How do I convert this quadratic equation into vertex form?

It was originally in intercept form as f(x) = (x + 6)(x - 2), but I now have it in standard form as f(x) = x^2 + 4x - 12.

I just need help converting to vertex form.

Question

How do I convert this quadratic equation into vertex form?

It was originally in intercept form as f(x) = (x + 6)(x - 2), but I now have it in standard form as f(x) = x^2 + 4x - 12.

I just need help converting to vertex form.

anonymous · Answer

I know it has something to do with "completing the square" but I'm not quite sure how to do that.

hartnn · Answer

what is the co-efficient of x ?
(we are starting completing the square)

anonymous · Answer

4... right?

hartnn · Answer

correct!
divide it by 2 and square it, what u get ?

anonymous · Answer

4/2 = 2, and 2 squared is 4 again

hartnn · Answer

right , so add and subtract 4

hartnn · Answer

(x^2+4x+4)-4-12
got what i did?

hartnn · Answer

my final aim is to bring it in vertex form 
$\Large f(x)=a(x-h)^2+k $

anonymous · Answer

Oh yeah I see now, took me a minute lol

anonymous · Answer

So (x^2 + 4x + 4) - 4 - 12 is simplified into (x^2 + 4x + 4) - 16... and (x^2 + 4x + 4) can be factored into (x + 2)^2?

anonymous · Answer

I was told vertex form is y - k = a(x - h)^2

hartnn · Answer

same thing. 
and yes
f(x)=(x+2)^2-16 is correct vertex form :)

anonymous · Answer

okay thank you (:

anonymous · Answer

so the x-value of the vertex would be -2?

hartnn · Answer

yup.

hartnn · Answer

welcome ^_^