Answer
See graph and explanations.
Work Step by Step
Step 1. Given the first derivative
$y'=\begin{cases} -x^2,\ x\leq0 \\x^2,\ x\gt0 \end{cases}$,
we have $y''=\begin{cases} -2x,\ x\leq0 \\2x,\ x\gt0 \end{cases}$
Step 2. A possible critical point is at $x=0$. Determine the increasing ($y'\gt0$) and decreasing ($y'\lt0$) regions by signs of $y'$: $..(-)..(0)..(+)..$; thus the function decreases on $(-\infty,0)$ and increases on $(0,\infty)$. A local and absolute minimum can be found at $x=0$. There is no local or absolute maximum.
Step 3. Use signs of $y''$ to determine concavity and possible inflection points at $x=0$: $..(+)..(0)..(+)..$. Thus the function is concave up on $(-\infty,0)$ and $(0,\infty)$, and $x=0$ is not an inflection point.
Step 4. Sketch the function based on the above results as shown.
