Answer
See graph and explanations.
Work Step by Step
(a) Step 1. Given $y'=6x(x+1)(x-2)$, we have $y''=6(x+1)(x-2)+6x(x-2)+6x(x+1)=6(3x^2-2x-2)$
Step 2. Letting $y'=0$, we have $6x(x+1)(x-2)=0$, which gives $x=-1,0,2$ as possible critical points.
Step 3. Check the signs of $y'$ to get $..(-)..(-1)..(+)..(0)..(-)..(2)..(+)..$; thus the function decreases on $(-\infty,-1)$, $(0,2)$ and increases on $(-1,0)$, $(2,\infty)$ . This means that $x=-1,2$ give a local minima and $x=0$ gives a local maximum.
Step 4. Letting $y''=0$, we have $x=\frac{2\pm\sqrt {4+24}}{6}=\frac{1\pm\sqrt {7}}{3}$. Check the signs of $y''$ to get $..(+)..(\frac{1-\sqrt {7}}{3})..(-)..(\frac{1+\sqrt {7}}{3})..(+)..$, indicating $x=\frac{1\pm\sqrt {7}}{3}$ are inflection points with the function concave up for $x\lt\frac{1-\sqrt {7}}{3}, x\gt\frac{1+\sqrt {7}}{3}$, and concave down for $\frac{1-\sqrt {7}}{3}\lt x\lt \frac{1+\sqrt {7}}{3}$.
(b) Graph the function based on the above information as shown in the figure.
