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So i have a function

$y=\frac{x^3+2x+9}{\sqrt{4x^2+3x+2}}$.

At wolfram alpha they say that it has a non linear asymptote $y=\frac{x^2}{2} - \frac{3x}{16}+ \frac{251}{256}$.

How do you predict this?

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closed as off-topic by Did, Arnaldo, user223391, Dave, Shailesh Jul 18 '17 at 0:07

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  • $\begingroup$ is it $$\frac{x^3+2x+9}{\sqrt{4x^2+3x+2}}$$? $\endgroup$ – Dr. Sonnhard Graubner Jul 17 '17 at 9:42
  • $\begingroup$ Sorry , i added braces to make it more readable . Guess its not serving its purpose. $\endgroup$ – user33699 Jul 17 '17 at 9:43
  • $\begingroup$ Question has been edited $\endgroup$ – user33699 Jul 17 '17 at 9:44
  • $\begingroup$ $ y= \frac{x^3+2x+9}{2x} (1+\frac{3}{4x}+\frac{1}{2x^2})^{\frac{-1}{2}} $ ... binomially expand the second bracket ... $\endgroup$ – Donald Splutterwit Jul 17 '17 at 9:51
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    $\begingroup$ @Fakemistake well just write " y={x^3+2x+9}/{root(4x^2+3x+2)} asymptotes " :) $\endgroup$ – user33699 Jul 17 '17 at 10:32
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$ y= \frac{x^3+2x+9}{2x} (1+\frac{3}{4x}+\frac{1}{2x^2})^{\frac{-1}{2}} $ ... binomially expand the bracket ... \begin{eqnarray*} \left(1+\frac{3}{4x}+\frac{1}{2x^2}\right)^{\frac{-1}{2}} &=&1+ \frac{-1}{2} \left(\frac{3}{4x}+\frac{1}{2x^2}\right) +\frac{\frac{-1}{2}(\frac{-1}{2}-1)}{2} \left(\frac{3}{4x}+\frac{1}{2x^2}\right)^2 +\cdots \\ &=&1+\frac{-3}{8x}+\frac{-5}{128x^2}+\cdots \end{eqnarray*} Now multiply this by $\frac{x^3+2x+9}{2x}$ ... we get $y=\frac{x^2}{2} - \frac{3x}{16}+ \frac{251}{256} +o(\frac{1}{x})$.

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    $\begingroup$ Thanks a lot . I really appreciated your help. $\endgroup$ – user33699 Jul 17 '17 at 10:21

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