Try this idea: g(t) = (1-t) f(a) + t f(b) where t goes from 0 to 1. At the end points, we get g(0) = (1) f(a) + (0) f(b) = f(a) g(1) = (0) f(a) + (1)

Try this idea: g(t) = (1-t) f(a) + t f(b) where t goes from 0 to 1. At the end points, we get g(0) = (1) f(a) + (0) f(b) = f(a) g(1) = (0) f(a) + (1) f(b) = f(b). This can be used to linearly blend any two things, as for t in between 0 and 1, we'll get part of f(a) and part of f(b). Now, if we want to use x instead of t, where x goes from a to b, we need to convert: x = a <=> t = 0 x = b <=> t = 1 The distance that x has travelled from a is (x-a). The total distance from a to b is (b-a). t is the proportion of that total distance, so t is the distance travelled over the total distance: x-a t = --- b-a Now (renaming g to f as we're changing parameterisation* ) f(x) = (1 - (x-a)/(b-a)) f(a) + ((x-a)/(b-a)) f(b) = (1 + (x-a)/(a-b)) f(a) + ((x-a)/(b-a)) f(b) //since -(b-a) = a-b = ((a-b)/(a-b) + (x-a)/(a-b)) f(a) + ((x-a)/(b-a)) f(b) //since 1 = (a-b)/(a-b) = ((a-b+x-a)/(a-b)) f(a) + ((x-a)/(b-a)) f(b) //combining terms with common denominator = ((x-b)/(a-b)) f(a) + ((x-a)/(b-a)) f(b) Does that explain it? * g(x) is something different: g(x) = (1-x) f(a) + x f(b), just by substituting in.