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anonymous

  • 5 years ago

The tangent to the cubic function that is defined by y = x^3 - 6x^2 + 8x at point A(3,-3) intesects the curve at another point, B. Find the coordinates of point B. Illustrate with a sketch. y' = 3x^2 - 12x + 8 y' = 3(3)^2 - 12(3) + 8 y = -1 <-- slope of the line of A This line also has the point B. y = mx + b -3 = -1(3) + b 0 = b Therefore, line equation is y = -1x. So, I equate: -1x = x^3 - 6x^2 + 8x 0 = x^3 - 6x^2 + 9x 0 = x(x^2 - 6x + 9) 0 = x(x-3)(x-3) x = 3, or x = 0. X = 3 we already have, so x = 0 should be point B. I sub back in X = 0 into the main cubic function and I retrieve

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