Answer
Step 1: $y=-x^5$
Step 2: $x=-2,0,2,5$, $f(0)=-10$.
Step 3: $x=-2,2,5$ (multiplicity 1, crosses the x-axis), $x=0$ (multiplicity 2, touches the x-axis),
Step 4: $4$.
Step 5: See graph.
Work Step by Step
Step 1: Given $f(x)=-x^5+5x^4+4x^3-20x^2=-x^4(x-5)+4x^2(x-5)=-x^2(x-5)(x^2-4)=-x^2(x-5)(x+2)(x-2)$, we can determine the end behavior as similar to $y=-x^5$
Step 2: For x-intercept(s), let $f(x)=0$, we have $x=-2,0,2,5$, for y-intercept(s), let $x=0$, we have $f(0)=-10$.
Step 3: We can determine the zeros $x=-2,2,5$ (multiplicity 1, crosses the x-axis), $x=0$ (multiplicity 2, touches the x-axis),
Step 4: The maximum number of turning points is $n-1=5-1=4$.
Step 5: See graph.
