Answer
Step 1: $y=-x^3$.
Step 2: x-intercepts $x=0,\pm1$; y-intercepts $y=0$.
Step 3: $x=0,\pm1$ (all multiplicity 1, cross the x-axis)..
Step 4: $2$.
Step 5: See graph.
Work Step by Step
Step 1: Given $f(x)=x-x^3=-x(x+1)(x-1)$, 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 $x=0,\pm1$; to find the y-intercepts, let $x=0$ to get $y=0$.
Step 3: We can find the zeros as $x=0,\pm1$ (all multiplicity 1, cross the x-axis)..
Step 4: The maximum number of turning points is $n-1=2$.
Step 5: See graph.
