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