Answer
General shape:
Work Step by Step
Step 2
$y'=x^{2}-x-6=(x-3)(x+2)$
$y''=2x-1$
Step 3
$y'=0$ when $ x=-2,3\qquad$... critical points.
Step 4
The graph of $y'$ is a parabola that opens up, so it is negative between the zeros, and positive on the other two intervals.
$ \left[\begin{array}{ccccccc}
y': & & ++ & | & -- & | & ++ & \\
& (-\infty & & -2 & & 3 & & \infty)\\
y: & & \nearrow & & \searrow & & \nearrow &
\end{array}\right]$
Step 5
For concavity,
$y''=2x-1,\ \qquad y''=0$ for $x=\displaystyle \frac{1}{2}$
$ \left[\begin{array}{ccccc}
y': & & -- & | & ++ & \\
& (-\infty & & 1/2 & & \infty)\\
y: & & \cap & infl & \cup &
\end{array}\right]$
