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 top half of the circle, so we take the positive:
$y=\sqrt{9-x^{2}}$
Thus the function is:
$f(x)=\sqrt{9-x^{2}}, -3\leq x\leq 3$
