Answer
See graph and explanations.
Work Step by Step
Step 1. Given $y=x^2-\frac{x^3}{6}$, we have $y'=2x-\frac{x^2}{2}$ and $y''=2-x$
Step 2. Letting $y'=0$, we have $2x-\frac{x^2}{2}=0$ and $x^2-4x=0$ which gives $x=0, 4$ as critical points.
Step 3. We have $y(0)=0, y(4)=\frac{16}{3}$. Check $y''(0)=2\gt0, y''(4)=-2\lt0$; thus $y(0)$ is a local minimum and $y(4)$ is a local maximum.
Step 4. Check signs of $y'$ to get $..(-)..(0)..(+)..(4)..(-)..$; thus the function decreases on $(-\infty,0)$, $(4,\infty)$ and increases on $(0,4)$
Step 5. Letting $y''=0$, we have $x=2$ and check the signs of $y''$ to get $..(+)..(2)..(-)..$ indicating $x=2$ is an inflection point with the function concave up for $x\lt2$ and concave down for $x\gt2$.
Step 6. The x-intercept can be found by letting $y=0$, which gives $x=0, 6$, and the y-intercept can be found by letting $x=0$ to give $y=0$.
Step 7. The end behaviors of the functions are $x\to-\infty, y\to\infty$ and $x\to\infty, y\to-\infty$
Step 8. Graph the function based on the above information as shown in the figure.
