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Kaiser
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RE: y=a(x-h)^2+k

ChIcKeN BallZ Wrote:but (x-h)(x-h) is the factored form to solve, First.Outer.Inner.Last.

you cannot factor (x-h)² except with special products, since it isn't a multiplication inside, you can't just square both terms

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17/12/2008 07:37 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 - roberth - 18/12/2008, 06:10 AM

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