Answer
General shape:
.
Work Step by Step
Step 2
$y'=x(x^{2}-12)=x^{3}-12x$
$y''=3x^{2}-12=3(x^{2}-4)=3(x+2)(x-2)$
Step 3
$y'=0$ when $x=0,\ x=\pm\sqrt{12}=\pm 2\sqrt{3}$
$ x=0,\ \pm 2\sqrt{3}\qquad$... critical points.
Step 4
$\left[\begin{array}{ccccccccccccc}
y': & -- & | & ++ & | & -- & | & ++\\
& & -2\sqrt{3} & & 0 & & 2\sqrt{3} & \\
y: & \searrow & \min & \nearrow & \max & \searrow & \min & \nearrow
\end{array}\right]$
Step 5
For concavity, $y''=0$ for $x^{2}=\pm 2$
$\left[\begin{array}{cccccccc}
y': & ++ & | & -- & | & ++ & \\
& & -2 & & 2 & & \\
y: & \cup & i.p. & \cap & i.p. & \cup &
\end{array}\right]$
