# Math

find a rational function that satisfies the given conditions vertical asymptotes x=-6,x=7 horizontal asymptotes y=10/9 x intercepts (6,0)

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1. vertical asymptotes:

y = 1/((x+6)(x-7))
y(6) = 0, so
y = (x-6)/((x+6)(x-7))

horizontal asymptote at y = 10/9 needs a quadratic on top, so we could have

y = (10/9)(x-6)^2/((x+6)(x-7))

but that does not really cross the x-axis at x=6, but just touches it. So, how do we get that quadratic on top with no other x-intercepts? We could have a quadratic factor on top that is never zero, but then we need another linear factor underneath, such as

y = (10/9)(x-6)(x^2+1)/((x+6)^2(x-7))

See

http://www.wolframalpha.com/input/?i=(10%2F9)(x-6)(x%5E2%2B1)%2F((x%2B6)%5E2(x-7))

kind of a strange beast, but it does the job.

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