Completing the square is a way of factoring, and it is also a way to put a quadratic into vertex form to determine the vertex of the parabola. To begin completing the square, set the equation equal to 0 and then move the constant over by subtraction to get [tex] 2x^2-6x=-4 [/tex]. The rule for completing the square is that the leading coefficient has to be a 1. Ours is a 2. So we have to factor it out. [tex] 2(x^2-3x)=-4 [/tex]. We will take half the linear term, square it, and add it to both sides. Our linear term is 3. Half of 3 is 3/2 and 3/2 squared is 9/4. We add 9/4 inside the parenthesis. But outside is the 2 we factored out, hanging out front there, refusing to be ignored. It is a multiplier. That means what we have actually added in is 2*9/4 which is 18/4. That looks like this: [tex] 2(x^2-3x+\frac{9}{4})=-4+\frac{18}{4} [/tex]. On the right we will do the addition there by finding a common denominator, but on the left we have, in this process, created a perfect square binomial, the whole point of this nightmare. That looks like this: [tex] 2(x-\frac{3}{2})^2=\frac{1}{2} [/tex]. The only left to do, if you need to, is move the 1/2 back over to the other side of the equals sign and set the quadratic back equal to y. [tex] 2(x-\frac{3}{2})^2-\frac{1}{2}=y [/tex]. The vertex of our parabola is (3/2, -1/2).
