Answer
Step 1: $y=x^4$.
Step 2: x-intercepts $x=-4,\pm2$; y-intercepts $y=32$.
Step 3: $x= 2$ (multiplicity 2, touch the x-axis) and $x=-4,-2$ (multiplicity 1, cross the x-axis)..
Step 4: $3$.
Step 5: See graph.
Work Step by Step
Step 1: Given $f(x)=(x-2)^2(x+2)(x+4)$, we have $n=4, a_4\gt0$. Thus the end behavior of the graph is rise to the right and rise to the left, similar to $y=x^4$.
Step 2: To find the x-intercepts, let $f(x)=0$ to get $x=-4,\pm2$; to find the y-intercepts, let $x=0$ to get $y=32$.
Step 3: We can find the zeros as $x= 2$ (multiplicity 2, touch the x-axis) and $x=-4,-2$ (multiplicity 1, cross the x-axis)..
Step 4: The maximum number of turning points is $n-1=3$.
Step 5: See graph.