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