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November 1, 2014

November 1, 2014

Posted by **I Need Help** on Sunday, April 11, 2010 at 12:47pm.

- Math -
**Damon**, Sunday, April 11, 2010 at 1:30pmWell first lets get one with three real zeros.

y = k(x-x1)(x-x2)(x-x3)

but one happens twice

y = k(x-x1)(x-x2)(x-x2)

denominator = 0 at x=-2 and x = 3

y = c[(x-x1)(x-x2)(x-x2)]/[(x+2)(x-3)]

when x = 0, y = 1

y =1 = c[-x1 x2^2]/-6

6 = c x1 x2^2

when x and y are big, y = 2x

big x

y = c x^3/x^2 = c x

so c = 2

so

x1 x2^2 = 3 or x1 = x2^2/3

y = 2[(x-x1)(x-x2)(x-x2)]/[(x+2)(x-3)]

y = 2[(x-x2^2/3)(x-x2)(x-x2)]/[(x+2)(x-3)]

hmmm, unless I am missing something I am still free to choose x2

- error -
**Damon**, Sunday, April 11, 2010 at 1:43pmx1 x2^2 = 3 or x1 = 3/x2^2

y = 2[(x-x1)(x-x2)(x-x2)]/[(x+2)(x-3)]

y = 2[(x - 3/x2^2)(x-x2)(x-x2)]/[(x+2)(x-3)]

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