# Rewrite the equation in vertex form. Then find the Vertex`y=4x^2+4x+1`

txmedteach | Certified Educator

In trying to find the vertex here, you are trying to get the equation into the form:

`y = a(x-h)^2+k`

Where "a" is the coefficient of the `x^2` term and (h,k) is the coordinates of the vertex. Note that sometimes the equation replaces a with 4p. This has to do with alternate definitions of the parabola that don't concern us for this problem.

So, that's not bad, but we have to go from your equation to the equation we put right here. So..how do we do that?

First, start by factoring from each term a 4, but leave the constant out of the parentheses. This will sometimes make some horrible fractions, but it's where you start:

`y = 4(x^2 + x) + 1`

Now, you want to make what's inside the parentheses a perfect square trinomial. In order to do that, look at the coefficient of the x (just x, not `x^2`). In our case this is 1. To make the perfect square, we divide this number by 2, and then we square it. In our case:

`1-:2 = 1/2`

`(1/2)^2 = 1/4`

So, what we'll do is both add and subtract this term from inside the parentheses (if we just add it, then we change the equation!):

`y = 4(x^2+x+1/4 - 1/4) +1`

Now, we want to remove that `-1/4` because it's not part of our perfect square (if you don't get why, trust me for now, I'll show you later why you do this). To do this, you recognize that you can multiply it by 4 to get it out of the parentheses (again, the distributive property at work):

`y = 4(x^2+x+1/4) -4*1/4 + 1`

`y = 4(x^2 + x + 1/4) - 1 + 1`

`y = 4(x^2+x+1/4)`

Hooray! that got rid of some of the equation (doesn't always happen this way!). Now, we see why we added that fraction in there. Well, it turns out that the above equation can be factored readily into this:

`y = 4(x+1/2)^2`

If you don't believe me, FOIL out `(x+1/2)(x+1/2)` to get `x^2 + 1/2x + 1/2x + 1/2*1/2` which is `x^2 + x + 1/4`.

Now, we have the equation in the form we need: vertex form. Just a reminder, vertex form:

`y = a(x-h)^2 + k`

Our equation:

`y = 4(x+1/2)^2`

Hopefully, you can see that our equation is the same as the vertex form, just with k=0.

Now, to find the vertex, we just find (h,k). Now, we just said k=0. But what about h? The vertex form specifies (x-h) in the parentheses, but we have `x+1/2`! Well, remember:

`x+1/2 = x-(-1/2)`

So our h, then, is -1/2.

Therefore, our vertex will be:

`(h,k) = (1/2, 0)`

Hope that helps some!