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

Algebra ->  Test -> 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       Log On


   

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) About Me  (Show Source):
You can put this solution on YOUR website!
First, derive y=x%5E3-6x%5E2 to get dy%2Fdx=3x%5E2-12x. Derive dy%2Fdx to get the second derivative d%5E2y%2Fdx%5E2=6x-12. 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 6x-12=0. 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 dy%2Fdx=3x%5E2-12x 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 y-y%5B1%5D=m%28x-x%5B1%5D%29, where 'm' is the slope and is the given point the line goes through, to find the equation of the tangent line.