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Slushba132 · 17/12/2008 05:41 PM

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

Mickey · 17/12/2008 06:09 PM

must use logs!!! too much work, sorry slushie

roberth · 17/12/2008 06:27 PM

i used to be able to do these, but i havent done maths in two years, and i hated it

Also, this is probably slushbas homework but he doesn't want to do it himself :P

Mickey · 17/12/2008 06:36 PM

y=a(x-h)(x-h)+k
a(x-h)(x-h)=y+k/(x-h)(x-h)
a=(y+k)/[(x-h)(x-h)]

my quick 10 second guess

feinicks · 17/12/2008 07:28 PM

answer:

Spoiler:

Kaiser · 17/12/2008 07:30 PM

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

y=a(x-h)²+k

(y-k)/(x-h)²= a

too easy

by the way, that is a quadratic function

Mickey · 17/12/2008 07:32 PM

Kaiser 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

y=a(x-h)²+k

(y-k)/(x-h)²= a

too easy

by the way, that is a quadratic function

I WAS RIGHT Inluv

Kaiser · 17/12/2008 07:32 PM

ChIcKeN BallZ Wrote:a=(y+k)/[(x-h)(x-h)]

my quick 10 second guess

(x-h)² is not the same as X²-h²

x²-h² = (x+h)(x-h)
(x-h)² = X² - 2xh + h²

Mickey · 17/12/2008 07:34 PM

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

(x*x)+(x*h)+(h*x)+(h*h)
x^2+2hx+h^2

Kaiser · 17/12/2008 07:37 PM

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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