Answer
Step 1: $y=-2x^3$.
Step 2: $x=0,2$, $f(0)=0$,
Step 3: $x=2$ (multiplicity 1, crosses the x-axis), $x=0$ (multiplicity 2, touches the x-axis).
Step 4: $2$
Step 5: See graph.
Work Step by Step
Step 1: For $f(x)=-2x^3+4x^2=-2x^2(x-2)$, we can determine the end behavior of the graph as similar to $y=-2x^3$.
Step 2: We can find the x-intercept(s) (let $y=0$) $x=0,2$, y-intercept(s) (let $x=0$) $f(0)=0$,
Step 3: We can determine the zero(s) $x=2$ (multiplicity 1, crosses the x-axis), $x=0$ (multiplicity 2, touches the x-axis).
Step 4: We can determine the maximum number of turning points $n-1=3-1=2$
Step 5: See graph.
