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