Okay I've been working on this for some times now, and I simply can't figure out if my assumption is correct or not, pic related.
Let's say I want to find f1(x) where x=-1.
Do I simply use this calculation
a=ˣ2-ˣ1√⎺y2/y1
y2 or y1
b=————
aˣ
and use the point on the chart as (x1, y1) = (2, 0) and (x2, y2) = (1, 6), then use the f(x)=b∙aˣ calculation where x=-1?
>>8490158
No-one can help you unless you provide enough information to explain wtf you're talking about.
>>8490176
I'm making an assumption that the exponential function can be found using the presented calculation and wanted confirmation on whether or not I'm doing it correctly.
>>8490201
Never mind figured it out.
>>8490201
What are f1(), f2() etc?
There's no way that you can identify a function from two values. Or any number of values, for that matter.
At most, if you're given a function definition which includes N coefficients, you can (usually) deduce those coefficients given N values.
E.g. if you know that the function is linear f(x)=m*x+c, then you can determine m and c from f(x1)=y1 and f(x2)=y2.
>>8490158
are all these functions of the form [math] b•a^x[/math] ?
if so you have that
f(0)=b and f(1)=ba and so on