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 I know that

Question 261766: 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
I know that the point of inflection relates to the second derivative, so do I just take the double derivative of the given equation?
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
First, derive to get . Derive to get the second derivative . Recall that the point of inflection occurs when the second derivative is zero. So set the second derivative equal to zero to get the equation . Solve for 'x' and use this 'x' value to find the corresponding 'y' value of the point of inflection. Then use the 'x' value and plug it into to find the slope of the tangent line at that x value.

Since you have the slope of the line (ie the slope of the tangent) and the point that the line goes through (the inflection point), you can the formula , where 'm' is the slope and

is the given point the line goes through, to find the equation of the tangent line.
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