Give an example of a rational function that has vertical asymptote x = 3 and x = -3, horizontal asymptote y = 2 and y-intercept is (0, 4)

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  1. f(x) = (x-4)/(x^2-9) + 2

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  2. Hmmm. f(0) = -4/-9 + 2 ≠ 4

    How about

    y = -18/(x^2-9) + 2 = 2(x^2-18)/(x^2-9)

    That also has two x-intercepts. If you don't want any of those, then you will need to have y>0 for all x.

    y = 2*9^2/(x^2-9)^2 + 2

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