Answer
Step 1: $y=-\frac{1}{2}x^4$
Step 2: $x=-4,1$, $f(0)=2$.
Step 3: $x=-4$ (multiplicity 1, crosses the x-axis), $x=1$ (multiplicity 3, crosses the x-axis)
Step 4: $3$.
Step 5: See graph.
Work Step by Step
Step 1: Given $f(x)=-\frac{1}{2}(x+4)(x-1)^3$, we can determine the end behavior as similar to $y=-\frac{1}{2}x^4$
Step 2: For x-intercept(s), let $f(x)=0$, we have $x=-4,1$, for y-intercept(s), let $x=0$, we have $f(0)=2$.
Step 3: We can determine the zeros $x=-4$ (multiplicity 1, crosses the x-axis), $x=1$ (multiplicity 3, crosses the x-axis)
Step 4: The maximum number of turning points is $n-1=4-1=3$.
Step 5: See graph.
