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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.

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Did you forget about the denominator or something? By your reasoning, $x$ is $x^2/x$ so $x$ grows like $x^2$! –  anon Sep 6 '11 at 2:01
Thank you for making explicit what you are not understanding. It makes it easier to give an appropriate answer, as I think has been done. I hope they help. –  Ross Millikan Sep 6 '11 at 4:33
@anon: well, $x^2!$ really does grow quite a bit... :) (Yes, I know what you meant to say.) –  Ｊ. Ｍ. Sep 6 '11 at 5:50

2 Answers 2

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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HINT $\ $ The rational function $\rm\ f\: \sim\: a\ x+ b\ \iff\ f\ =\ a\ x + b + g/h,\ $ with $\rm\: deg\ g\ <\ deg\ h\:.\:$ Therefore it is equivalent to computing the polynomial ("integral") and fractional parts of $\rm\:f\:$ using the Division Algorithm. Your reasoning fails due to the fact that it does not correctly compute the polynomial part of the quotient. But it is easy to compute the polynomial part without division:

$\ $ the coef of $\rm\ x^\:$ in $\rm\ 2\ x^2 + 3\ x + 6\ =\ (2\ x + 1)\ (x + q) + r\ \ \Rightarrow\ \ 3\ =\ 2\:q + 1\ \ \Rightarrow\ \ q = 1$

Therefore we deduce that the polynomial part is $\rm\ x + q\ =\ x + 1\ $ so $\rm\ y = x+1\ $ is an asymptote.

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So in this case there´s no horizontal assymptote, but a slant assymptote? Am I right? –  Vinicius L. Beserra Dec 31 '12 at 12:46

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