Answer
Step 1: $y=-x^3$.
Step 2: x-intercepts $x=-4,1$; y-intercepts $y=16$.
Step 3: $x=-4$ (multiplicity 2, touch the x-axis) and $x=1$ (multiplicity 1, cross the x-axis).
Step 4: $2$.
Step 5: See graph.
Work Step by Step
Step 1: Given $f(x)=(x+4)^2(1-x)$, we have $n=3, a_3\lt0$. Thus the end behavior of the graph is fall to the right and rise to the left, similar to $y=-x^3$.
Step 2: To find the x-intercepts, let $f(x)=0$ to get $x=-4,1$; to find the y-intercepts let $x=0$ to get $y=16$.
Step 3: We can find the zeros as $x=-4$ (multiplicity 2, touch the x-axis) and $x=1$ (multiplicity 1, cross the x-axis).
Step 4: The maximum number of turning points is $n-1=2$.
Step 5: See graph.
