Answer
(a) $x=-\frac{1}{2}$ (multiplicity 2) and $x=-4$ (multiplicity 3).
(b) $x=-4$ crosses the x-axis, $x=-\frac{1}{2}$ touches the x-axis.
(c) $4$.
(d) $y=-2x^5$.
Work Step by Step
(a) For $f(x)=-2(x+\frac{1}{2})^2(x+4)^3$, we can list real zero as $x=-\frac{1}{2}$ (multiplicity 2) and $x=-4$ (multiplicity 3).
(b) At $x=-4$ the graph crosses the x-axis, while at $x=-\frac{1}{2}$ the graph touches the x-axis.
(c) The maximum number of turning points on the graph is given by $n-1=5-1=4$.
(d) As $n=5, a_5\lt0$, the end behaviors are fall to the right and rise to the left similar to $y=-2x^5$.