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If 3x^2-2y^2=4xy + 20, find the point(s) at which the graph from Problem 28 has vertical tangents. I found that dy/dx= (3x-2y)/(2x+2y), but now I don't know what to do. Do I set dy/dx to 0?

  • Calculus -

    For a vertical line, the slope would be undefined, that is,
    (3x-2y)/(2x+2y) has to be undefined.
    When does that happen?
    When the denominator is zero, so
    2x + 2y = 0
    y = -x

    sub into the origianal
    3x^2 - 2(-x)^2 = 4x(-x) + 20
    5x^2 = 20
    x = ±2

    if x=2, y = -2,
    if x = -2, y = 2

    points where tangent is vertical are (2,-2) and (-2,2)

  • Calculus -

    thanks! that was really helpful ;)

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