# 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 ).

• Use of calculus to find abslute maxima and minima is the easier way. Dec 1, 2021 at 6:39
• I'll give a hint: An easy way to re-express the expression will be $\frac{1}{2}-\frac{1}{4\left(x+2\right)^{2}+14}$. Think of what circumstance would maximise the expression, which would be by minimizing the term subtracted. Dec 1, 2021 at 6:41
• @NikolaAlfredi Can you please show how. Because I am not versed with using the approach of calculus in such questions. Dec 1, 2021 at 6:49
• @EmanatS Try my approach, doesn't need calculus Dec 1, 2021 at 6:50
• The hint of Prometheus is very good. It gives the result quickly. Dec 1, 2021 at 6:51

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

• Isn't very obvious on the first glance when it comes to the minimum, but to explain it, $2x^2+4x+9 = 2(x+1)^2+7$ Dec 1, 2021 at 6:54
• How did you split it just like that? Basically, how did you reduce that expression? Dec 1, 2021 at 6:55
• @Prometheus It's true... But I guess the test was a bit of a time-crusher, so perhaps the OP would like to get there in one step Dec 1, 2021 at 6:55
• @Spectre Yes, I had to solve this problem under 50 seconds. And Prometheus, yes I know the method of completing the square. I just do not understand how to apply it to this problem. Please explain this approach to get a one step form like I'm five. Dec 1, 2021 at 6:59
• @EmanatS I have explained Nikola's step below, if you ever didn't get how he came to that simplification. Dec 1, 2021 at 7:04

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 $$y$$).

• Got it. I appreciate your kind efforts, and now it is easily clear to me. Thanks to both you, and Nikola. Also, can you please delve on calculus based approach like someone mentioned? I want to learn how to work this out with calculus too. Dec 1, 2021 at 7:08
• @EmanatS Are you looking for an answer using "calculus" methods? Dec 1, 2021 at 7:09
• @TeresaLisbon Not really. But I am open to learning through it too. Dec 1, 2021 at 7:09

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 $$[3/7, 1/2)$$.

• Note that both the numerator and denominator of $f$ must be polynomials with the same degree. This means that the expression could be linear/linear, quadratic/quadratic, cubic/cubic, and so on. Dec 1, 2021 at 7:12
• Woah, you're awesome. And so is @NikolaAlfredi I love this forum now, you both explained this problem so beautifully to me putting in all efforts to explain it to me. Thank you for a generalized solution to it as well. So, "limit of f as x increases/decreases without bound is an/bn" is always an upper limit of the expression right? Any places where I need to watch out this rule for? Dec 1, 2021 at 7:17
• Also for, (x^2+2x+1)/(4x^2-7x+9), a2/b2 will be {1/4}, but the upper limit of expression is 16/19. Shouldn't it be 1/4 according to your assertion? Edited Dec 1, 2021 at 7:21
• No. This is not always true. Consider the simple case $\frac{x}{x + 1}$. You'll see that it attains all real values except $y = 1$, and it is defined for all real $x$ except $x = -1$. In general, the limit is not the upper bound. It is only for the case where $p(x) < q(x)$ and $\deg p = \deg q$. ($p$ and $q$ are the polynomials in $f$, that is, $f(x) = p(x)/q(x)$). Dec 1, 2021 at 7:23

$$y = \frac{x^2 + 2x + 4}{2x^2 + 4x + 9} = \frac12 - \frac {1/2}{2x^2 + 4x + 9} \implies \frac {dy}{dx} = \frac {4x + 4}{\text{whatever}} \text{ Let } \color{green}{\frac{dy}{dx} = 0 \implies x = -1}, y(x = -1) = \color{blue}{\frac37} \text{ also } y(x\to \infty) = \color{blue}{\frac 12}$$