Answer
$y(1)=2$ is a local maximum and $y(3)=6$ is a local minimum; no absolute extrema;
no inflection points;
concave down on $(-\infty,2)$ and concave up on $(2,\infty)$.
See graph.
Work Step by Step
Step 1. Given the function $y=\frac{x^2-3}{x-2}, x\ne2$, we have $y'=\frac{2x(x-2)-(x^2-3)}{(x-2)^2}=\frac{x^2-4x+3}{(x-2)^2}$ and $y''=\frac{(x-2)^2(2x-4)-2(x^2-4x+3)(x-2)}{(x-2)^4}=\frac{2(x-2)(x^2-4x+4-x^2+4x-3)}{(x-2)^4}=\frac{2}{(x-2)^3}$
Step 2. The extrema happen when $y'=0$, undefined, or at endpoints. We have $x=1,3$ as critical points with $ y(1)=2, y(3)=6$
Step 3. Examine the sign change of $y'$ across the critical points: $..(+)..(1)..(-)..(2)..(-)..(3)..(+)..$; we can identify $y(1)=2$ as a local maximum and $y(3)=6$ as a local minimum. The function has no absolute extrema as when $x\to\pm\infty, y\to\pm\infty$
Step 4. The inflection points can be found when $y''=0$ or when it does not exist. We do not have any inflection points.
Step 5. To identify the intervals on which the functions are concave up and concave down, we need to examine the sign of $y''$ on different intervals. We have $..(-)..(2)..(+)..$,
Step 6. The function is concave down on $(-\infty,2)$ and concave up on $(2,\infty)$.
Step 7. See graph.
