We begin with #y=2x^2+4x-5#. We need to rewrite this into vertex form. The easiest way to do that is to complete the square. This is a very useful way to factor, and I don't want to spend too much time explaining it right now, but I'll cover it as I go.
In order to begin completing the square we need to factor the equation until the leading coefficient (meaning #a# from #ax^2+bx+c#) is #1#. Sometimes we get lucky and the leading coefficient is already #1#, but in this case we'll need to factor. So, we factor out the #2#, which leaves us with #2(x^2+2x-5/2)#. Now, I realize that fraction might look scary, but don't worry about it. Fractions are no big deal, and I advise you not to convert it into decimal form. I don't want to lose any value based on rounding.
So, our next step is to take the middle component (#b#, or in this case #2#) and divide it in half. That gives us #1#. We take that value and square it. #1^2# is just #1#. We add that to our equation, BUT we need to subtract it. IF we add #1# and then subtract #1#, we haven't changed the value of the equation, which is important. We could add #523, 650, 320#, as long as we also subtract it.
We have #2(x^2 +2x +1 -1 -5/2)#. Now, please notice how lovely #x^2+2x+1# looks.T hat is a perfect square, and we can combine it to this: #(x+1)^2#. If you don't belive me, just multiply #(x+1)*(x+1)#. Now, we still need to deal with #-1+5/2#. Let's combine them and see what we have to work with.
If we put it all together, we'll have #2((x+1)^2-7/2)#. I want to clean this up a bit, so let's mutiply that #2# out. #-7/2*2# leaves us with just #-7#. We are left with #2(x+1)^2-7#, which is in vertex form . Nice work, we're done!
