Answer
(a) $x=-\frac{1}{2}$ (multiplicity 2), $x=-4$ (multiplicity 3).
(b) crosses the x-axis at $x=-4$, touches at $x=-\frac{1}{2}$.
(c) $4$.
(d) $y=-2x^5$.
Work Step by Step
(a) Given $f(x)=-2(x+\frac{1}{2})^2(x+4)^3$, we can find zero(s) $x=-\frac{1}{2}$ (multiplicity 2), $x=-4$ (multiplicity 3).
(b) The graph crosses the x-axis at $x=-4$, touches at $x=-\frac{1}{2}$.
(c) The maximum number of turning points on the graph is $n-1=5-1=4$.
(d) The end behavior is similar to those of $y=-2x^5$.
