At what point on the curve y = x - 3x^2 + 2 is the tangent parallel to the x-axis.

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The slope of a tangent drawn at any point (a, f(a)) to a curve described by y = f(x) is given by the value of f'(a).

For the curve y = x - 3x^2 + 2, the derivative y' = 1 - 6x. If a line is parallel to the x-axis, the slope of the line is 0.

To determine the point required in the problem solve y' = 0 for x.

1 - 6x = 0

=> x = 1/6

y = 1/6 - 3*(1/36) + 2 = 25/12

At the point `(1/6, 25/12)` the tangent to the curve y = x - 3x^2 + 2 is parallel to the x-axis.

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