x^2-6x+y^3 -12 y = 11. The derivative of the equation is y' = (6-2x)/(3y^2 - 12). Find the coordinates of all points on the curve where the line tangent to the curve is vertical.

Question

loser66 · Answer

y' is slope of the tangent line, and when it is vertical, its denominator =0. It means 3y^2 -12=0 --> y^2 =4 or y =$\pm 2$
replace each of the value of y to original one to find out x's

anonymous · Answer

Thank you so much! This helps me out sooo much :)!