Answer
See graph and explanations.
Work Step by Step
Step 1. Identify the domain of the function $y=\frac{x-1}{x^2(x-2)}$ as $(-\infty,0)\cup(0,2)\cup (2,\infty)$.
Step 2. Rewrite the function as $y=\frac{1}{4(x-2)}-\frac{1}{4x}+\frac{1}{2x^2}$ (this decomposition may take a few extra steps, but makes it easier to calculate derivatives). Take derivatives to get $y'=-\frac{1}{4(x-2)^2}-\frac{1}{4x^2}-\frac{1}{x^3}$ and $y''=\frac{1}{2(x-2)^3}+\frac{1}{2x^3}+\frac{3}{x^4}$.
Step 3. We can test the possible critical points at $x=0,1,2$. Check the signs of $y'$ across these points: $..(+)..(0)..(-)..(1)..(-)..(2)..(-)..$; thus the function increases on $(-\infty, 0)$ and decreases on $(0,2),(2,\infty)$, and there are no extrema.
Step 4. Check concavity across $x=0,1,2$ with signs of $y''$:$..(+)..(0)..(+)..(1)..(-)..(2)..(+)..$; thus the function is concave up on $(-\infty,0), (0,1), (2,\infty)$ and concave up on $(1,2),$, and $x=1$ is an inflection point.
Step 5. We can identify vertical asymptotes as $x=0$ and $x=2$, and a horizontal asymptote as $y=0$.
Step 6. The function has no y-intercepts but has a x-intercept at $x=1$.
Step 7. Based on the above results, we can graph the function as shown in the figure.
