Answer
$y(1)=1$ is a local maximum and $ y(0)= y(2)=0$ is a local minima;
absolute minimum is at $y(0)=y(2)=0$ and there is no absolute maximum;
concave up on $(-\infty,0),(2,\infty)$,
concave down on $(0,2)$.
See graph.
Work Step by Step
Step 1. Given the function $y=|x^2-2x|=\begin{cases} x^2-2x,\ x\geq2\ or \ x\leq0 \\ 2x-x^2, \ -0\lt x\lt2\end{cases}$, we have $y'=\begin{cases} 2x-2,\ x\geq2\ or \ x\leq0 \\ 2-2x, \ -0\lt x\lt2\end{cases}$ and $y''=\begin{cases} 2,\ x\geq2\ or \ x\leq0 \\ -2, \ -0\lt x\lt2\end{cases}$.
Step 2. The extrema happen when $y'=0$, undefined, or at endpoints. We have $x=0,1,2$ as critical points with $ y(0)=0, y(1)=1, y(2)=0$.
Step 3. Examine the sign change of $y'$ across the critical points: $..(-)..(0)..(+)..(1)..(-)..(2)..(+)..$; we can identify $y(1)=1$ as a local maximum and $ y(0)= y(2)=0$ as local minima. The absolute minimum is at $y(0)=y(2)=0$ and there is no absolute maximum.
Step 4. The inflection points can be found when $y''=0$ or when it does not exist. We do not have possible inflection points within the domain as $y'$ is not continuous at $x=0,2$.
Step 5. To identify the intervals on which the functions are concave up and concave down, we need to examine the sign of $y''$ on different intervals. We have $..(+)..(0)..(-)..(2)..(+)..$,
Step 6. The function is concave up on $(-\infty,0),(2,\infty)$ and concave down on $(0,2)$.
Step 7. See graph.
