#### Answer

$f(x)=-\sqrt{9−x^2}, -3\leq x\leq 3$

#### Work Step by Step

We are given
$y=x^2+y^2=9=3^2$
This describes a circle with radius 3 centered at (0,0).
We solve for y:
$y^2=9−x^2$
$y=\pm\sqrt{9−x^2}$
We want the bottom half of the circle, so we take the negative:
$y=-\sqrt{9−x^2}$
Thus the function is:
$f(x)=-\sqrt{9−x^2}, -3\leq x\leq 3$