Finding the range of $y =\frac{x^2+2x+4}{2x^2+4x+9}$ (and $y=\frac{\text{quadratic}}{\text{quadratic}}$ in general) I had this problem in an exam I recently appeared for:

Find the range of
$$y =\frac{x^2+2x+4}{2x^2+4x+9}$$

By randomly assuming the value of $x$, I got the lower range of this expression as $3/7$. But for upper limit, I ran short of time to compute the value of it and hence couldn't solve this question.
Now, I do know that one way to solve this expression to get its range is to assume the whole expression as equals to K, get a quadratic in K, and find the maximum/minimum value of K which will in turn be the range of that expression. I was short on time so avoided this long winded method.
Another guy I met outside the exam center, told me he used an approach of $x$ tending to infinity in both cases and got the maximum value of this expression as $1/2$. But before I could ask him to explain more on this method, he had to leave for his work.
So, will someone please throw some light on this method of $x$ tending to infinity to get range, and how it works. And if there exists any other efficient, and quicker method to find range of a function defined in the form of a ( quadratic / quadratic ).
 A: The question can be easily solved by this technique:
As $\displaystyle y = \frac {x^2 + 2x + 4}{2x^2 + 4x + 9} \implies 2y = \frac {2x^2 + 4x + 9 - 1}{2x^2 + 4x + 9}$.
Thus, $\displaystyle 2y = 1-\frac {1}{2(x + 1)^2 + 7} $
Squares can never be less than zero so the minimum value of the function : $\displaystyle 2(x + 1)^2 + 7 $ would be $7$ , or Maximum value of $\displaystyle \frac {1}{2(x + 1)^2 + 7} $ is $\displaystyle \frac {1}{7} $.
This tells that minimum value of $y $ will be $\displaystyle \frac{3}{7}$.
And so on.. check for $x \rightarrow \infty$.
From here you can easily tell the maximum and minimum values : $\displaystyle y \in \left [ \frac {3}{7}, \frac {1}{2} \right ) $
A: As a follow-up to @NikolaAlfredi's answer:
$ 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}) \implies 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 $y$).
A: In general, if $\deg f = 0$ where $$f(x) = \frac{a_nx^n + a_{n - 1}x^{n - 1} + \cdots + a_1x + a_0}{b_nx^n + b_{n - 1}x^{n - 1} + \cdots + b_1x + b_0},$$ the limit of $f$ as $x$ increases/decreases without bound is $a_n/b_n$.
In your case, $a_2 = 1$ and $b_2 = 2$. Hence, $a_2/b_2 = 1/2$.

We'll factor $f$ as $$\frac{x^2+2x+4}{2x^2+4x+9} = \frac{(x + 1)^2 + 3}{2(x + 1)^2 + 7}.$$
Notice that for all $x \in \mathbb{R}$, $f > 0$. Also, we can see that $(x+1)^2 + 2 < 2(x + 1)^2 + 7$. This means that the range should be a part of $(0,1/2)$. Since both numerator and denominator have $(x + 1)^2$ without any remaining $x$'s, we can see that this will be at its minimum when $x = -1$. Then, $$f(-1) = \frac{(-1 + 1)^2 + 3}{2(-1 + 1)^2 + 7} \\ = \frac{(0)^2 + 3}{2(0)^2 + 7} \\ \frac{3}{7}$$
Therefore, the range is $[3/7, 1/2)$.
A: $$y = \frac{x^2 + 2x + 4}{2x^2 + 4x + 9} = \frac12 - \frac {1/2}{2x^2 + 4x + 9} \implies \frac {dy}{dx} = \frac {4x + 4}{\text{whatever}} \text{ Let } \color{green}{\frac{dy}{dx} = 0 \implies x = -1}, y(x = -1) = \color{blue}{\frac37} \text{ also } y(x\to \infty) = \color{blue}{\frac 12}$$
