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