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January 30, 2015

January 30, 2015

Posted by **Brianna** on Sunday, October 16, 2011 at 11:01pm.

- ****college algebra…radical functions**** -
**Steve**, Monday, October 17, 2011 at 1:39pmYou know that the denominator has to be zero at x=5 and x=-7

The hole means that the numerator and denominator are both zero at x=2

so, we can start with

f(x) = (x-2)/[(x-2)(x-5)(x+7)]

Now, we need a horizontal asymptote at y=14. That means that the numerator and denominator must have the same degree, x^n, with the highest degree having a coefficient in the numerator 14 times that in the denominator. So, the simplest one I can think of is

f(x) = 14x^2(x-2)/[(x-2)(x-5)(x+7)]

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