Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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
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 $$f(x):=100\cdot p_0(x)+b\cdot p_a(x)+0\cdot p_{100}(x)\,.$$

share|cite|improve this answer
Thank you, Meun freund! – astralmaster Nov 24 '13 at 13:53

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.