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Here is what y(x) = -x+100 graph looks like:

enter image description here

I am trying to change this equation, so that the graph becomes a curve which still crosses the axis Y at (0,100) and axis X at (100,0):

enter image description here

Can anybody give me a hint on this?

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That depends on how curved you want. There's a whole family of functions you could construct passing through points $(0,100)$ and $(100,0)$. – Mark Fantini Nov 24 '13 at 13:23
It doesn't matter, however curved it is - there will be a variable to make it more/less curved. I just can't remember the equation to make it curved. – astralmaster Nov 24 '13 at 13:25
There are an infinite number of curves which would fit your requirement. I just give you the first coming to my mind : y = 100 - 10 Sqrt[x]. – Claude Leibovici Nov 24 '13 at 13:33
2… – Carsten S Nov 24 '13 at 13:39
1… – Carsten S Nov 24 '13 at 13:40

1 Answer 1

up vote 3 down vote accepted

Probably the easiest way is to choose a third point $(a,b),\ \ b\in(0,100)$, what your curve also contains, and then construct a quadratic function to fit these $3$ points, using $$ \begin{aligned} p_0(x) &:=\frac{(x-a)(x-100)}{(-a)(-100)} \\ p_a(x) &:=\frac{x(x-100)}{a(a-100)} \\ p_{100}(x) &:=\frac{x(x-a)}{100(100-a)} \end{aligned}$$ These satisfy $p_i(j)=1$ if $i=j$ and $p_i(j)=0$ if $i\ne j$ for $i,j\in\{0,a,100\}$. So, a quadratic curve can be obtained as $$f(x):=100\cdot p_0(x)+b\cdot p_a(x)+0\cdot p_{100}(x)\,.$$

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Thank you, Meun freund! – astralmaster Nov 24 '13 at 13:53

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