Answer
Step 1: $y=x^3$.
Step 2: $x=-4,2$, $f(0)=16$,
Step 3: $x=-4$ (multiplicity 1, crosses the x-axis), $x=2$ (multiplicity 2, touches the x-axis).
Step 4: $2$
Step 5: See graph.
Work Step by Step
Step 1: For $f(x)=(x-2)^2(x+4)$, we can determine the end behavior of the graph as similar to $y=x^3$.
Step 2: We can find the x-intercept(s) (let $y=0$) $x=-4,2$, y-intercept(s) (let $x=0$) $f(0)=16$,
Step 3: We can determine the zero(s) $x=-4$ (multiplicity 1, crosses the x-axis), $x=2$ (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.
![](https://gradesaver.s3.amazonaws.com/uploads/solution/736743bb-1571-4e89-ac61-4bca11c72b0a/steps_image/small_1630507286.jpg?X-Amz-Algorithm=AWS4-HMAC-SHA256&X-Amz-Credential=AKIAJVAXHCSURVZEX5QQ%2F20240908%2Fus-east-1%2Fs3%2Faws4_request&X-Amz-Date=20240908T003423Z&X-Amz-Expires=900&X-Amz-SignedHeaders=host&X-Amz-Signature=d2904bc57bf671e60abe03cb9d3db8c90656b8db62a2e0fc9542bbee18d32113)