Answer
See graph and explanations.
Work Step by Step
Step 1. Given $y=x\sqrt {3-x}=x(3-x)^{1/2}, x\leq3$, we have $y'=(3-x)^{1/2}-\frac{1}{2}x(3-x)^{-1/2}=(3-x)^{-1/2}(3-\frac{3}{2}x)$ and $y''=-\frac{3}{2}(3-x)^{-1/2}+\frac{1}{2}(3-x)^{-3/2}(3-\frac{3}{2}x)=(3-x)^{-3/2}(\frac{3}{4}x-3)$
Step 2. Letting $y'=0$, we have $(3-x)^{-1/2}(3-\frac{3}{2}x)=0$, which gives $x=2$; also $x=0$ is undefined in $y'$ and $x=3$ is an endpoint. Thus $x=0,2,3$ are possible critical points. We have $y(0)=0, y(2)=2, y(3)=0$.
Step 3. Check signs of $y'$ to get $..(+)..(0)..(+)..(2)..(-)..(3)$; thus the function increases on $(-\infty,0)$, $(0,2)$ and decreases on $(2,3)$ . This also means that $y(2)=2$ is a local maximum (it is also an absolute maximum) and $y(3)=0$ is a local minimum.
Step 4. Letting $y''=0$, we have $x=4$, which is outside the domain. We also need to consider $x=0$, where $y''$ is undefined. Check the signs of $y''$ to get $..(-)..(0)..(-)..(3)$, indicating no inflection points with the function concave down for the entire domain.
Step 5. The end behaviors of the functions are $x\to-\infty, y\to-\infty$.
Step 6. The x-intercept can be found by letting $y=0$, which gives $x=0,3$. The y-intercept can be found by letting $x=0$ which gives $y=0$.
Step 7. Graph the function based on the above information as shown in the figure.
