Answer
See graph and explanations.
Work Step by Step
Step 1. Given $y=-x^3+6x^2-9x+3$, we have $y'=-3x^2+12x-9$ and $y''=-6x+12$
Step 2. Letting $y'=0$, we have $-3x^2+12x-9=0$ and $x^2-4x+3=0$, which gives $x=1, 3$ as critical points.
Step 3. We have $y(1)=-1, y(3)=3$. Check $y''(1)=6\gt0, y''(3)=-6\lt0$; thus $y(1)$ is a local minimum and $y(3)$ is a local maximum.
Step 4. Check signs of $y'$ to get $..(-)..(1)..(+)..(3)..(-)..$; thus the function decreases on $(-\infty,1)$, $(3,\infty)$ and increases on $(1,3)$
Step 5. Letting $y''=0$, we have $x=2$ 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 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.
