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y=a(x-h)^2+k
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roberth
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RE: y=a(x-h)^2+k
Assassinator Wrote:
Slushba132 Wrote:solve for "a"...

or at least find x, y, h and k in terms of a

so
a = ""

50 e-pigs for the person who does it correctly

You mean... find a in terms of x, y, h and k... right?

Heh, way too easy.

osnap1584 Wrote:Anyone want to find the integral of (x+5)/[(x-1)(x+2)] ?

Harder than Slushie's one. But pretty trivial if you know what you're doing.

Use partial fraction decomposition.

∫(x+5)/[(x-1)(x+2)]= ∫[A/(x-1)+B/(x+2)]

Ax+2A+Bx-B = x+5
=> A=2, B=-1.

∫(x+5)/[(x-1)(x+2)]
=∫[2/(x-1)-1/(x+2)]
=2ln(x-1)-ln(x+2)+C.

Edit: Stupid typo. Wrote "+" instead of "-".

wait..WHAT!!!

(This post was last modified: 18/12/2008 06:14 AM by Assassinator.)
18/12/2008 06:10 AM
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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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