Proving that a function is asympotically linear I have a function:
$$
y = \frac{20x^2 + 14x + 2}{7m+11xm-m\sqrt{37+70x+x^2}}
$$
Note: In this case, m is simply some parameter
When I plot out this graph, it looks asymptotically linear.
How can I go about checking:

*

*If the function is, in fact, asymptotically linear

*If so, what is the asymptotic function's formula.

 A: 
When I plot out this graph, it looks asymptotically linear.

For large $x$, we have
$$\sqrt{37+70x+x^2} ~\approx~ |x|\tag 1$$
in the sense that the quotient of the functions on either side approach $1$ as $x\to\pm\infty$. $\def\sign{\operatorname{sign}}$
This means
$$\begin{align}
y &= \frac{20x^2 + 14x + 2}{7m+11xm-m\sqrt{37+70x+x^2}}\\
&\stackrel{(1)}\approx \frac{20x^2 + 14x + 2}{7m+11xm-m|x|}\\
&\stackrel{(2)}= \frac{20x + 14 + 2/x}{7m/x+11m-m\sign x}\\
&\stackrel{(3)}\approx \frac{20x+14}{m\cdot(11-\sign x)} \\
\end{align}$$
provided $m\neq0$. Step (2) divides all terms by $x$ and uses $|x|=x\sign x$.  Step (3) neglects all terms of the form $\mathit{const}/x$ because they tend to 0 as $x\to\pm\infty$.
Taking it all together, we have that
$$
y\approx \begin{cases}
\dfrac{10x+7}{5m}, &\text{ for large positive } x \\
\dfrac{10x+7}{6m}, &\text{ for large negative }x
\end{cases}$$
Again, "$\approx$" implies that the quotient of the functions on either side tend to 1.  And the even stronger statement holds that their difference tend to 0 as their values tend to $+\infty$ resp. $-\infty$.
A: You can proceed in the same way you act when you study if a function has asymptotes. Asympotes are straight lines of the form $r: Y = ax + b$, where
$$a = \lim_{x\to +\infty} \frac{f(x)}{x}$$
$$b = \lim_{x\to +\infty} f(x) - ax$$
Provided that $a\in\mathbb{R}\backslash\{0\}$ and $b\in\mathbb{R}$.
In your case:
$$a = \lim_{x\to +\infty} \frac{\frac{20x^2 + 14x + 2}{7m+11xm-m\sqrt{37+70x+x^2}}}{x} = \frac{2}{m}$$
(I let you to do the math, as an exercise).
$$b = \lim_{x\to +\infty} \frac{20x^2 + 14x + 2}{7m+11xm-m\sqrt{37+70x+x^2}} - \frac{2x}{m} = \frac{7}{m}$$
So you get your asymptote:
$$Y = \frac{2}{m}x +\frac{7}{m}$$
Which shows you, being $m$ a constant, that $y$ behaves linearly.
