<s>[INST] Finding asymptotes to $y = \frac{2x^2 + 3x - 6}{2x + 1}$ I need to find the asymptotes of 
$y = \frac{2x^2 + 3x - 6}{2x + 1}$.
The asymptote at $x = -1/2$ is clear. If one long divides they can easily see that there is an asymptote of $y = x + 1$ as $x$ goes to infinity.
However, what is wrong with this reasoning? I claim that as $x$ goes to infinity, the $2x^2$ term will dominate, so the graph will be on the order of $y = 2x^2$, which has no asymptote. So $y = x + 1$ is not an asymptote.
 [/INST] As $x \to \infty$, $2x^2$ is the  dominant term in the numerator, while $2x$ is the dominant term in the denominator.  So the leading term in $y$ is $\frac{2x^2}{2x} = x$.  You still need the constant term, which you can do this way: $$\frac{2x^2 + 3x - 6}{2x+1} \approx 
\frac{2x^2 (1 + 3/(2x))}{2x (1 + 1/(2x))} \approx x \left(1 + \frac{3}{2x}\right) \left(1 - \frac{1}{2x} \right) \approx x \left(1 + \frac{3/2 - 1/2}{x}\right) = x + 1$$
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