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Basically, I wanted to create a "loser" graph. Along the $x$ axis we'd have time and along the $y$ axis is how much of a loser someone is. I want the graph to be an upside down bell shaped graph meaning that apart from a specific point in time, how big a loser you are will increase. This specific moment in time will correspond to the minimum point of the graph and from here, with (lets say exponential) gradient, the graph will take shape.

This graph will effectively need to be able to have a minimum at every point yet at the same time it can only have one minimum point.

I'm quite confused about what kind of formula would give this. I know that you could probably change the minimum point bit at every point by maybe adding in some constant that translates the graph with respect to the time co-ordinate I want, but I'm not sure what kind of formula an "initial" graph would have. It can't just be $y = -(x - a)^2 + b$ with $a$ some constant that I can change to give me the minimum point corresponding to my current time and $b > 0$, just because no-one can be $0$ loser.

Kaish
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  • "This graph will effectively need to be able to have a minimum at every point yet at the same time it can only have one minimum point." ????? I don't understand this question at all. Why don't you want $y=(x-a)^2+b$ again? –  Feb 06 '13 at 01:41
  • Instead of a prose description, it would probably be easier if you could post a freehand sketch of the shape you want your graph to have. That would make it easier to suggest formulas that will look like it. – hmakholm left over Monica Feb 06 '13 at 01:42
  • Do you want a graph of a single function $y=f(x)$, or do you want a family of functions $y=f_t(x)$ where $t$ is time? – Gerry Myerson Feb 06 '13 at 05:08
  • You could include a picture of how you expect it to behave in some interval, and describe what happens outside this window. – Allawonder May 14 '18 at 17:44

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Let $a$, $b$, and $c$ be minimum, birth, and maximum loserness, respectively. Let $d$ be the age of minimum loserness. Then the Cauchy distribution $\frac1{\pi(1+t^2)}$ has the following transformation according to the parameters $a,b,c,d$.

Matt S
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