- 2 \le - 2 + \sqrt{16 - {x}^{2}}$
Since $- {x}^{2} \le 0$ we get

$\sqrt{16 - {x}^{2}} \le 4$
or
$- 2 + \sqrt{16 - {x}^{2}} \le 2$