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I am trying to find an equation for the graph that crosses the Y axis at (0,100), X axis at (100, 0), is a curve with adjustable degree of "bending" and has an axis of symmetry y(x) = x. Here are few examples of such graphs:

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I've tried quadratic, cubic, quartic equations but I can't adjust the degree at which the curve is bent. Any hints/tips appreciated.

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1 Answer 1

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I know a function like the second one. Have a look at it to find if it works or not. That is $$x^{\frac{2}3}+y^{\frac{2}3}=100^{\frac{2}3}$$

enter image description here

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Thank you, I played a bit with this one but any type of change involves displacing the crossing points with the axises. –  astralmaster Nov 24 '13 at 17:12
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Salam my friend, have a nice time +) –  Sami Ben Romdhane Nov 24 '13 at 17:21
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@astralmaster One thing that doesn't change crossing points is to replace all three exponents, $2/3$ in the above example, by any number $c$ with $0<c<1$, i.e. $x^c+y^c=100^c$, which meets the axes at $(100,0)$ and $(0,100)$ and has different degrees of bending. As $c$ is taken closer to $0$ the bending increases. –  coffeemath Nov 24 '13 at 17:44
    
+1 my friend! I hope you are well! –  amWhy Nov 24 '13 at 17:58
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Me too, Babak! $\ddot \smile$ –  amWhy Nov 24 '13 at 18:05

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