Only a graph is not a prove for what type of funtion it is, but it can give you an idea what type it might be. It will also not work. An exponential function is a function of the form f(x)=a*b^x where b is a positive real number. Here f(x) the increase and x = number of boosts. So let's assume it's an exponential function. We then have f(0) = 0, so it follows that a = 0 as b^0 = 1. So we have f(x) = 0, which I would really like, but I don't think CG will do this XD. Now you could say that we start at x = 1 and don't bother with x = 0 as that isn't really relevant for us. Although I think for coding it will be (not sure about that). We have f(1) = a*b^1 = 20 and f(2) = a*b^2 = 60. 2 functions and 2 unknowns (a and b), so we can solve this system of equations. We have ab = 20, this gives a = 20/b. We substitute a = 20/b in a*b^2 = 60 and get 20b = 60, so b = 3. So we get f(x) = (20/3)*3^x. But f(3) = 180 =/ 120, so exponential function also doesn't work here. Contradiction with assumption that the correct function was exponential.
You can write it as a quadratic function as was shown already by @Ragnarok_COTF. Or what I think CG most likely have in their code: an arithmetic series. Which would be x_{n} = x_{n-1} + 20*n, with n = number of boosts, x_{n} the increase at n boosts and x_{0} = 0 (the starting point). Then we get
x_{0} = 0
x_{1} = 0 + 20*1 = 20 (correct)
x_{2} = 20 + 20*2= 60 (correct)
x_{3} = 60 + 20*3 = 120 (correct)
x_{4} = 120 + 20*4 = 200 (correct)
x_{5} = 200 + 20*5 = 300 (correct).
I also agree the increase is way too high.
