Answer
Step 1: $y=-x^3$
Step 2: $x=-2,-1,3$, $f(0)=6$.
Step 3: $x=-2,-1,3$ (multiplicity 1, crosses the x-axis),
Step 4: $2$.
Step 5: See graph.
Work Step by Step
Step 1: Given $f(x)=(3-x)(2+x)(x+1)=-(x-3)(x+2)(x+1)$, we can determine the end behavior as similar to $y=-x^3$
Step 2: For x-intercept(s), let $f(x)=0$, we have $x=-2,-1,3$, for y-intercept(s), let $x=0$, we have $f(0)=6$.
Step 3: We can determine the zeros $x=-2,-1,3$ (multiplicity 1, crosses the x-axis),
Step 4: The maximum number of turning points is $n-1=3-1=2$.
Step 5: See graph.
