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