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