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Math Help - Parametric Tangent Problem

  1. #1
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    Parametric Tangent Problem

    Te parametric form for the tangent line to the graph of y = 2x^(2)+2x-1 at x = -1 is

    Ok, So I got the slope at x=-1, which is -2. Then I found a points along the line, (-1,-1).


    So I have the slope of the tangent line, y= -2x-3.

    How do I translate this into parametric form?

    Thanks
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  2. #2
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    If a line has equation y=\frac{p}{q}(x - x_0)+y_0,~q\not= 0
    the parametric form is  \left\{ \begin{gathered}  x = qt + x_0  \hfill \\  y = pt + y_0  \hfill \\ \end{gathered}  \right.
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  3. #3
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    So, how would I find p and q?
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  4. #4
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    Quote Originally Posted by r2d2 View Post
    So, how would I find p and q?
    \dfrac{p}{q} is the slope of the line.
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  5. #5
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    so P=-2 and q=-1?

    So then the parametric equation should be,

    <-2t-1,t-1>?
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  6. #6
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    Quote Originally Posted by r2d2 View Post
    so P=-2 and q=-1?
    So then the parametric equation should be,
    <-2t-1,t-1>?
    No indeed.

    \dfrac{p}{q}=\dfrac{-2}{1} so <t-1,-2t-1>.
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