Answer
See graph and explanations.
Work Step by Step
Step 1. Identify the domain of the function: rewriting the function as $y=\frac{x^2-1+1}{x^2-1}=1+\frac{1}{x^2-1}=1+\frac{1}{(x+1)(x-1)}$, we can identify the domain as $(-\infty,-1)\cup (-1,1)\cup (1,\infty)$. The function is an even function and symmetric with respect to the $y$-axis ($f(-x)=f(x)$).
Step 2. Take derivatives to get $y'=-\frac{2x}{(x^2-1)^2}$ and $y''=\frac{-2(x^2-1)^2+4x(x^2-1)(2x)}{(x^2-1)^4}=\frac{2(3x^2+1)}{(x^2-1)^3}$.
Step 3. We can find the possible critical points as $x=0, \pm1$. Check the signs of $y'$ across the critical points: $..(+)..(-1)..(+)..(0)..(-)..(1)..(-)..$; thus the function increases on $(-\infty, -1),(-1,0)$ and decreases on $(0,1),(1,\infty)$. A local maximum can be found at $y(0)=0$. There is no absolute maximum or minimum.
Step 4. Check concavity across $x=\pm1$ with signs of $y''$:$..(+)..(-1)..(-)..(1)..(+)..$; thus the function is concave down on $(-1,1)$ and concave up on $(-\infty,-1),(1,\infty)$, but there is no inflection point as the function is not defined at $x=\pm1$..
Step 5. We can identify vertical asymptotes as $x=\pm1$, and a horizontal asymptote as $y=1$.
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.
