Answer
Step 1: $y=x^3$
Step 2: $x=-3,1$, $f(0)=-9$.
Step 3: $x=1$ (multiplicity 1, crosses the x-axis), $x=-3$ (multiplicity 2, touches the x-axis)
Step 4: $2$.
Step 5: See graph.
Work Step by Step
Step 1: Given $f(x)=(x-1)(x+3)^2$, we can determine the end behavior as similar to $y=x^3$
Step 2: For x-intercept(s), let $f(x)=0$, we have $x=-3,1$, for y-intercept(s), let $x=0$, we have $f(0)=-9$.
Step 3: We can determine the zeros $x=1$ (multiplicity 1, crosses the x-axis), $x=-3$ (multiplicity 2, touches the x-axis)
Step 4: The maximum number of turning points is $n-1=3-1=2$.
Step 5: See graph.
