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Calculus

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determine the point(s)at which the graph of y^4 = y^2 -x^2 has a horizontal tangent.

  • Calculus -

    y^4 = y^2 - x^2
    4y^3 y' = 2y y' - 2x
    y' = -2x/(4y^3-2y)

    Now, assuming that y ≠ 0 or ±1/√2,
    y'=0 when x=0

    so, y^4 = y^2, and y = ±1, so
    (0,1) and (0,-1)

    extra credit: what happens at y = 0 or ±1/√2?

  • Calculus -

    how did you know it was + or - l? I got what x equals but I was confused on how to find the y values?

  • Calculus -

    if x=0,

    y^4 = y^2
    y^2(y^2-1) = 0
    y = 0 or ±1
    but y=0 is not a possibility.

  • Calculus -

    where did you get y^2 -1 from? Did you subtract y^2 from the right side? if so where did the -1 come from?

  • Calculus -

    Hello? This is calculus. Have you forgotten you algebra I?

    y^4 = y^2
    y^4 - y^2 = 0
    y^2(y^2-1) = 0
    ...

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