Answer
See graph and explanations.
Work Step by Step
Step 1. Given $y=\frac{1}{8}(x^3+3x^2-9x-27)$, we have $y'=\frac{3}{8}(x^2+2x-3)$ and $y''=\frac{3}{4}(x+1)$
Step 2. Letting $y'=0$, we have $x^2+2x-3=0$, which gives $x=-3,1$ as critical points.
Step 3. We have $y(-3)=0, y(1)=-4$. Check $y''(-3)=-\frac{3}{2}\lt0, y''(1)=\frac{3}{2}\gt0$; thus $y(-3)$ is a local maximum and $y(1)$ is a local minimum.
Step 4. Check signs of $y'$ to get $..(+)..(-3)..(-)..(1)..(+)..$; thus the function increases on $(-\infty,-3)$, $(1,\infty)$ and decreases on $(-3,1)$
Step 5. Letting $y''=0$, 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 $x\lt-1$ and concave up for $x\gt-1$.
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 $x^2(x+3)-9(x+3)=0$ or $(x+3)^2(x-3)=0$, which gives $x=-3,3$. The y-intercept can be found by letting $x=0$, which gives $y=-\frac{27}{8}$
Step 8. Graph the function based on the above information as shown in the figure.
