Answer
See graph and explanations.
Work Step by Step
Step 1. Rewrite the function as $y=\frac{2x^2+x-1}{x^2-1}=\frac{(2x-1)(x+1)}{(x-1)(x+1)}=\frac{2x-1}{x-1}=2+\frac{1}{x-1}, x\ne1$, we can identify a hole at $x=-1$, a vertical asymptote at $x-1$ and a horizontal asymptote at $y=2$. The domain can be found to be all real numbers except $x\ne\pm1$.
Step 2. Take derivatives to get $y'=-\frac{1}{(x-1)^2}$ and $y''=\frac{2}{(x-1)^3}$. Since $y'\lt0$ for all values in the domain, the function will be decreasing for all $x$. With $y''\gt0$ the function will be concave up when $x\gt1$, For $y''\lt0$ the function will be concave down when $x\lt1$, and $x=1$ is not an inflection point as the function is not continuous at the point.
Step 3. The y-intercepts can be found by letting $x=0$ to get $y=1$, and the x-intercepts can be found by letting $y=0$ to get $x=1/2$
Step 4. Based on the above results, we can graph the function as shown in the figure.
