Answer
General shape:
.
Work Step by Step
Step 2
$y'=(x^{2}-2x)(x-5)^{2}=x(x-2)(x-5)^{2}$
$y''=(2x-2)(x-5)^{2}+(x^{2}-2x)\cdot 2(x-5)$
$= 2(x-5)[(x-1)(x-5)+(x^{2}-2x) ]$
$= 2(x-5)[x^{2}-5x-x+5+x^{2}-2x]$
$= 2(x-5)[2x^{2}-8x+5]$
Step 3
$y'=0$ when $ x=0,\ x=2 ,\ x=5 \qquad$... critical points.
Step 4
$\left[\begin{array}{cccccccc}
y': & ++ & | & -- & | & ++ & | & ++\\
& & 0 & & 2 & & 5 & \\
y: & \nearrow & \max & \searrow & \min & \nearrow & & \nearrow
\end{array}\right]$
Step 5
For concavity, $y''=0$ for $x=5,\ $and
$x=\displaystyle \frac{8\pm\sqrt{64-40}}{4}=\frac{8\pm 2\sqrt{6}}{4}=\frac{4\pm\sqrt{6}}{2}$
$\left[\begin{array}{ccccccccccc}
y': & -- & | & ++ & | & -- & | & ++\\
& & \frac{4-\sqrt{6}}{2} & & \frac{4+\sqrt{6}}{2} & & 5 & \\
y: & \cap & i.p. & \cup & i.p. & \cap & i.p. & \cup
\end{array}\right]$
