Answer
See graph and explanations.
Work Step by Step
Step 1. Given $y=x^3(8-x)=8x^3-x^4$, we have $y'=24x^2-4x^3$ and $y''=48x-12x^2$
Step 2. Letting $y'=0$, we have $6x^2-x^3=0$, which gives $x=0,6$ as critical points.
Step 3. We have $y(0)=0, y(6)=432$. Check $y''(0)=0, y''(6)=-144\lt0$; thus only $y(6)$ is a local minimum.
Step 4. Check signs of $y'$ to get $..(+)..(0)..(+)..(6)..(-)..$; thus the function increases on $(-\infty,0)$, $(0,6)$ and decreases on $(6,\infty)$
Step 5. Letting $y''=0$, we have $x=0,4$. Check the signs of $y''$ to get $..(-)..(0)..(+)..(4)..(-)..$, indicating $x=0,4$ are inflection points with the function concave down for $x\lt0, x\gt4$ and concave up for $0\lt x\lt4$.
Step 6. The end behaviors of the functions are $x\to-\infty, y\to-\infty$ and $x\to\infty, y\to-\infty$
Step 7. The x-intercept can be found by letting $y=0$ which gives $x^3(8-x)=0$ which gives $x=0,8$. The y-intercept can be found by letting $x=0$ which gives $y=0$
Step 8. Graph the function based on the above information as shown in the figure.
