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y=x^2-4x-12 factored is (x-6) (x+2), how do i determine vertex?

Mathematics
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vertex form is y = a(x-h)^2 +k where vertex is (h,k)
so how do i put this equation into vertex form?
i just dont see where the h and k come in from

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Other answers:

(x^2-h)^2+k?
'h' and 'k' are just place holders. They, like 'a', can be anything.
so how do i find the vertex now? what do i have to put in for h and k
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You broke down 12 into a 4 and 8? I see that
@loser66 why did you add 4 and subtract 4?
As @Loser66 pointed out, by adding (+4) and subtracting (-4) you are adding 0 to the equation.
Is this completing the square?
Similar in principle but not the exact same thing. You are just massaging the equation to fit the form you need, without changing anything.
It's like multiplying two fractions to get a common denominator, or changing both sides of an equation equally, they are all about getting the numbers to "look" like what you need.
I see what you guys did now
Alright i understand, so the vertex is -2,-16 ?
I just checked my answer sheet and thats right! Thanks alot!!
Yes, good job everyone. :D
You guys are awesome!

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