Answer
See graph and explanations.
Work Step by Step
Step 1. Given $y=\frac{x^2-4}{x^2-3}=1-\frac{1}{x^2-3}$, we have $y'=\frac{2x}{(x^2-3)^2}$
Step 2. Letting $y'=0$, we have $x=0$; the function is undefined at $x=\pm\sqrt 3$, so the critical points are $x=0, \pm\sqrt 3$
Step 3. We have the $y'$ sign changes as $..(-)..(-\sqrt 3)..(-)..(0)..(+)..(\sqrt x)..(+)..$ which gives decreasing regions of $(-\infty,-\sqrt 3), (-\sqrt 3,0)$ and increasing regions $(0,\sqrt 3),(\sqrt 3,\infty)$. There is a local minimum at $(0,\frac{4}{3})$.
Step 4.The function has three asymptotes: $x=\pm\sqrt 3, y=1$
Step 5. Graph the function based on the above information as shown in the figure (red curve).
