SOLUTION: The equation of the tangent line to the curve y=x^3-6x^2 at its point of inflection is
(A) y=-12x+8
(B) y=-12x+40
(C) y=12x-8
(D) y=-12x+12
(E) y=12x-40
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-> SOLUTION: The equation of the tangent line to the curve y=x^3-6x^2 at its point of inflection is
(A) y=-12x+8
(B) y=-12x+40
(C) y=12x-8
(D) y=-12x+12
(E) y=12x-40
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Question 263008: The equation of the tangent line to the curve y=x^3-6x^2 at its point of inflection is
(A) y=-12x+8
(B) y=-12x+40
(C) y=12x-8
(D) y=-12x+12
(E) y=12x-40
You can put this solution on YOUR website! The equation of the tangent line to the curve y=x^3-6x^2 at its point of inflection is
(A) y=-12x+8
(B) y=-12x+40
(C) y=12x-8
(D) y=-12x+12
(E) y=12x-40
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y' = 3x^2 - 12x
y'' = 6x - 12 = 0
x = 2 at the point of inflection
f(2) = -16 --> (2,-16) is the point of inflection
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f'(2) = -12 = m at the point
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y = mx + b
-16 = -12*2 + b
b = 8
y = -12x + 8 is tangent at the point of inflection.
(A)
