Find the points on the curve y = 2x3 + 3x2 − 12x + 8 where the tangent line is horizontal.

points (-2,28) and (1,1) will have a horizontal tangent line. This problem involves the first derivative of the function and solving it for those cases where the derivative is 0. So take the expression y = 2x3 + 3x2 â’ 12x + 8 And for each exponent, multiply the coefficient by the exponent and then subtract 1 from the exponent. Giving. y = 6x2 + 6x â’ 12 Now you have a quadratic equation, use the quadratic formula to find it's solutions where a = 6, b = 6, and c = -12 The roots are at x = -2 and x = 1 So for the curve y = 2x^3 + 3x^2 â’ 12x + 8 the tangent line is horizontal at x=-2 and x = 1. The points will be y = 2(-2)^3 + 3(-2)^2 â’ 12(-2) + 8 y = -16 + 12 + 24 + 8 = 28 (-2,28) y = 2x^3 + 3x^2 â’ 12x + 8 y = 2 + 3 - 12 + 8 = 1 (1,1)

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