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Math Help - Find Vertical and Horizontal Tangent

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    Find Vertical and Horizontal Tangent

    Parametric

    x= t^4 + 4t^3 - 8t^2

    y= 2t^2 - t
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by camherokid View Post
    Parametric

    x= t^4 + 4t^3 - 8t^2

    y= 2t^2 - t
    dy/dt = 4t - 1
    dx/dt = 4t^3 + 12t^2 - 16t


    Horizontal tangents occur when dy/dt = 0, provided dx/dt is not zero at that point
    so for horizontal tangent:
    4t - 1 = 0
    => t = 1/4
    so a horizontal tangent occurs when t = 1/4


    Vertical tangents occur when dx/dt = 0
    so for vertical tangent:
    4t^3 + 12t^2 - 16t = 0
    => 4t(t^2 + 3t - 4) = 0
    => 4t(t + 4)(t - 1) = 0
    => t = 0, t = -4, t = 1 .........the vertical tangents occur at these values of t.

    You may want to see Tangents with Parametric Equations

    to find the points that the tangents occur, just plug in the values we found for t into the original x and y equations, you will get coordinate points for each value of t that way
    Last edited by ThePerfectHacker; April 7th 2007 at 08:42 PM.
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