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find all points on the curve where x=2. Show there is a horizontal tangent to the curve at one of the points.

  • calculus -

    sub x = 2 into your equation to get

    4 - 4y = y^2 - 1
    0 = y^2 + 4y - 5
    (y+5)(y-1) = 0

    so when x=2, y =-5 or when x=2, y = 1

    so the points are (2,-5) and (2,1)

    check them by subbing them back into the original equation, they work

  • calculus -

    I found the derivative of your equation implicitly and got
    dy/dx = (2x+2xy)/(2y = x^2)

    for a horizontal tangent this has to be zero, which means the numerator should be zero

    subbing in (2,-5) makes the top zero, so there is a horizontal tangent at the point (2,_5)

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