Question 261766
First, derive {{{y=x^3-6x^2}}} to get {{{dy/dx=3x^2-12x}}}. Derive {{{dy/dx}}} to get the second derivative {{{d^2y/dx^2=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/dx=3x^2-12x}}} 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[1]=m(x-x[1])}}}, where 'm' is the slope and *[Tex \LARGE \left(x_1,y_1\right)] is the given point the line goes through, to find the equation of the tangent line.