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Homework Help: Rational Functions Question

  1. Nov 23, 2007 #1
    1. The problem statement, all variables and given/known data
    Find two constants for 'a' and 'b' such that the verticle asymptote will be [tex]\pm[/tex] [tex]\frac{3}{5}[/tex]

    y=[tex]\frac{ax^2+7}{9-bx^2}[/tex]


    I rearranged so that it becomes [tex]-bx^2+8[/tex] in the denominator since i know that there are two roots that are [tex]\pm[/tex] it must be a square and since 3 is the numerator of the root it must -9 .... so i rearranged again to get

    y=[tex]\frac{-ax^2-7}{bx^2-9}[/tex]

    in which case i found the constant for a (-1) and [tex]5^2[/tex] is 25 so i found b as well so the equation would be

    y=[tex]\frac{-x^2-7}{25x^2-9}[/tex]

    is this right??? I have no way to check my answer so i just want to make sure :D
     
  2. jcsd
  3. Nov 23, 2007 #2

    rock.freak667

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    well for a vertical asymptote...the denominator of the function should be zero
    in your case [itex]9-bx^2=0[/itex]
    so that [itex]x=\pm\fract{3}{\sqrt{b}}[/itex]
    so then equate that to [itex]\pm\frac{3}{5}[/tex] and find b
     
  4. Nov 23, 2007 #3
    not gonna lie i don't get it...
    would 25x^2-9 give u two values of x that equate to 0???
    (5x-3)(5x+3).

    I just don't understand what your doing there lol.
     
  5. Nov 23, 2007 #4

    symbolipoint

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    Another try: you want [itex]9-bx^2=0[/itex]
    when x = +3/5 and x = -3/5; so if you want to do this in a crude way, just find the expression for b. This is b=(-9)/(x^2). So what is x ? You were already given the x values, since you want the vertical asymtotes at x=+3/5 and x=-3/5. Find b for both of these values by substituting. ...
    b=25.

    I see no particular big restrictions on a, except that a is not equal to zero.
     
  6. Nov 23, 2007 #5

    rock.freak667

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    Can't it be for all values of a? as the value of 'a' doesn't affect the vertical asymptotes
     
  7. Nov 23, 2007 #6
    yah all values can be 'a'. i just made it -1 so that the denominator would have better form.... i think i did that right i wasn't EXACTLY sure if i could change all the signs in the equation by multiplying thru by -1 to move the negative to the top ??? :S
     
  8. Nov 24, 2007 #7

    Dick

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    The value of 'a' can affect the asymptotes if it happens to be -7*5^2/3^3.
     
  9. Nov 24, 2007 #8
    the 'a' value affects horizontal asymptote if the powers are the same. The only factor for the vertical asymptotes is that it makes the denominator 0 without making the numerator 0.
     
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