Finding the range of y=x2+2x+42x2+4x+9 (and y=quadraticquadratic in general)

The question can be easily solved by this technique: As $y=\frac{x^2+2x+4}{2x^2+4x+9}\Rightarrow 2y=\frac{2x^2+4x+9-1}{2x^2+4x+9}$. Thus, $2y=1-\frac{1}{2(x+1{)}^2+7}$ Squares can never be less than zero so the minmum value of the function: $2(x+1{)}^2+7$ would be 7, or Maximum value of $\frac{1}{2(x+1{)}^2+7}\text{is}\frac{1}{7}$. This tells that miminum value of y will be $\frac{3}{7}$. From here you can easily tell the maximum and minimum values: ${\displaystyle y\in [\frac{3}{7},\frac{1}{2})}$

${\displaystyle y=\frac{x^2+2x+4}{2x^2+4x+9}=\frac{2x^2+4x+8}{2(2x^2+4x+9)}=\frac{2x^2+4x+9-1}{2(2x^2+4x+9)}=\frac{1}{2}(1-\frac{1}{2x^2+4x+9})\Rightarrow 2y=1-\frac{1}{2x^2+4x+9}}$.
Now find the extremes of the range of the expression in the RHS of the above equation (which I believe you can; if not someone else or I myself shall try and add it) and divide them by 2 to get the required extremes(taking half since we get values for 2y and not y).