Answer
Step 1: $y=-x^3$
Step 2: $x=0,\pm2$, $f(0)=0$.
Step 3: $x=0,\pm2$ (multiplicity 1, crosses the x-axis),
Step 4: $2$.
Step 5: See graph.
Work Step by Step
Step 1: Given $f(x)=4x-x^3=-x(x^2-4)=-x(x+2)(x-2)$, we can determine the end behavior as similar to $y=-x^3$
Step 2: For x-intercept(s), let $f(x)=0$, we have $x=0,\pm2$, for y-intercept(s), let $x=0$, we have $f(0)=0$.
Step 3: We can determine the zeros $x=0,\pm2$ (multiplicity 1, crosses the x-axis),
Step 4: The maximum number of turning points is $n-1=3-1=2$.
Step 5: See graph.
