## College Algebra 7th Edition

$f(x)=\sqrt{9-x^{2}}, -3\leq x\leq 3$
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: $f(x)=\sqrt{9-x^{2}}, -3\leq x\leq 3$