Answer
See graph and explanations.
Work Step by Step
Step 1. Identify the domain of the function: rewriting the function as $y=-\frac{x^2-1-3}{x+1}=-x+1+\frac{3}{x+1}$, we can identify the domain as $(-\infty,-1)\cup (-1,\infty)$.
Step 2. Take derivatives to get $y'=-1-\frac{3}{(x+1)^2}$ and $y''=\frac{6}{(x+1)^3}$.
Step 3. We can find the possible critical points as $x=-2, -1,2$. Check the signs of $y'$ across the critical points: $..(-)..(-2)..(-)..(-1)..(-)..(2)..(-)..$; thus the function decreases for all x in the domain. This function has no extrema.
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 $y(0)=4$ and the x-intercepts can be found at $x=\pm2$.
Step 7. Based on the above results, we can graph the function as shown in the figure.
