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