Finding values for $a$ and $b$ such that the function is continuous everywhere. $$f(x)=\begin{cases}
\frac{x^2-1}{x+1},&\text{if }x<-4\\\\
ax^2+2x+b,&\text{if }-1\le x<0\\\\
|x+a+2|,&\text{if }x\ge 0\;.
\end{cases}$$
I usually can solve these kinds of questions but the fact that the function is not defined on $(-4,-1)$ throws me off.

Note that $$\frac{x^2-1}{x+1}=x-1$$ if $x\ne-1$, so the definition of $f$ can be simplified to 
$$f(x)=\begin{cases}
x-1,&\text{if }x<-4\\
ax^2+2x+b,&\text{if }-1\le x<0\\
|x+a+2|,&\text{if }x\ge 0\;.
\end{cases}$$
This function is certainly continuous on the open ray $(\leftarrow,-4)$. We don’t have to worry about continuity at any point of $[-4,-1)$, because those points aren’t in the domain of the function. Thus, the problem really boils down to choosing $a$ and $b$ so that the function
$$g(x)=\begin{cases}
ax^2+2x+b,&\text{if }-1\le x<0\\
|x+a+2|,&\text{if }x\ge 0
\end{cases}$$
is continuous on the closed ray $[-1,\to)$. That’s not hard to do, but there are infinitely many pairs $\langle a,b\rangle$ that work. Consequently, I strongly suspect that there may be a misprint in the problem, and that the first case of the definition of $f$ was supposed to be $x<-1$.
