Answer
See graph and explanations.
Work Step by Step
Step 1. Identify the domain of the function: rewrite the function as $y=\frac{x}{x^2-1}=\frac{1}{2(x+1)}+\frac{1}{2(x-1)}$ (this decomposition is easy to see backwards); we can identify the domain as $(-\infty,-1)\cup(-1,1)\cup (1,\infty)$.
Step 2. Take derivatives to get $y'=-\frac{1}{2(x+1)^2}-\frac{1}{2(x-1)^2}$ and $y''=\frac{1}{(x+1)^3}+\frac{1}{(x-1)^3}$.
Step 3. We can find the possible critical points as $x=\pm1$. Check the signs of $y'$ across the critical points: $..(-)..(-1)..(-)..(1)..$; thus the function decreases on $(-\infty, -1),(-1,1),(1,\infty)$,and there are no extrema.
Step 4. Check concavity across $x=0,\pm1$ with signs of $y''$:$..(-)..(-1)..(+)..(0)..(-)..(1)..(+)..$; thus the function is concave down on $(-\infty,-1), (0,1)$ and concave up on $(-1,0),(1,\infty)$, and $x=0$ is an inflection point.
Step 5. We can identify a vertical asymptote as $x=\pm1$, and a horizontal asymptote as $y=0$.
Step 6. The y-intercepts can be found by letting $x=0$ to get $y(0)=0$.
Step 7. Based on the above results, we can graph the function as shown in the figure.
