Answer
Step 1: $y=-x^5$.
Step 2: x-intercepts $x=0,\pm2, 5$; y-intercept $y=0$.
Step 3: $x=0$ (multiplicity 2, touch the x-axis) and $x=\pm2, 5$ (all with multiplicity 1, cross the x-axis).
Step 4: $4$.
Step 5: See graph.
Work Step by Step
Step 1: Given $f(x)=-x^5+5x^4+4x^3-20x^2=-x^4(x-5)+4x^2(x-5)=-x^2(x-5)(x^2-4)=-x^2(x-5)(x+2)(x-2)$,
the end behavior of the graph is similar to $y=-x^5$.
Step 2: To find the x-intercepts, let $f(x)=0$ to get $x=0,\pm2, 5$; to find the y-intercepts, let $x=0$ to get $y=0$.
Step 3: We can find the zeros as $x=0$ (multiplicity 2, touch the x-axis) and $x=\pm2, 5$ (all with multiplicity 1, cross the x-axis).
Step 4: The maximum number of turning points is $n-1=4$.
Step 5: See graph.