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Graphing Rational Functions

  1. May 22, 2010 #1
    Ok so, This summer I will be taking a Pre-calc/trig course intensive, to get ready to take calculus in the fall, to start up my track for physics.

    I got a Pre Calculus Workbook For Dummies and I have to say so far I'm not too pleased.

    I have already found a bunch of typos, and when there are typos in the direction of the inequality, or the sign of an answer... that's bad.

    anyways, I'm up to graphing rational functions and they're doing a horrible job explaining it.

    I guess, I'm slowly learning the rules and what to do:

    1. set the top of the fraction =0 and solve to find the x intercept.
    2. set x to 0 to find y intercept
    3. set denominator to 0 to find verticle asymptotes

    I understand that, but I guess the whole idea is wishy washy to me. I am just plugging in numbers randomly and drawing out the graph at that point?

    I see a lot of graphs of rational functions that have random twists and turns, and what if I just happen to not plug in a point at that exact spot to find it?

    is there a systematic way to work on these? or is it literally find those 3 things I said above, and then just plug away?
     
  2. jcsd
  3. May 22, 2010 #2

    LCKurtz

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    Pretty good. 3 gives possible vertical asymptotes, but they might not be if the numerator is zero for the same value of x. But that is something you will learn about in calculus.

    Another think you might look at is what is called the extent. If you can factor the numerator and denominator, the signs of the factors determine the sign of the fraction. Here's an example of what I mean:

    [tex] y = \frac {(x-2)}{x(x+3)}[/tex]

    The sign of y is determined by the signs of the three factors, and each factor changes sign at its corresponding root. The numerator changes sign as x passes through x = 2 and the denominator factors change sign at x = 0 and x = -3, respectively. Those three values are the only places where y can change sign:

    ---------(-3)--------------(0)------(2)--------------

    If x > 2, all three factors are positive, so the y is positive and you can shade out the area under the x axis for x > 2. Now think of x sliding from the right to the left across x = 2. The numerator factor changes sign there so you have one negative factor and y is negative. So you can shade out the positive y values when x is between 0 and 2. When x moves left across x = 0 another factor changes sign so the fraction becomes positive again.

    Over each interval it is convenient to put a little symbol like this to indicate the signs (I have just shown the left one where they are all negative because I can't make them space properly, but hopefully you get the idea):

    -
    ___
    - -

    ---------(-3)--------------(0)------(2)--------------

    Once you have shaded where the graph can't be and use you other info, you can almost guess what the graph looks like.
     
  4. May 22, 2010 #3
    dude that was a lot of help, That makes things make alot more sense

    So once I shade appropriately, I'm plugging points to verify specifically where the function is?
     
  5. May 22, 2010 #4

    LCKurtz

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    Yes. Once you combine your extent info with your roots and asymptotes, you won't need too many more points. And remember, if a factor is squared like (x-2)2, it will still cause its zero or asymptote depending on whether it is in the numerator or denominator, but it won't change sign as x passes through 2.
     
  6. May 23, 2010 #5
    One more little thing that you may have already guessed. If you have the same factor in the numerator and denominator, then the graph looks like the simplified fraction except for the fact that it is undefined where the factor is 0.
    Ie., y = (x^2 - 2x + 2)/(x - 1) has the same graph as y = x - 2 except that the point (1, 0) is removed.
    Once you learn calculus, you will be able to draw more accurate graphs of the function without relying on plugging in lots of points.
     
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