Intersection point in graphs

All,
I have 2 sets of data. One set is directly proportional to the data on
x-axis and the other is is inversely proportional. I want to find out at
which point on the x-axis do the 2 data sets intersect. I can see it
graphically but I need some kind of a formula to spit out that intersection
point.
Any help, suggestions will be appreciated.
Thanks,
R.K.

General answer:

If we have two functions y1=f(x) and y2=g(x), we find the point(s) of intersection by setting y1=y2 [f(x)=g(x)]and solving for x.

Specific case:

directly proportional function y1=f(x)=ax
indirectly proportional funciton y2=g(x)=b/x
set them equal ax=b/x -> x^2=b/a -> x=+/- sqrt(b/a)

I leave it to you to decide if you want the positive root or the negative root, or if I have correctly/incorrectly interpreted the form of your functions.

Thanks Mr. Shorty. Problem I am facing is that thought the relationships are
directly and inversely proportional but they are not perfect linear
equations. In that case how do I solve for intersection point. Is there a
code or formulae?
Thanks again for your help.
RK

"MrShorty" wrote:

>
> General answer:
>
> If we have two functions y1=f(x) and y2=g(x), we find the point(s) of
> intersection by setting y1=y2 [f(x)=g(x)]and solving for x.
>
> Specific case:
>
> directly proportional function y1=f(x)=ax
> indirectly proportional funciton y2=g(x)=b/x
> set them equal ax=b/x -> x^2=b/a -> x=+/- sqrt(b/a)
>
> I leave it to you to decide if you want the positive root or the
> negative root, or if I have correctly/incorrectly interpreted the form
> of your functions.
>
>
> --
> MrShorty
> ------------------------------------------------------------------------
> MrShorty's Profile: http://www.excelforum.com/member.php...o&userid=22181
> View this thread: http://www.excelforum.com/showthread...hreadid=480125
>
>

In many cases, it is difficult or impossible to find an algebraic solution like I suggested above. Back to the general case:

Set spreadsheet up so you have a cell calculating f(x)-g(x) at some x that is reasonable. Then you can use Goal Seek or Solver (Goal Seek is easier to use, but Solver is more robust) to numerically find the value of x that yields f(x)-g(x)=0. There are hazards associated with such a numerical solution (may converge to the wrong solution or may not converge at all), but this approach is general enough to work in many cases.