Answer
Step 1: $y=x^5$.
Step 2: x-intercepts $x=0,-4$; y-intercepts $y=0$.
Step 3: $x= 0$ (multiplicity 2, touch the x-axis) and $x=-4$ (multiplicity 1, cross the x-axis)..
Step 4: $4$.
Step 5: See graph.
Work Step by Step
Step 1: Given $f(x)=x^2(x^2+1)(x+4)$, we have $n=5, a_5\gt0$. Thus the end behavior of the graph is rise to the right and fall to the left, similar to $y=x^5$.
Step 2: To find the x-intercepts, let $f(x)=0$ to get $x=0,-4$; to find the y-intercepts, let $x=0$ to get $y=0$.
Step 3: We can find the zeros as $x= 0$ (multiplicity 2, touch the x-axis) and $x=-4$ (multiplicity 1, cross the x-axis)..
Step 4: The maximum number of turning points is $n-1=4$.
Step 5: See graph.