Is there any way to find the function, given only a few points? I know that it is an exponential function, f(x)=ab^x, but I'm not sure what a and b are.
I'm not completely sure I understand, but I'll try to help. Say you're given the points (4,2), and (2, 5). Plug them into your formula as so: (4,2): 2=ab^4 (2,5): 5=ab^2 So then you have a system of two equations: 2=a * b^4 5=a * b^2 You should now be able to solve it, either through substitution or adding the two equations. For the purposes of this example, I'll keep going. Let's look at the second one, which is easier. Isolate any variable (b is easier). A percentage sign is a root sign (%4 would be the square root of four, 3%4 would be the cubed root of 4.) 5=a * b^2 5/a=b^2 %5/%a=b So then you have b relative to a: b is equal to the square root of 5 over a. Substitute this into your other equation. 2=ab^4 2=a * (%5/%a)^4 2=a * 25/a^2 2=25a/a^2 <--simplify by taking out an a from top and bottom 2=25/a a=25/2 Now just plug a into any of your equations, I chose the second one. 5=ab^2 5=(25/2) * b^2 10/25=b^2 <--simplify the left side 2/5=b^2 <--square root both sides b=%2/%5 Tada!
Thanks for trying to help, but its not really what I'm looking for. I should rephrase my problem. I am only given a few points, but I can tell that it is similar to an exponential decay. I have to create a model by creating a similar equation to the given points. Is there a way of doing that?
Mhm. The process is called regression. If you have a set of data points and you know that they are more or less exponential, you can use exponential regression to find an exponential function that will approximate the points. You could do this by hand (trust me, you don't wanna, it would take me a year to explain over the internet) or a calculator (some of the more advanced ones have functions that will calculate the exponential regression for you). Here's an example of linear regression (finding the equation of a line that approximates a set of data points): http://en.wikipedia.org/wiki/Linear_regression The same idea can be applied to exponential regression, simply by taking the log of the y coordinates, finding the linear regression, then exponentiate the equation. If that's way over your head (I didn't explain that very well), ignore this paragraph.
Yay, thanks a lot. Found the equation by using expReg() on my ti-84. Now I just have to finish the paper -_-