# Thread: Exponential curve help

1. ## Exponential curve help

hey people,

Working on an awkward problem, need help solving it.

I have a curve that is +21 on the X, +9 on the Y, assuming your starting at 0/0 for the first point. At 14x, Y=3.11. At 7x, Y=.51. I'm not entirely sure the 3.11Y and .51Y is 100% right on, but I know the 7x and 14x. Is making sense?

I need to get a slope formula for this, and I can't remember how to do any of this from high school days....

2. It is awkward mostly because your notation is not useful. Have you never seen simple function notation?

I believe you might mean this:

"I have a curve that is +21 on the X, +9 on the Y"

f(21) = 9

"assuming your starting at 0/0 for the first point"

f(0) = 0

"At 14x, Y=3.11."

f(14) = 3.11

"At 7x, Y=.51."

f(7) = 0.51

This gives four points: (0,0),(7,0.51),(14,3.11),(21,9)

Please observe these substantially more compact and useful notational conventions.

Now you must decide...

Why do you think an exponential model would be good?
Are you SURE you want it to go through (0,0)?

3. It makes no difference to me where the point of origin is. you have to understand, it's been a while since I've done this, but it is slowly coming back.

Basically I was given a graphic of a curve that appears exponential, and took a couple measurements and thats what I have. But I have to be able to represent it in a way that someone else can read it and use the dimensions, so I figured an exponential curve would be best.

4. It also would be important for you to understand that basic communication is required in order for anything useful to come of this discussion.

1) You did not answer my question. Have you seen basic function notation? Are you able to use it? It would be very beneficial.

2) You said "assuming your starting at 0/0 for the first point" and then retracted to "It makes no difference to me where the point of origin is." It is impossible to get a useful model without a clear idea of what is wanted.

A simple least squares exponential model, $y = 0.1375*e^{1.4353*x}$ doesn't quite strike me as useful as may be desired, particular since it does not actually match any of the three given points.

A quadratic least squares model matches the three points nicely, $y = 0.033571x^{2} - 0.333571x+1.2$, but has the unfortunate side effect of turning around and INcreasing as you head toward zero from five. This may be unacceptable.

A very simple power model, $y = 0.003159*x^{2.6214}$ is pretty close to the given values and does not have the increasing problem of the quadratic near zero.

If it were me, I'd go with the "power model". In any case, you must be VERY wary of extrapolation. I thought a little about what happens between zero and seven. If you are trying to project behavior at 50, I would be very cautious indeed about relying on the results of any of these three models.