dinak23
y=x^2-4x-12 factored is (x-6) (x+2), how do i determine vertex?
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Compassionate
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vertex form is y = a(x-h)^2 +k where vertex is (h,k)
dinak23
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so how do i put this equation into vertex form?
dinak23
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i just dont see where the h and k come in from
dinak23
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(x^2-h)^2+k?
eSpeX
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'h' and 'k' are just place holders. They, like 'a', can be anything.
dinak23
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so how do i find the vertex now? what do i have to put in for h and k
Loser66
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|dw:1372118543905:dw|
dinak23
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You broke down 12 into a 4 and 8? I see that
dinak23
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@loser66 why did you add 4 and subtract 4?
eSpeX
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As @Loser66 pointed out, by adding (+4) and subtracting (-4) you are adding 0 to the equation.
dinak23
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Is this completing the square?
eSpeX
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Similar in principle but not the exact same thing. You are just massaging the equation to fit the form you need, without changing anything.
eSpeX
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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.
dinak23
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I see what you guys did now
dinak23
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Alright i understand, so the vertex is -2,-16 ?
dinak23
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I just checked my answer sheet and thats right! Thanks alot!!
eSpeX
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Yes, good job everyone. :D
dinak23
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You guys are awesome!