Answer
See graph and explanations.
Work Step by Step
Step 1. Given $y=x^3-3x^2+3$, we have $y'=3x^2-6x$ and $y''=6x-6$
Step 2. Letting $y'=0$, we have $3x^2-6x=0$ and $x^2-2x=0$ which gives $x=0, 2$ as critical points.
Step 3. We have $y(0)=3, y(2)=-1$. Check $y''(0)=-6\lt0, y''(2)=6\gt0$; thus $y(0)$ is a local maximum and $y(2)$ is a local minimum.
Step 4. Check signs of $y'$ to get $..(+)..(0)..(-)..(2)..(+)..$; thus the function increases on $(-\infty,0)$, $(2,\infty)$ and decreases on $(0,2)$
Step 5. Let $y''=0$; we have $x=1$. Check the signs of $y''$ to get $..(-)..(1)..(+)..$ indicating $x=1$ is an inflection point with the function concave down for $x\lt1$ and concave up for $x\gt1$.
Step 6. The end behaviors of the functions are $x\to-\infty, y\to-\infty$ and $x\to\infty, y\to\infty$
Step 7. Graph the function based on the above information as shown in the figure.
