<s>[INST] Let $y= \frac{x^2+3x+1}{x^2+x+1} \forall x\in R$. Find the range of $y$ 
Let $y= \dfrac{x^2+3x+1}{x^2+x+1} \forall x\in R$. Find the range of $y$

I have reached: 
$x^2(y-1) + x(y-3) + (y-1) = 0$
I also know that the denominator of $y= f(x)$ is greater than $0$. 
How do I continue from here? I am unable to find a suitable condition to proceed.
Edit: 
vertex of function in the numerator is at $(\frac{-3}{2}, \frac{-5}{4})$
and of the function in the denominator is at: $(\frac{-1}{2},\frac{3}{4})$
but that ain't much helpful, is it? 
 [/INST] $$y=1+\frac{2x}{x^2+x+1}$$
The function $y(x)$ is continuous and differentiable in $\Bbb R$. Moreover
$\lim_{x\to\pm\infty}y(x)=1$, so $y(x)$ is bounded. This means that the range of $y$ is an interval $I$.
Then
$$y'=\frac{2x^2+2x+2-4x^2-2x}{(x^2+x+1)^2}=\frac{2-2x^2}{(x^2+x+1)^2}$$
which vanishes at $x=\pm1$.
Compute $y(1)$ and $y(-1)$ to find the endpoints of $I$.
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