Answer
(a) See graph
(b) domain $(-\infty,\infty)$, range $[-16,\infty)$.
(c) decreasing on $(-\infty,0)$, increasing on $(0,\infty)$.
Work Step by Step
(a) For $y=\frac{1}{4}x^2-16$, we can determining that the graph opens up, vertex $(0,-16)$, axis of symmetry $x=0$, y-intercept $(0,-16)$, x-intercept(s) $(\pm8,0)$. See graph
(b) We can determine the domain $(-\infty,\infty)$, range $[-16,\infty)$.
(c) We can determine the function is decreasing on $(-\infty,0)$, increasing on $(0,\infty)$.
