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Math Help - question about graphing

  1. #1
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    question about graphing

    How do I graph g(x) = (x - 3)/(x^2 -1)

    Thanks in advance
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  2. #2
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    Hi marie7

    These are the steps :
    1. find x and y intercept
    2. find vertical and horizontal asymptotes
    3. find stationary points
    4. find the intersection between horizontal asymptote and the graph

    ^^
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  3. #3
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    Would I be able to get step by step to this question as it is the first time I've done it please. I know how to sketch quadratic functions but not this kind.
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  4. #4
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    Hi marie7

    1. I'm sure you can find the x and y intercept ( intersection between the x and y axis)

    2. To find vertical asymptote : set the denominator = 0 and solve for x

    3. To find horizontal asymptote : horizontal asymptote = y = \lim_{x\rightarrow\infty} f(x)

    4. To find stationary points : set \frac{dy}{dx}=0 and solve for x. Then subs. the value of x obtained to f(x) to get the coordinate of stationary point(s).

    5. To find the intersection between horizontal asymptote and the graph, just substitute the value of horizontal asymptote to f(x) and solve for x to get the coorodinate

    ^^
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  5. #5
    Member eXist's Avatar
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    Quote Originally Posted by songoku View Post
    4. To find stationary points : set \frac{dy}{dx}=0 and solve for x. Then subs. the value of x obtained to f(x) to get the coordinate of stationary point(s).
    ^^
    I believe this is pre-calc, correct me if I'm wrong, but I don't think they teach derivation that early.
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  6. #6
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    I haven't done the question yet because I've been watching some math videos as I'm still quite lost, but in this chapter of my textbook it doesn't mentioned what you've said - the textbook really isn't good though. Still not sure what to do... the most it mentions is the horizontal and vertical asymptote and that's it. The instructions are terrible and the examples in the textbook doesn't really help me with this question.
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  7. #7
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    Hi eXist and marie7

    Sorry I'm not aware this is pre-calculus sub forum.

    @eXist : maybe there is another way to find the stationary points?

    @marie7 :
    1. Intersection with the x-axis means that the value of y = 0, so

    \frac{x-3}{x^2-1}=0


    x-3 =0

    x=3

    Hence, the x-intercept = (3,0)

    2. Intersection with y-axis means that the value of x = 0, so

    y=\frac{-3}{-1}

    y = 3

    Hence, the y-intercept = (0,3)


    Have you been taught about asymptotes, marie? ^^
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  8. #8
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    Hi there Songoku,
    Thanks for your help by the way!
    No, I'm not sure how to handle asymptotes. I also need to graph the function. Could you show me the steps? That's all the question is asking, for me to graph the function. Thanks so much
    Marie
    Last edited by marie7; August 7th 2009 at 09:30 PM.
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  9. #9
    Member eXist's Avatar
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    Quote Originally Posted by songoku View Post
    @eXist : maybe there is another way to find the stationary points?
    There is. The most common way (Before derivatives) that I've seen done is to use the mid point theorem. Take the zeroes of the function and find the mid points between them.

    For example:

    f(x) = -x^2 +4

    The zeroes are at x=2 and x=-2

    So you find the mid point between (2, 0) and (-2, 0). The x value that is, which comes out to be x=0.

    Then you substitute this value back into the original equation to get (0, 4). So therefor, the stationary point for f(x) = -x^2 +4 is (0, 4).
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  10. #10
    Senior Member furor celtica's Avatar
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    are you sure this girl hasnt studied derivatives? bcos i have had questions similar to this one adn i have studied derivatives. but i'm taking a cambridge course, so then again maybe the american program is different.
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  11. #11
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    Are you talking about me? I'm Australian, not American. lol. Derivatives is in the next chapter of my textbook, so this question is just before that.
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  12. #12
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    y = 0 when x = 3 therefore, y intercept at x = 3

    x does not = 0 at any point, therefore horizontal asymptote at y = 0

    x cannot equal positive or negative 1, as this would mean the slop of the curve is undefined at these points, therefore vertical asymptote at x = -1 and +1

    the curve is convex from -1<x<1, and concave for x<-1 and x>1

    Is this right? And if so, can someone please show me what the graph is supposed to look like? just to check. because i'm still a little confused. This is a rectangular hyperbola, am I right? Thank you.
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  13. #13
    Member eXist's Avatar
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    The graph I came up with is:



    However it is very late for me at the moment, so please excuse me if it isn't correct/perfect.
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  14. #14
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    AHHHH actually I understand it now. Thanks for your help eXist!!
    Last edited by marie7; August 8th 2009 at 01:26 AM.
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  15. #15
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    Talking

    Quote Originally Posted by marie7 View Post
    I haven't done the question yet because I've been watching some math videos as I'm still quite lost, but in this chapter of my textbook it doesn't mentioned what you've said - the textbook really isn't good though. Still not sure what to do...
    To learn how to find intercepts, try here.

    To learn how to find asymptotes, try here.

    To learn how to graph rational functions, here.

    Graphing rational functions is not generally hard, but the process can be tiresome. Do enough practice exercises that you start getting a "feel" for how the process works.
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