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Determine a function whose graph has vertical asymptotes at $x=+2$ and $x=-2$, and a horizontal asymptote at $y=0$?

I don't know to satisfy these conditions.

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    $\begingroup$ What does having vertical and horizontal asympotes mean? Think of appropriate denominator $\endgroup$ – Angelo Rendina Nov 30 '14 at 21:43
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$$\frac{1}{(x+2)(x-2)}$$

Can you see why it works?

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  • $\begingroup$ How did you get it? $\endgroup$ – Rich Nov 30 '14 at 21:45
  • $\begingroup$ @Rich If it has to have an vertical asymptote it means the the denominator vanishes at $x=2$ and $x=-2$. This is the simplest function that satisfies the requests. You can see that is also has an horizontal asymptote at $y=0$. If you wanted, say, and horizontal asymptote at $y=2$, this would not work. Can you see how to fix this function so that it has a astympote in $y=2$ instead of $y=0$? What the numerator would look like in this case= $\endgroup$ – Ant Nov 30 '14 at 21:47
  • $\begingroup$ I get it now, thank you! $\endgroup$ – Rich Nov 30 '14 at 21:53
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There are countless examples of such functions: $$f(x)=\begin{cases}c\cdot\tan^k\bigg(\dfrac\pi4x\bigg),\qquad x\in(-2,2)\\\\\dfrac a{(x+2)^n},~\qquad\qquad~ x<-2\\\\\dfrac b{(x-2)^p},~\qquad\qquad~ x~>~2\end{cases}\qquad\qquad\qquad\qquad\qquad\begin{align}&a,b,c\in\mathbb R^*\\\\&k,n,p\in\mathbb N^*\end{align}$$

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