Answer
Step 1: $y=x^3$.
Step 2: x-intercepts $x=-4,-1,2$; y-intercepts $y=-8$.
Step 3: $x=-4$ (multiplicity 1, cross the x-axis), $x=-1$ (multiplicity 1, cross the x-axis), and $x=2$ (multiplicity 1, cross the x-axis).
Step 4: $2$.
Step 5: See graph.
Work Step by Step
Step 1: Given $f(x)=(x+1)(x-2)(x+4)$, we have $n=3, a_3\gt0$. Thus the end behavior of the graph is rise to the right and fall to the left, similar to $y=x^3$.
Step 2: To find the x-intercepts, let $f(x)=0$ to get $x=-4,-1,2$; to find the y-intercepts let $x=0$ to get $y=-8$.
Step 3: We can find the zeros as $x=-4$ (multiplicity 1, cross the x-axis), $x=-1$ (multiplicity 1, cross the x-axis), and $x=2$ (multiplicity 1, cross the x-axis).
Step 4: The maximum number of turning points is $n-1=2$.
Step 5: See graph.
