Completing the Square
For Leading Coefficients that aren't 1

We looked at completing the square for equations who's leading coefficients are 1.  If the leading factor is not 1 we factor the number out of the leading coefficient and the second term so that the leading coefficient is one and then we proceed to complete the square.

The technique is very easy.

Lets look at a quadratic equation in the standard form.

Y=AX²+BX+C 
We factor A out of both the A and B term.
Y=A(X²+[B/A]X)+C 
Notice how we don't factor the A term out of C

To write this equation in vertex form, we take 1/2 of the B/A value and write our equation like this:

Y=(X+[1/2][A/B])²+k

It looks kind of messy but lets use an example.
Y=2X²+4X-3

Our equation now looks like this:

Y=2(X²+[4/2]X)-3 =2(X²+2X)-3
Y=2(X+1)²-3-(2)(1)²=2(X+1)²-5
Vertex is at X=-1 Y=-5 or (-1,-5)

So watch what we did with the k term.  
Lets expand 2(X+1)²

2X²+4X+2So we gained a 2 in the expansion of our modified equation so we have to also subtract a 2.
So our form becomes A(X-h)²+k
k is calculated by taking C-A(h)²

So lets do another example:
Y=-3X²-12X+4
Factor the -3 out of the first and second term:
-3(X²+4X)+4
Complete the square:
-3(X+2)²+4-(-3)(2)²=-3(X+2)²+16
The vertex is at:  X=-2  Y=16 or (-2,16)

Practice:
Y=6X²+18X+2
Y=-2X²+5X+3
Y=-7X²-21X+14

Answers posted in next post: