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Math Help - [SOLVED] Finding polynomials from a diagram...

  1. #1
    PTL
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    [SOLVED] Finding polynomials from a diagram...

    Find polynomials p(x) and q(x) such that f(x) = p(x)/q(x) has vertical asymptotes x=-2, x=5, horizontal asymptote y=1 and f(-1) = f(4)=0.


    Any hints on how to do this?
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  2. #2
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    as x\neq -2,5 then q(x) = (x+2)(x-5)

    A horizontal asymptote could imply a constant term of +1 in f(x)

    The points f(-1) = f(4)=0 will be used to solve any arbitrary constants.
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  3. #3
    PTL
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    Does that mean we have f(x) = \frac{p(x)}{x^{2}-3x-10}

    So p(-1) = 0
    p(4) = 0
    and p(x)/q(x) has a local max at (1.5, almost(1))?

    So the derivative of \frac{p(x)}{x^{2}-3x-10} at 1.5 = 0? (1.5 because it's half-way between -2 and 5..

    Is that how I'm supposed to solve for p(x)?
    Last edited by PTL; August 2nd 2009 at 03:03 PM. Reason: oops... latex...
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  4. #4
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    There's a lot of Latex errors there my friend.
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  5. #5
    PTL
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    Quote Originally Posted by pickslides View Post
    There's a lot of Latex errors there my friend.
    Fixed that, sorry...
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  6. #6
    Senior Member DeMath's Avatar
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    Quote Originally Posted by PTL View Post
    Find polynomials p(x) and q(x) such that f(x) = p(x)/q(x) has vertical asymptotes x=-2, x=5, horizontal asymptote y=1 and f(-1) = f(4)=0.

    Any hints on how to do this?
    If \left[ \begin{gathered}f\left( { - 1} \right) = \frac{{p\left( { - 1} \right)}}{{q\left( { - 1} \right)}}=0, \hfill \\f\left( 4 \right) = \frac{{p\left( 4 \right)}}{{q\left( 4 \right)}} = 0; \hfill \\ \end{gathered}  \right. then \left[ \begin{gathered}p\left( { - 1} \right) = 0, \hfill \\p\left( 4 \right) = 0; \hfill \\ \end{gathered}  \right. \Rightarrow p\left( x \right) = a\left( {x + 1} \right)\left( {x - 4} \right).

    So, we have

    f\left( x \right) = a \cdot \frac{{\left( {x + 1} \right)\left( {x - 4} \right)}}{{\left( {x + 2} \right)\left( {x - 5} \right)}} = a \cdot \frac{{{x^2} - 3x - 4}}{{{x^2} - 3x - 10}}.

    If y=1 is a horizontal asymptote, then  \mathop {\lim }\limits_{x \to \infty } f\left( x \right) = 1.

    Find the limit

    \mathop {\lim }\limits_{x \to \infty } f\left( x \right) = a \cdot\mathop {\lim }\limits_{x \to \infty } \frac{{{x^2} - 3x - 4}}{{{x^2} - 3x - 10}} = a \cdot\mathop {\lim }\limits_{x \to \infty } \frac{{1 - {3 \mathord{\left/{\vphantom {3 x}} \right.\kern-\nulldelimiterspace} x} - {4 \mathord{\left/{\vphantom {4 {{x^2}}}} \right.\kern-\nulldelimiterspace} {{x^2}}}}}{{1 - {3 \mathord{\left/{\vphantom {3 x}} \right.\kern-\nulldelimiterspace} x} - {4 \mathord{\left/{\vphantom {4 {{x^2}}}} \right.\kern-\nulldelimiterspace} {{x^2}}}}} = a \cdot1.

    So a = 1

    Finally f\left( x \right) = \frac{{{x^2} - 3x - 4}}{{{x^2} - 3x - 10}}.
    Last edited by DeMath; August 4th 2009 at 03:46 AM. Reason: typo
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  7. #7
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    Hello, PTL!



    Find polynomials p(x) and q(x) such that: . f(x) \:=\:\frac{p(x)}{q(x)} has vertical asymptotes x=-2, x=5,
    horizontal asymptote y=1 and f(-1) \:=\: f(4)\:=\:0
    For vertical asymptotes, the denominator must go to zero for x = -2\text{ and }x = 5
    . . Hence: . q(x) \:=\:(x+2)(x-5)
    So we have: . f(x) \;=\;\frac{p(x)}{(x+2)(x-5)}

    For the horizontal asymptote, the function goes to 1 as x \to \infty
    . . Hence, p(x) has the same degree as q(x), and has a leading coefficient of 1.
    That is: . p(x) \:=\:x^2 + ax + b

    And we have: . f(x) \;=\;\frac{x^2+ax+b}{(x+2)(x-5)}


    Since f(\text{-}1) \:=\:0\!:\quad f(\text{-}1) \;=\;\frac{(\text{-}1)^2 + a(\text{-}1) + b}{(\text{-}1+2)(\text{-}1-5)} \:=\:0 \quad\Rightarrow\quad a - b \:=\:1 .[1]

    Since f(4) \:=\:0\!:\quad f(4) \;=\;\frac{4^2 + a(4) + b}{(4+2)(4-5)} \:=\:0 \quad\Rightarrow\quad 4a + b \:=\:-16 .[2]

    Add [1] and [2]: . 5a \:=\:-15 \quad\Rightarrow\quad \boxed{a \,=\,-3} \quad\Rightarrow\quad\boxed{ b \,=\,-4}


    Therefore: . f(x) \;=\;\frac{x^2 - 3x - 4}{(x+2)(x-5)}



    Edit: Ah ... DeMath beat me to it.
    Oh well, my explanation is slightly different.
    .
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  8. #8
    Senior Member DeMath's Avatar
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    Also see picture

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