Answer
Step 1: $y=4x^3$
Step 2: $x=-\frac{5}{2},\pm1$, $f(0)=-10$.
Step 3: $x=-\frac{5}{2},\pm1$ (multiplicity 1, crosses the x-axis),
Step 4: $2$.
Step 5: See graph.
Work Step by Step
Step 1: Given $f(x)=4x^3+10x^2-4x-10=2x^2(2x+5)-2(2x+5)=2(2x+5)(x^2-1)=2(2x+5)(x+1)(x-1)$, we can determine the end behavior as similar to $y=4x^3$
Step 2: For x-intercept(s), let $f(x)=0$, we have $x=-\frac{5}{2},\pm1$, for y-intercept(s), let $x=0$, we have $f(0)=-10$.
Step 3: We can determine the zeros $x=-\frac{5}{2},\pm1$ (multiplicity 1, crosses the x-axis),
Step 4: The maximum number of turning points is $n-1=3-1=2$.
Step 5: See graph.
