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y=a(x-h)^2+k
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osnap1584
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RE: y=a(x-h)^2+k
just playing you'll learn it when you take calculus. the answer to my question has logs by the way. Basically it is used o find the area under the curve of a graph. I haven't really grasped the concept itself but it's used a lot in physics as well.

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17/12/2008 09:38 PM
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Messages In This Thread
y=a(x-h)^2+k - Slushba132 - 17/12/2008, 05:41 PM
RE: y=a(x-h)^2+k - Mickey - 17/12/2008, 06:09 PM
RE: y=a(x-h)^2+k - roberth - 17/12/2008, 06:27 PM
RE: y=a(x-h)^2+k - Mickey - 17/12/2008, 06:36 PM
RE: y=a(x-h)^2+k - feinicks - 17/12/2008, 07:28 PM
RE: y=a(x-h)^2+k - Kaiser - 17/12/2008, 07:30 PM
RE: y=a(x-h)^2+k - Mickey - 17/12/2008, 07:32 PM
RE: y=a(x-h)^2+k - Kaiser - 17/12/2008, 07:32 PM
RE: y=a(x-h)^2+k - Mickey - 17/12/2008, 07:34 PM
RE: y=a(x-h)^2+k - Kaiser - 17/12/2008, 07:37 PM
RE: y=a(x-h)^2+k - Mickey - 17/12/2008, 07:46 PM
RE: y=a(x-h)^2+k - Kaiser - 17/12/2008, 07:57 PM
RE: y=a(x-h)^2+k - Mickey - 17/12/2008, 07:59 PM
RE: y=a(x-h)^2+k - Slushba132 - 17/12/2008, 08:06 PM
RE: y=a(x-h)^2+k - diego - 17/12/2008, 08:18 PM
RE: y=a(x-h)^2+k - roberth - 17/12/2008, 08:35 PM
RE: y=a(x-h)^2+k - osnap1584 - 17/12/2008, 09:05 PM
RE: y=a(x-h)^2+k - Slushba132 - 17/12/2008, 09:28 PM
RE: y=a(x-h)^2+k - osnap1584 - 17/12/2008 09:38 PM
RE: y=a(x-h)^2+k - Assassinator - 18/12/2008, 12:38 AM
RE: y=a(x-h)^2+k - roberth - 18/12/2008, 06:10 AM

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