Modeling mathematical functions Here's something that's been bothering me for a while now, what I don't understand is, if I have a function and I wish to constrain it to specific values... And let's say I have three pairs of x,y values, why do I have to have three constraints? Can someone elaborate on that?
$$f(x) = k_1e^{k_2x} + k_3$$
where
$$f(0) = 0$$
$$f(0.1) = 1$$
$$f(1) = 100$$
For example, the $x = 0$ simply "cancels" the $e$ term. Therefore:
$k_1+k_3= f(0) = 0$
And so forth... But why do I need three of them to model it properly?
 A: 
...let's say I have three pairs of x,y values, why do I have to have three constraints?

Solving a system of equations requires one to have as many variables as there are equations. If not, your problem is underdetermined (more unknowns than equations) or overdetermined (more equations than unknowns), and those are a bit more complicated to handle in general...
Getting back to the matter of fitting $k_1e^{k_2x} + k_3$, I presume you've already assembled your equations like so:
$$\begin{align*}
0&=k_1e^{0\cdot k_2} + k_3\\
1&=k_1e^{0.1\cdot k_2} + k_3\\
100&=k_1e^{1\cdot k_2} + k_3
\end{align*}$$
You said you've managed to turn the first equation into $k_1+k_3=0$. That's now one less degree of freedom. You can now replace all the $k_1$'s appearing in the other two equations with $-k_3$, or replace all the $k_3$'s with $-k_1$. You call the shots. The solution is really a matter of replacing your variables one-by-one with "known" quantities, up until the point where you have one equation in one unknown. That bit is usually easier to solve than many equations in many unknowns...
