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How can I find equation for this function?

interesting function

the asymptotes on the sides can be for example at $x = -8$ and $x = 8.$

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What do you mean by equation? and algebraic expression? is your function continuous? if its not then given only a finite number of points on a graph, their could be more then one expression that interpolates them all. Also how does one know what those points are, there not clearly marked. – Ethan Dec 14 '12 at 22:37
user50222: could you give us a little context regarding where the graph originated, or what it represents? (E.g., is the graph a plot of data of some sort? Is this related your studies of a particular subject, and if so, what's the subject or topic?) – amWhy Dec 14 '12 at 22:59
Yes, I meant expression. Function has only three vertical asymptotes. Everywhere else function is continuous. Let's say, that function intersects axe x in 2,3,4,5,6,7 and -2,-3,-4,-5,-6,-7. And also in +/- infinity it has limit to 0 – user50222 Dec 14 '12 at 23:02
up vote 3 down vote accepted

Let's give it a try: since the function we look for $\,f(x)\,$ is even (according to the given graph), we can begin with $\,g(x)=(x^2-4)(x^2-9)(x^2-25)(x^2-36)(x^2-49)\,$ , and it has vertical asymptotes at $\,x=0\,,\,\pm 8\,$ , we can put


(not $\,x^2-64\,$ in the denominator as we need, for example $\,f(x)\xrightarrow [x\to -8^-]{} -\infty\,$

and since we need

$$\lim_{x\to \pm\infty}f(x)=0$$

we need the denominator's degree higher than the numerator's but in a way as to be sensible to sign changes around $\,\pm\,8\,$, so for example


Check carefully whether the above makes the cut.

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but at 8 and -8 limit is undefined. There will not be an asymptote, right? – user50222 Dec 15 '12 at 0:07
For a straight like $\,x=x_0\,$ to be a vertical asymptote it is sufficient and necessary that at least either one of the one-sided limits of the function when $\,x\to x_0^+\,\,\,or\,\,\,x\to x_0^-\,$ is $\,\infty\,\,\,or\,\,\,-\infty\,$, which is precisely what happens in two of the three vertical asymptotes. – DonAntonio Dec 15 '12 at 0:28

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