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Math Help - Finding a polynomial equation based off its graph

  1. #1
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    Finding a polynomial equation based off its graph



    So it wants me to find a formula. I got ((-9)(x-1))/((x+3)(x-3)). The last part of the graph is not coming out correctly though. How do I alter the formula so that the right side of the graph flips over the x-axis?
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  2. #2
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    Re: Finding a polynomial equation based off its graph

    Hello, DasRabbit!


    There are vertical asymptotes at x = 3 and x = \text{-}3.
    f(x) has (x-3)(x+3) in the denominator.

    With the "usual" vertical asymptote, the graph goes to +\infty on one side
    . . and to \text{-}\infty on the other side (as it does at x = \text{-}3).

    At x = 3, the graph goes to +\infty on both sides.
    This indicates that the factor (x-3) in the denominator is to an even power.
    . . Hence, the denominator has: . (x-3)^2(x+3)

    Since there is a horizontal asymptote (y = 0), the numerator has a degree less than 3.
    . . Assume (for simplicity) that the numerator is linear.

    We have: . f(x) \:=\:\frac{ax+b}{(x-3)^2(x+3)}

    We want: f(0) = \text{-}1\!:\;\frac{b}{(-3)^2(3)} \:=\:\text{-}1 \quad\Rightarrow\quad b \,=\,\text{-}27

    We want: f(1) = 0\!:\;\frac{a+b}{(\text{-}2)^2(4)} \:=\:0 \quad\Rightarrow\quad\frac{a-27}{16} \:=\:0 \quad\Rightarrow\quad a \:=\:27

    Therefore: . f(x) \:=\:\frac{27x - 27}{(x-3)^2(x+3)} \quad\Rightarrow\quad f(x) \:=\:\frac{27(x-1)}{(x-3)^2(x+3)}

    Test a few points to ensure that graph is on the correct sides of the x-axis.
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