Answer
See graph and explanations.
Work Step by Step
Step 1. Given $y=x^2(2x^2-9)=2x^4-9x^2$, we have $y'=8x^3-18x$ and $y''=24x^2-18$
Step 2. Letting $y'=0$, we have $8x^3-18x=0$, which gives $x=0,\pm\frac{3}{2}$ as critical points. We have $y(-\frac{3}{2})=-\frac{81}{8}, y(0)=0, y(\frac{3}{2})=-\frac{81}{8}$.
Step 3. Check signs of $y'$ to get $..(-)..(-\frac{81}{8})..(+)..(0)..(-)..(\frac{81}{8})..(+)..$; thus the function increases on $(-\frac{81}{8},0)$, $(\frac{81}{8},\infty)$ and decreases on $(-\infty,-\frac{81}{8}),(0,\frac{81}{8})$. This also means that $y(-\frac{3}{2})=-\frac{81}{8}, y(\frac{3}{2})=-\frac{81}{8}$ are local minima and $y(0)=0$ is a local maximum.
Step 5. Letting $y''=0$, we have $x=\pm\frac{\sqrt 3}{2}$. Check the signs of $y''$ to get $..(+)..(-\frac{\sqrt 3}{2})..(=)..(\frac{\sqrt 3}{2})..(+)..$, indicating $x=\pm\frac{\sqrt 3}{2}$ is an inflection point with the function concave down for $-\frac{\sqrt 3}{2}\lt x\lt\frac{\sqrt 3}{2}$ and concave up for $x\lt-\frac{\sqrt 3}{2}$ and $x\gt\frac{\sqrt 3}{2}$.
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 $x^2(2x^2-9)=0$ which gives $x=0,\pm\frac{3\sqrt 2}{2}$. 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.
