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At what point(s) does the curve defined by x^2-6x=6-6y^2 have horizontal tangents?

I took the derivative and solved for y'. I got y'=-1/6x/y+1/2y. hmm...what am I suppose to do?

  • Calculus -

    leave your derivative as

    dy/dx = (2x-6)/(-12y) after you differentiated implicitly.

    then (2x-6)/(-12y) = 0 can only be true if
    2x-6 = 0
    x = 3 now put that back into the original equation to get y = ...

    your equation will be y = that value

  • Calculus -

    would it by (2,sqrt(10)/2)?

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