Answer
See graph and explanations.
Work Step by Step
Step 1. Identify the domain of the function: rewrite the function as $y=\frac{(x-1)^3}{(x+2)(x-1)}=\frac{x^2-2x+1}{x+2}=\frac{x^2+2x-4x-8+9}{x+2}=x-4+\frac{9}{x+2}, x\ne1$; we can identify the domain as $(-\infty,-2)\cup(-2,1)\cup (1,\infty)$.
Step 2. Take derivatives to get $y'=1-\frac{9}{(x+2)^2}$ and $y''=\frac{18}{(x+2)^3}$.
Step 3. We can find the possible critical points as $x=-5,-2,1$ (set $y'=0$). Check the signs of $y'$ across the critical points: $..(+)..(-5)..(-)..(-2)..(-)..(1)..(+)..$, thus the function increases on $(-\infty, -5),(1,\infty)$ and decreases on $(-5,-2),(-2,1)$. A local maximum can be found at $y(-5)=--12$. There is no local minimum as the location is a hole $(1,0)$.
Step 4. Check concavity across $x=-2$ with signs of $y''$:$..(-)..(-2)..(+)..$; thus the function is concave down on $(-\infty,-2)$ and concave up on $(-2,\infty)$, but there is no inflection point as the function is not defined at $x=-2$..
Step 5. We can identify a vertical asymptote as $x=-2$, and a slant asymptote as $y=x-4$.
Step 6. The y-intercepts can be found by letting $x=0$ to get $y=1/2$.
Step 7. Based on the above results, we can graph the function as shown in the figure.
