Results 1 to 8 of 8

Thread: [SOLVED] Finding polynomials from a diagram...

  1. #1
    PTL
    PTL is offline
    Junior Member
    Joined
    Aug 2009
    Posts
    29

    [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?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,237
    Thanks
    33
    as $\displaystyle x\neq -2,5$ then $\displaystyle q(x) = (x+2)(x-5) $

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

    The points $\displaystyle f(-1) = f(4)=0$ will be used to solve any arbitrary constants.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    PTL
    PTL is offline
    Junior Member
    Joined
    Aug 2009
    Posts
    29
    Does that mean we have f(x) = $\displaystyle \frac{p(x)}{x^{2}-3x-10}$

    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 $\displaystyle \frac{p(x)}{x^{2}-3x-10}$ 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; Aug 2nd 2009 at 03:03 PM. Reason: oops... latex...
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,237
    Thanks
    33
    There's a lot of Latex errors there my friend.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    PTL
    PTL is offline
    Junior Member
    Joined
    Aug 2009
    Posts
    29
    Quote Originally Posted by pickslides View Post
    There's a lot of Latex errors there my friend.
    Fixed that, sorry...
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Senior Member DeMath's Avatar
    Joined
    Nov 2008
    From
    Moscow
    Posts
    474
    Thanks
    5
    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 $\displaystyle \left[ \begin{gathered}f\left( { - 1} \right) = \frac{{p\left( { - 1} \right)}}{{q\left( { - 1} \right)}}=0, \hfill \\f\left( 4 \right) = \frac{{p\left( 4 \right)}}{{q\left( 4 \right)}} = 0; \hfill \\ \end{gathered} \right.$ then $\displaystyle \left[ \begin{gathered}p\left( { - 1} \right) = 0, \hfill \\p\left( 4 \right) = 0; \hfill \\ \end{gathered} \right. \Rightarrow p\left( x \right) = a\left( {x + 1} \right)\left( {x - 4} \right).$

    So, we have

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

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

    Find the limit

    $\displaystyle \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 }$$\displaystyle \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 $\displaystyle a = 1$

    Finally $\displaystyle f\left( x \right) = \frac{{{x^2} - 3x - 4}}{{{x^2} - 3x - 10}}.$
    Last edited by DeMath; Aug 4th 2009 at 03:46 AM. Reason: typo
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    12,028
    Thanks
    848
    Hello, PTL!



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

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

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


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

    Since $\displaystyle 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]: .$\displaystyle 5a \:=\:-15 \quad\Rightarrow\quad \boxed{a \,=\,-3} \quad\Rightarrow\quad\boxed{ b \,=\,-4} $


    Therefore: .$\displaystyle 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.
    .
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Senior Member DeMath's Avatar
    Joined
    Nov 2008
    From
    Moscow
    Posts
    474
    Thanks
    5
    Also see picture

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Venn Diagram & Logic: Help
    Posted in the Discrete Math Forum
    Replies: 2
    Last Post: May 27th 2009, 09:50 AM
  2. [SOLVED] venn diagram help!
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: Feb 23rd 2009, 12:25 AM
  3. [SOLVED] Vennn Diagram help..
    Posted in the Statistics Forum
    Replies: 6
    Last Post: Feb 14th 2009, 11:59 AM
  4. [SOLVED] Venn Diagram
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: Jan 31st 2009, 11:08 PM
  5. [SOLVED] [SOLVED] venn diagram etc
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: Nov 2nd 2005, 10:39 AM

Search Tags


/mathhelpforum @mathhelpforum