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Graphing Help Needed

  1. Feb 23, 2005 #1
    i need to graph f(x)=2^2+9/5x^2+2

    the vertical asym. would be the i squareroot10/2 ?

    if that is correct how do i graph an imaginary #
     
  2. jcsd
  3. Feb 23, 2005 #2
    You're telling me you need to graph

    [tex]9/5x^2+6[/tex]

    ?
     
  4. Feb 23, 2005 #3
    no in the numerator is: 2x^2+9
    denomorator is: 5x^2+2
     
  5. Feb 23, 2005 #4
    So it is

    [tex]\frac{2x^2+9}{5x^2+2}[/tex]

    and you have to sketch the graph?
     
  6. Feb 23, 2005 #5
    It has horizontal asymptote at y=2/5 and no vertical asymptotes since the denominator is never zero.
     
  7. Feb 24, 2005 #6
    thank you very much
     
  8. Feb 24, 2005 #7

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    Yes,asymptotes (especially vertical ones) refer to REAL values for numbers...

    Daniel.
     
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