Answer
Step 1: $y=4x^3$.
Step 2: x-intercepts $x=-\frac{5}{2},\pm1$; y-intercepts $y=-10$.
Step 3: $x=-\frac{5}{2},\pm1$ (all with multiplicity 1, cross the x-axis)..
Step 4: $2$.
Step 5: See graph.
Work Step by Step
Step 1: Given $f(x)=4x^3+10x^2-4x-10=2x^2(2x+5)-2(2x+5)=2(2x+5)(x^2-1)=2(2x+5)(x+1)(x-1)$,
the end behavior of the graph is similar to $y=4x^3$.
Step 2: To find the x-intercepts, let $f(x)=0$ to get $x=-\frac{5}{2},\pm1$; to find the y-intercepts, let $x=0$ to get $y=-10$.
Step 3: We can find the zeros as $x=-\frac{5}{2},\pm1$ (all with multiplicity 1, cross the x-axis)..
Step 4: The maximum number of turning points is $n-1=2$.
Step 5: See graph.
