Let "f" represent the function in that first box. Since you then add 2 you are looking for a function so that

f(4)+ 2= 3, f(12)+ 2= 6, and f(28)+ 2= 10.

That is the same as saying that f(4)= 1, f(12)= 4, and f(28)= 8. You can, at worst, fit a parabola through three points so try f(x)= . Putting in the given values, , , and . That gives you three equations to solve for a, b, c.

There are, of course, an infinite number of correct solutions to this problem.