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Math Help - HELP!!! With graphs (It might be hyberbolic--unsure!)

  1. #1
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    HELP!!! With graphs (It might be hyberbolic--unsure!)

    I need to know how to sketch graphs of the form y= 1/ x+c:

    (i) y= 1/ x+2
    (ii) y= 1/ x-4
    (iii) y= 1/ x+1

    Please help!! I'm clueless. I think they are hyberbolas in the xy=k thing but i cant draw or work it out. Pls show the steps

    Thanks.
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  2. #2
    Super Member PaulRS's Avatar
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    I suppose you know how f(x)=\frac{1}{x} looks like

    The cases you posted are equal to this one in shape, mind you they have another position (they are horizontal translations).

    The function f(x)=\frac{1}{x+c} has a vertical asymptote (the only one) in x=-c, so what you have to do is to move (horizontally) \frac{1}{x} so that it has its aymptote there

    Remember that they have horizontal asymptotes on both sides

    Attached Thumbnails Attached Thumbnails HELP!!! With graphs (It might be hyberbolic--unsure!)-graph-1x.gif  
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  3. #3
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    Quote Originally Posted by PaulRS View Post
    I suppose you know how f(x)=\frac{1}{x} looks like

    The cases you posted are equal to this one in shape, mind you they have another position (they are horizontal translations).

    The function f(x)=\frac{1}{x+c} has a vertical asymptote (the only one) in x=-c, so what you have to do is to move (horizontally) \frac{1}{x} so that it has its aymptote there

    Remember that they have horizontal asymptotes on both sides

    But how come the x+3 is situated on top of the x-intercept 2?? and what if it is y= 1/ x-4

    the minus sign.....
    im so sleepy
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  4. #4
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    PLS HELP!!!!!!!
    i dunno./.............

    im gonna b in big truoouble...omg its so late..........
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  5. #5
    Super Member PaulRS's Avatar
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    Let's go over the cases.

    First f(x)=\frac{1}{x+2}

    It has a vertical asymptote in x=-2 (what would f(-2) be? )

    Apart from that, it has the same shape the ones in the other post have.

    Now g(x)=\frac{1}{x-4}

    The vertical asymptote is x=4(what would f(4) be? )


    Finally h(x)=\frac{1}{x+1}

    Asymptote x=-1

    The important thing is that you realise that: f(x)=g(x+6) for example
    Just for you to see f(1)=\frac{1}{1+2} and g(7)=\frac{1}{7-4}
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