Friday, February 19, 2010

Completing the Square

"Completing the square" is a method of expressing a quadratic function, and it is an especially useful form for graphing the expression.

If we have a function:

f(x) = (x+1)^2 + 3,

from the rules of graphing (1, 2, 3) that I have posted already, we can tell that this is a second degree polynomial (upright parabola), shifted to the left 1 unit, and up 3 units. However, if I were to ask you to graph:

f(x) = x^2 + 2x + 4,

that appears to be much more complicated. However, it describes the exact same curve, it is just written differently. "Completing the square" is the process used to convert this complicated and nasty form into the simpler form above.

To do this, you basically have to ignore the constant number (4, in this case) at first. So, to start, we look at x^2 + 2x. Now, we essentially do FOIL backwards to get it in the form (x+y)(x+y), or (x + y)^2... a little tricky, but with practice, you will get better at it and recognize certain patterns.

I will write as step by step below, and hopefully that demo will explain what I mean with all of this:

f(x) = x^2 + 2x + 4... re-writing to focus what we're doing:

f(x) = [x^2 + 2x] + 4... now reverse FOIL, determine the constant required in the square brackets to make it a perfect square. To keep things balanced, whatever number you add, you have to also take away:

f(x) = [x^2 + 2x + 2] + 4 - 2... simplify:
f(x) = (x+1)^2 +2

To get an x^2 and a 2x, we can see that the First term has to be just x, and the Outside and Inside terms have to be the same (because it's a perfect square) and add to 2x, which is 1x plus 1x, and which means that the y term in the general statement has to be 1. So, to get the x^2 + 2x correct, we have determined that is corresponds to (x+1)^2. Then, as noted, whatever number we have added to the expression to make it a perfect square, we also have to subtract from the number that we started with.

Hopefully this makes some sense to you. Let's try another example:

f(x) = x^2 + 10x + 37
f(x) = [x^2 + 10x] + 37
f(x) = [x^2 + 10x + 25] + 37 - 25
f(x) = (x+5)^2 + 12

Much easier to graph this than when it started. :) Here's another:

f(x) = x^2 + 14x + 20
f(x) = [x^2 + 14x] + 20
f(x) = [x^2 + 14x + 49] + 20 - 49
f(x) = (x+7)^2 -29

Let's try one more, a bit harder this time. Same logic applies.

f(x) = 4x^2 + 4x + 19
f(x) = [4x^2 + 4x] + 19
f(x) = [4x^2 + 4x + 1] + 19 -1
f(x) = (2x + 1)^2 + 18

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