So today in class, for those of you who were not there, or were there and are still lost about the new concept we went over which was Completing a Square, here is a little summarization of what had happened.

We were first given equations that needed to be factored out, these were the examples:

x² + 2x + 1 --> (x-1)²
x² - 2x + 1 --> (x-1)²
x² + 4x + 4 --> (x-2)²
x² + 10x + 25 --> (x+5)²
x² -8x +16 --> (x-4)²

the solutions that are on the right hand side of the equation is the result you get from using Completing a Square method. By using the method which is take the middle term divide it by two and then square it.

Notice that square root of the third term in the equation is result you get when you factor using completing the square. For example, for the third equation the square root of 4 is 2.

Now, converting Y = ax² + bx + c into Y= a(x-h)² + k

For example:
Y = x² + 4x -5
Y = x² + 4x + ___ - 5
^ Notice when the equation is rewritten there is a blank space, that is for the unknown value to complete the square.
Y = (x² + 4x + _4_) -5-4
^ since you added 4 in the brackets you would need to subtract 4 from the outside of the bracket so it is balanced, and it won't change the outcome of the equation. Having said that, the answer that you would get would be:
Y = (x+2)² - 9

Now since your equation is converted into the Y = a(x-h)² + k, it will be easier for you to find these:

Vertex : (-2, -9)
A.O.S. : x = -2
Domain : (-oo, oo)
Range: [-9, oo)

Oh btw, the "oo" are supposed to represent infinity, sorry I'm new at this . okay anyways, back to the topic.

Okay, now let's try this example:

Y = 3x² - 12x + 16

Since the leading coefficient is more than 1, you need to factor this out. Which will give you an equation of:

Y = 3(x² - 4x + ___ ) + 16
Now use the completing the square method, so you can fill in the blank space.
Y = 3(x² - 4x + _4_) 16 - 12
The same method applies, and since you've added 12 in the brackets you would subtract 12 from the outside to balance it. The end result will give you,
Y = 3(x-2)² + 4

Now everything will be easier, when asked to find the vertex, axis of symmetry, domain and range of this equation.

The last example which was given to us was a tricky one, and this is how it looked like

Y = -2x² -5x + 3 ; if you already noticed as 5 being the middle term, it is not easy to divide it by 2. Since the leading coefficient of this equation is negative you would need to factor it out. Giving you this:

Y = -2(x² + 5/2x + ___) + 3 .

Now you would divide 5/2 by 2. By doing this is it the same as multiplying it which looks like this:

5/2 * ½ = 5/4
(5/4 * 5/4) = 25 / 16
Now you would multiply the 25 / 16 by 2/1 . Or if it is easier for you, you would right away notice that if you were to cross the two 16 divided by 2 would give you 8 . and now it would look like this:

25 / 8 * 1 / 1 which is 25 / 8. Now you would plug everything that you had done into the equation.

Y = -2(x² + 5/2+ _25/16_) + 3 + 25/8

24/8 + 25/8 = 49/8

you would end up with the equation looking like this:

Y = -2(x² + 5/4)² +48/8.

So far, I hope all of you reading this are getting me so far, If not I apologize because I'm not sure if I summarized it good enough. But, Basically it is taking the middle term , dividing it by 2 and squaring it. Then whatever you get you would put that as the third term in the equation inside of the bracket. Whatever you had added in the bracket you would do the opposite of it outside, just to balance the equation. Well, I think this about wraps about what we had learnt today in class. Yes, I know its quite long. Oh and due for tomorrow in class is Excercise 5, Questions 1 and 2 .

And the Scribe for tomorrow is ROSLYN =)=)