Answer
See graph and explanations.
Work Step by Step
Step 1. Rewrite the function as $y=\frac{x}{2}-\frac{2}{x}; x\ne0$, we can identify a vertical asymptote at $x=0$, a slant asymptote $y=\frac{x}{2}$, and the domain can be found to be all real numbers except $x\ne0$.
Step 2. Take derivatives to get $y'=\frac{1}{2}+\frac{2}{x^2}$ and $y''=-\frac{4}{x^3}$. Critical points can be found as $x=0, \pm2$. Check the signs of $y'$ across the critical points: $..(+)..(-2)..(-)..(0)..(-)..(2)..(+)..$; thus the function decreases on $(-2,0),(0,2)$ and increases on $(-\infty,-2),(2,\infty)$. A local maximum can be found as $y(-2)=-2$ and a local minimum can be found as $y(2)=2$.
Step 3. Check concavity across $x=0$ with signs of $y''$:$..(-)..(0)..(+)..$; thus the function is concave down on $(-\infty,0)$ and concave up on $(0,\infty)$, but there is no inflection point as the function is not defined at $x=0$.
Step 4. There are no x- or y-intercepts as $x\ne0$ and $y\ne0$.
Step 5. Based on the above results, we can graph the function as shown in the figure.
