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Graphs of Fractional Functions

  1. Aug 10, 2013 #1
    Hello,
    I have a problem that has me stumped, please help.
    Question:
    Find the equations of the asymptotes of the following fractional functions., and then, draw the graph and label x- and y-axis points of intersection.

    The equation:
    y= x/(x + 1)

    My try at it:
    I first divided x by (x + 1) to get the equations of the asymptotes:
    x=-1, y=1 from y= 1 + 1/(x+1).

    Then I drew the asymptotes equations on the graph:
    https://docs.google.com/drawings/d/1p2Lfkqab9qAVxzx9z0LFyWAm8-WeoP30yBSn9bHsDKM/edit (red is the asmptotes equations, black is the graph).

    Then I got stumped at the points of intersection with the x-axis:
    x-axis: (?, 0)
    y-axis: (0, 2)

    I got the y-axis by substituting 0 in for x in the previous equation:
    y = 1 + 1/(x + 1)
    y = 1 + 1/(0 + 1)
    y = 1 + 1/1
    y = 1 + 1
    y = 2

    But for x:
    y = 1 + 1/(x+1)
    because if I do negative one:
    0 = 1 + 1/(-1+1)
    0 = 1 + 1/0
    0 = 1 is incorrect.
     
  2. jcsd
  3. Aug 10, 2013 #2

    SteamKing

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    These are usually called 'rational functions' instead of 'fractional functions'.

    In your attempt at finding the x-intercept, you have used y = 1 + 1/(x+1) instead of the original function.

    y = 1 + 1/(x+1) is not equal to y = x / (x + 1)
     
  4. Aug 10, 2013 #3
    but if I use the original equation, wouldn't it be a no-solution?
     
  5. Aug 10, 2013 #4

    Mentallic

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    It wouldn't be a "no solution", but it would be difficult to find the solution. Your technique is correct, but your answer is wrong. If you want to find the x-intercepts of the equation [itex]y=x^2-1[/itex], you wouldn't want to be solving the incorrectly factored form [itex]y=(x-2)(x-3)[/itex] would you? You'd want to factor it correctly, and then go from there.

    Similarly in this case, you want to find the correct values of A and B such that

    [tex]\frac{x}{x+1}=A+\frac{B}{x+1}[/tex]

    And once you do, check your answer! For example, using the values A=1 and B=1 that you gave,

    [tex]1+\frac{1}{x+1}=\frac{1(x+1)+1}{x+1}=\frac{x+2}{x+1}\neq \frac{x}{x+1}[/tex]

    Hence those values of A and B were not correct.
     
  6. Aug 10, 2013 #5
    I understand your explanation, but then how would I get it into the y= A + B/(x+1) format?
     
  7. Aug 10, 2013 #6

    vanhees71

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    2016 Award

    Just write
    [tex]\frac{x}{x+1}=\frac{(x+1)-1}{x+1}=\ldots[/tex]
     
  8. Aug 10, 2013 #7

    Mentallic

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    vanhees71's method is the easiest and quickest, and with a little practice, you'll quickly learn to master it.

    The alternative method of finding A and B is called partial fraction decomposition and it may arise in your later math studies, so understanding its technique could be useful to you.

    [tex]\frac{x}{x+1}=A+\frac{B}{x+1}[/tex]

    the right hand side needs to be all on one fraction, so that we can compare it to the left,

    [tex]=\frac{A(x+1)+B}{x+1}[/tex]

    Now, since both sides have the same denominator, we just need to equate the numerators,

    [tex]x=A(x+1)+B[/tex]

    [tex]x=Ax+A+B[/tex]

    for these equations to be equal, the coefficient (number in front of) the x terms must be equal and the constants must be equal. It might help to look at the equality like this

    [tex]1x+0=Ax+(A+B)[/tex]

    So we have that A=1 and A+B=0, hence B=-1.

    If there is anything you don't understand, just ask.
     
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