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