What is 'completing the square' and how can I use it to find the minimum point of a quadratic curve?

Completing the square is a process used to put a quadratic curve into a particularly nice form. The form we want is y(x) = r(x+s)^2 + t for some numbers r,s, and t. A 'general' quadratic we would expect to be given as y(x) = ax^2 + bx + c. (Obviously we assume 'a' is not 0, otherwise we don't have a quadratic but a sneaky linear equation disguised as one, so we can divide by a without worrying about it). How do we move between them? If we expand our first equation and collect the terms we get y(x) = rx^2 + 2rsx + rs^2 + t. If we want the two curves to be equal we need that: r = a, 2rs = b, and rs^2 + t = c. (This is called equating coefficients). We can solve these equations to get: r = a, s = b/2a, and t = c - b^2/4a. Now we know how to complete the square for a general quadratic, but let's do an example to make sure. Suppose we are given y(x) = x^2 + 4x + 5. Using the notation above, we would say a = 1, b = 4, and c = 5. So we can fill into our formulas and get r = 1, s = 2, and t = 1. So we get y(x) = (x+2)^2 + 1. We can always multiple out as a check to see if our answer is correct. Rather than always remembering the formulas, it is not too hard to sort of 'reverse engineer' the answer once you know what you are doing, especially if a = 1. The trick is to remember that you will always end with (x + b/2)^2 with some correction at the end, which you could find out by multiplying out the equation. Now how can we use this form to find the minimum point on a quadratic curve? The key fact is that a square is always positive or 0. So if we have an equation in the right form y(x) = r(x+s)^2 + t. We will assume 'r' is positive so we get a positive quadratic and therefore a minimum. (Were 'r' negative we would get a negative quadratic so a maximum point, but the method is almost identical). Since r is positive, and (x+s)^2 is positive or 0, the smallest their product can be is 0, precisely when x = -s. Therefore the smallest y can be is 't', so the minimum point is (-s,t). As an exercise, you might like to check that you get the same result using differentiation.