Answer
Step 1: $y=-2x^4$
Step 2: $x=-2,2$, $f(0)=-32$.
Step 3: $x=-2$ (multiplicity 1, crosses the x-axis), $x=2$ (multiplicity 3, crosses the x-axis)
Step 4: $3$.
Step 5: See graph.
Work Step by Step
Step 1: Given $f(x)=-2(x+2)(x-2)^3$, we can determine the end behavior as similar to $y=-2x^4$
Step 2: For x-intercept(s), let $f(x)=0$, we have $x=-2,2$, for y-intercept(s), let $x=0$, we have $f(0)=-32$.
Step 3: We can determine the zeros $x=-2$ (multiplicity 1, crosses the x-axis), $x=2$ (multiplicity 3, crosses the x-axis)
Step 4: The maximum number of turning points is $n-1=4-1=3$.
Step 5: See graph.
