2007-09-12

For #A6- be sure to draw your triangle.

Minilesson on completing the square

Let's take a trip back to why/when this completing the square stuff came from. Algebra 2, yes. So, let's say that y = x^2 + 6x - 7. This would be a parabola, and if you complete the square you'll get it in the form of y=a(x-h)^2 + k. By completing the square, you are putting it in that form. To complete the square, you want the x^2 to have a coefficient of 1. You're going to group your x^2 and x terms together: y = (x^2 + 6x ) - 7. You want to factor this thing so that it looks like y = ( )^2. So you want to factor it so that it's the same number, same term, you get to choose the number to put there. You have to look for two numbers that multiuply to give you the blank, and add to give you 6, so the numbers have to be the same, so in this case it's 3 & 3. And then square this "perfect square": 3^2 = 9. You can either add nine to both sides, or you can subtract nine as well on the end so it's y = (x^2 + 6x + 9) + 7 -9. Factor this, and you get (x+3)^2 - 16. The vertex of this parabola is (-3, -16). That's all you are doing when completing the square. We will two more of these problems. But what happens when you have a coefficient on your x^2? You get something like y = 3x^2 + 6x - 7, and then you get y = 3(x^2 + 2x + 1) - 7 - 3. You do not subtract 1, you have to remember that you had that giant coefficient. Now you can rewrite/factor and get something that looks better.

To find the vertex of a parabola, take the derivative and find where it is zero, and so you can find your x value, instead of remembering the quadratic equation.