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Graphs of Fractional Functions

  1. Aug 10, 2013 #1
    I have a problem that has me stumped, please help.
    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


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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


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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


    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


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    Just write
  8. Aug 10, 2013 #7


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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.


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


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



    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


    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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