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The coordinates of the point where the normal to the curve y= 1/3 x^3 + 1/2 x^2 + x at x=1 intersects the y-axis are what?

  • Calc -

    dy/dx = x^2 + x + 1
    at x=1 ,slope of tangent is 1+1+1 = 3
    so slope of normal is -1/3
    when x=1, y = 1/3 + 1/2 + 1 = 11/6

    so equation of normal:
    y = (-1/3)x + b, at (1, 11/6)
    11/6 = -1/3 + b
    b = 3/2
    y = (-1/3)x + 13/6
    for y-intercept, let x = 0
    y = 13/6

    it cuts at (0,13/6)

    check my arithmetic

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