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