# Real World Problem... Using log to find points in a graph?

• Nov 5th 2008, 03:20 PM
dave2118
Real World Problem... Using log to find points in a graph?
I've been out of high school for 13 years so this may be an example to anyone viewing this that you do use these problems in the real world.

I've spent all day trying to figure this out and came home and found out I don't have any of my books from school (thanks wife).

I'm trying to write an equation to find points on a graph. It starts at (0,0), goes to (13,720) then to (78, 2160)

I found a few equations online but I just can't figure it out. I'd like to plug in at any point x to get y using those constants, but I've been at a loss all day.

Thanks for any help in advance. This is for a marketing internet project.
• Nov 6th 2008, 12:18 PM
Laurent
Given only three points, if they don't lie on a line, it is quite hard to guess what the whole graph should be like... It is like you would be given the color of three pixels on a picture: if the three pixels are black, a natural guess (but not necessarily a good one) would be that the whole picture is black, but if the colors are different, we have no clue and there's not much we can guess. I can give you one thousand formulas of graphs that go through the three points, but this would be pointless.

Perhaps it could help if you tell us what these points correspond to. Or give a few more points.
• Nov 6th 2008, 12:21 PM
dave2118
Basically I want the 3rd point to be the apex, so we can basically take the 4th and 5th and invert them:

(0,0)
(13,720)
(78, 2160)
(135, 720)
(156, 0)

I thought that y = ax^2 + bx + c would to this but I can't get it to work in a spreadsheet