Answer
See graph and explanations.
Work Step by Step
Step 1. Identify the domain of the function: rewriting the function as $y=\frac{x^2-2-2}{x^2-2}=1-\frac{2}{x^2-2}=1-\frac{2}{(x+\sqrt 2)(x-\sqrt 2)}$, we can identify the domain as $(-\infty,-\sqrt 2)\cup (-\sqrt 2,\sqrt 2)\cup (\sqrt 2,\infty)$. The function is an even function and is symmetric with respect to the $y$-axis ($f(-x)=f(x)$).
Step 2. Take derivatives to get $y'=\frac{4x}{(x^2-2)^2}$ and $y''=\frac{4(x^2-2)^2-8x(x^2-2)(2x)}{(x^2-2)^4}=-\frac{4(3x^2+2)}{(x^2-2)^3}$.
Step 3. We can find the possible critical points as $x=0, \pm\sqrt 2$. Check the signs of $y'$ across the critical points: $..(-)..(-\sqrt 2)..(-)..(0)..(+)..(\sqrt 2)..(+)..$; thus the function decreases on $(-\infty, -\sqrt 2),(-\sqrt 2,0)$ and increasess on $(0,\sqrt 2),(\sqrt 2,\infty)$. A local minimum can be found at $y(0)=2$. There is no absolute maximum or minimum.
Step 4. Check concavity across $x=\pm\sqrt 2$ with signs of $y''$:$..(-)..(-\sqrt 2)..(+)..(\sqrt 2)..(-)..$; thus the function is concave up on $(-\sqrt 2,\sqrt 2)$ and concave down on $(-\infty,-\sqrt 2),(\sqrt 2,\infty)$, but there is no inflection point as the function is not defined at $x=\pm\sqrt 2$..
Step 5. We can identify vertical asymptotes as $x=\pm\sqrt 2$, and a horizontal asymptote as $y=1$.
Step 6. The x-intercepts can be found by letting $y=0$ to get $x=\pm2$, and the y-intercept is $y(0)=2$.
Step 7. Based on the above results, we can graph the function as shown in the figure.
