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Math Help - Quick question

  1. #1
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    Quick question

    y^2-2x^2 = 2y+1
    Need to find all points where the tangent line is horizontal.
    To do this you just find the derivative and set it to 0 and solve right? The only thing that is tripping me up is the equation. What form do I need to put it into?
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  2. #2
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    Quote Originally Posted by zsig013 View Post
    y^2-2x^2 = 2y+1
    Need to find all points where the tangent line is horizontal.
    To do this you just find the derivative and set it to 0 and solve right? The only thing that is tripping me up is the equation. What form do I need to put it into?
    Use implicit differentiation to get dy/dx as a function of y and x: \frac{dy}{dx} = \frac{2x}{y-1}.

    Solve \frac{dy}{dx} = 0: x = 0.

    Substitute x = 0 into y^2-2x^2 = 2y+1 and solve for the y-coordinate. Check that the solution(s) do NOT satisfy y - 1 = 0 (why?)
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Ok

    Quote Originally Posted by zsig013 View Post
    y^2-2x^2 = 2y+1
    Need to find all points where the tangent line is horizontal.
    To do this you just find the derivative and set it to 0 and solve right? The only thing that is tripping me up is the equation. What form do I need to put it into?
    Another way to do this is to solve for y in you original equation to get y=\pm\sqrt{2(x^2+1)}+1...so you get y'=\frac{\pm{\sqrt{2}x}}{\sqrt{x^2+1}}...so y'=0,iff x=0...now using this you can come to teh same conclusing Mr. Fantastic did...this is a little harder but a neat way of doing it
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