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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(x3)(x3)
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
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(x3)(x3) 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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