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Math Help - Turning points

  1. #1
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    Unhappy Turning points

    Hi

    For y=2/(x-1), how do I exactly find the turning pts?

    I tried following my teacher's way by first turning it into a quadratic equation and finding its discriminant:

    y(x-1) = 2
    yx - y - 2 = 0
    discriminant = y^2
    For x to be real then its discriminant must be greater or equal to 0
    i.e. y^2 is greater than or equal to 0
    This becomes y is greater than or equal to 0.
    When y = 0, equation becomes 0 = 2.
    That's where I'm stuck. For the other graphs, my teacher's method allowed me to find the turning points. For this one (the first question ), it doesn't work.

    Could someone please help and tell me how to find the turning pt here?
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  2. #2
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    Quote Originally Posted by xwrathbringerx View Post
    Hi

    For y=2/(x-1), how do I exactly find the turning pts?

    I tried following my teacher's way by first turning it into a quadratic equation and finding its discriminant:

    y(x-1) = 2
    yx - y - 2 = 0
    discriminant = y^2
    For x to be real then its discriminant must be greater or equal to 0
    i.e. y^2 is greater than or equal to 0
    This becomes y is greater than or equal to 0.
    When y = 0, equation becomes 0 = 2.
    That's where I'm stuck. For the other graphs, my teacher's method allowed me to find the turning points. For this one (the first question ), it doesn't work.

    Could someone please help and tell me how to find the turning pt here?
    y = \frac{2}{x-1} does not have any turning points. It's a rectangular hyperbola.
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  3. #3
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    Dubai, UAE
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    But yx - y - 2 = 0 is not a quadratic. What 'discriminant' are you talking about then?
    To get turning points (if any), find its derivative and equate it to 0.

    I hope that helps.

    ILoveMaths07.
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