# 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 ).

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 : $$\displaystyle y \in \left [ \frac {3}{7}, \frac {1}{2} \right )$$