## Functions Modeling Change: A Preparation for Calculus, 5th Edition

$$(2\sqrt{2},\:-\frac{\pi }{6})$$
Converting the Cartesian coordinates to polar coordinates, we find: $$r=\sqrt{\left(\sqrt{6}\right)^2+\left(-\sqrt{2}\right)^2} = 2\sqrt{2} \\ θ=\arctan \left(\frac{-\sqrt{2}}{\sqrt{6}}\right) = θ=-\frac{\pi }{6}$$ Thus, it follows: $$(2\sqrt{2},\:-\frac{\pi }{6})$$