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