# Math Help - Identifying the vertex of a Quadratic Function

1. ## Identifying the vertex of a Quadratic Function

I really don't mean to be a nusance, I am new here and hope to be more of a contributer of this site rather that bothering people, so I apologize. I seem to have a hard time with a specific step in taking a quadratic trinomial that cannot be factored and and converting it into a vertex formula [f(x)=(x-h)²+k].

In this case:

f(x)= 2x² + 8x + 7

I know that the first step is to group the x terms leaving the leading coefficient out of the parenthesis such as:

f(x)= 2(x² + 4x) + 7

Next step is to take the middle term and square half the coefficient: (4/2)² and then add and subtract by that number.

f(x)= 2(x² + 4x+4-4) + 7

Now here is where it gets fuzzy for me, I read this online and here is the further step according to what I read:

Now, move the second -4 to the outside of the parenthesis, only now we have to consider that the outside coefficient isn't just a simple positive one. There's a 2 out there.
f(x) = 2(x² + 4x + 4) - 2(4) + 7

Question, is that -2 that is underlined the -4 that was outside the parentesis? And if so where did that four to the right of it come from? I'm stumped as to how it got there because if the -2 was the -4 inside the parenthesis and we have the +4 inside the parenthesis, that (4) that is next to the -2 seems to have crashed the party. I just can't figure out how it got there according to there steps and there seems to be no sufficient information as to how it got there, please explain.

2. Alright, given: $f(x)=2x^2+8x+7$

You want it in the form $f(x)=a(x-h)^2+k$, which can also be written as $y-k=a(x-h)^2$

So, subtract over 7 and factor out a 2: $y-7=2(x^2+4x)$

You understand the step of completing the square. Divide 4 by 2, square, and add: $2(x^2+4x+4)$ Because we added 8 (you have to multiple $a$ through the quadratic) to one side, we add it to the other:

$y+1=2(x+2)^2$, or
$f(x)=2(x+2)^2-1$

Vertex = $(h,k)$, or $(-2, -1)$