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