Answer
Step 1: $y=x^3$.
Step 2: x-intercepts $x=0,-4,3$; y-intercepts $y=0$.
Step 3: $x=0,-4,3$ (all with multiplicity 1, cross the x-axis)..
Step 4: $2$.
Step 5: See graph.
Work Step by Step
Step 1: Given $f(x)=x^3+x^2-12x=x(x^2+x-12)=x(x+4)(x-3)$, the end behavior of the graph is similar to $y=x^3$.
Step 2: To find the x-intercepts, let $f(x)=0$ to get $x=0,-4,3$; to find the y-intercepts, let $x=0$ to get $y=0$.
Step 3: We can find the zeros as $x=0,-4,3$ (all with multiplicity 1, cross the x-axis)..
Step 4: The maximum number of turning points is $n-1=2$.
Step 5: See graph.
