Answer
General shape:
Work Step by Step
Step 2
$y'=x(x-3)^{2}=x(x^{2}-6x+9)=x^{3}-6x^{2}+9x$
$y''=3x^{2}-12x+9=3(x^{2}-4x+3)=3(x-1)(x-3)$
Step 3
$y'=0$ when $ x=0,3\qquad$... critical points.
Step 4
$\left[\begin{array}{cccccccccc}
y': & & -- & | & ++ & | & ++ & \\
& (-\infty & & 0 & & 3 & & \infty)\\
y: & & \searrow & min & \nearrow & & \nearrow &
\end{array}\right]$
Step 5
For concavity,
$y''=0$ for $x=1, 3$
$\left[\begin{array}{cccccccccccc}
y': & & ++ & & -- & | & ++ & \\
& (-\infty & & 1 & & 3 & & \infty)\\
y: & & \cup & i.p. & \cap & i.p. & \cup &
\end{array}\right]$
