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Math Help - graph sketching qn

  1. #1
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    graph sketching qn

    I shifted this qn to the uni section, apparently its too challenging for pre-uni lvl. . Hope you guys can enlighten me.

    The function g is defined by g(x)=\dfrac {1}{(x-5)(x+3)}, x not equal to -3 and 5.

    i) Find the coordinates of the turning point of the curve y=g(x).

    ii) Hence, sketch the graph of y=g(x) and write down the range of the function g.

    PS: Using calculus, I found the turning point to be (1, 1/-16).

    Extra stuff:graph sketching qn-untitled.jpg
    Last edited by stupidguy; September 11th 2010 at 11:41 AM.
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  2. #2
    MHF Contributor Unknown008's Avatar
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    Yes, the turning point is correct.

    Your graph however is not... what is the actual graph...

    Put x = 2. Then, y = -1/15, which is lower than the turning point. This means that the curve goes down from the turning point.

    And this being an 'inverted' square function (ie like y = 1/x^2), you have some sort of 'volcano' shape.

    See here:

    http://www.wolframalpha.com/input/?i=+y%3D+\frac{1}{(x-5)(x%2B3)}

    [And this is a pre-university question]
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  3. #3
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    my graphs from graphmatica are correct.
    the black one is y=(x-5)(x+3)
    the pink one is the reciprocal.

    I can determine that the turning point is a maximum point by finding the dy/dx of the points before and after x=1.
    since x not equal to 3 and 5, i learnt to sketch the 'n' shape below the x-axis with asymptotes at x=3 and 5.

    How do i get the L-shapes? How do I get the range of the function g?

    [Pre-uni do not have calculus, hence I post it here]
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  4. #4
    MHF Contributor Unknown008's Avatar
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    Pre-uni does have... I don't know whether there is a section for that though.

    You know that x cannot be equal to -3 or 5, because otherwise, you get y = infinity, hence the ever going up of the 'L' shapes and the ever going down ' \cap' shape.

    As x goes to -ve infinity, y tends to zero.
    As x tends to +ve infinity, y tends to zero again.

    In your diagram, maybe it's because I didn't zoom in... and didn't see the other graph. Sorry.
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  5. #5
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    is there a horizontal asymptote at y=0? if so, why?

    How do you even know that there are 2 "L"s at those particular positions?

    [there isn't a calculus section at pre-uni]
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  6. #6
    MHF Contributor Unknown008's Avatar
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    Yes, because the curve never meets the x-axis. You can try putting y = 0, and solve for x, but you will end up with imaginary roots, or no solution if you prefer.
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  7. #7
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    Since there is a horizontal asymptote at y=0 and vertical asymptotes at x=-3 and 5, there are 2 L shapes.

    To find out if the Ls are above of below the x-axis, we sub x= 6 and x=-4. Since the respective y values are positive. The Ls are above the x-axis.

    Is that the right way of reasoning?
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  8. #8
    MHF Contributor Unknown008's Avatar
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    Yes, you can think of this like that
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