Answer
(a) See the graph.
(b) $(4\sqrt 2,\frac{7\pi}{4})$
(c) $(-4\sqrt 2,\frac{3\pi}{4})$
Work Step by Step
(a) $x\gt0$ and $y\lt0$.
The point lies in Quadrant IV.
$r^2=x^2+y^2$
$r^2=(4)^2+(-4)^2=32$
$r=±4\sqrt 2$
(b) $r=4\sqrt 2$
$tan~θ=\frac{y}{x}=\frac{-4}{4}=-1$
$θ=\frac{7\pi}{4}$
(c) $r=-4\sqrt 2$
$tan~θ=\frac{y}{x}=\frac{-4}{4}=-1$
$θ=\frac{7\pi}{4}-\pi=\frac{3\pi}{4}$
