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#1
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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. |
#2
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Intersection point in graphs
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 |
#3
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Intersection point in graphs
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 |
#4
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Intersection point in graphs
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. -- MrShorty ------------------------------------------------------------------------ MrShorty's Profile: http://www.excelforum.com/member.php...o&userid=22181 View this thread: http://www.excelforum.com/showthread...hreadid=480125 |
#5
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Intersection point in graphs
I believe rk0909 does not has f(x) or g(x) only has two series of X-Y values It's possible to build up two equaltion,that just exactly as Excel's bezier curve for both series and solve them however it's too complex,and not worth to do it if you want to do this intersection-find job only few times -- Pan ------------------------------------------------------------------------ Pan's Profile: http://www.excelforum.com/member.php...o&userid=28144 View this thread: http://www.excelforum.com/showthread...hreadid=480125 |
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