posted by Sammy .
For the parametric curve defined x=2t^3-12t^2-30t+9 and y=t^2-4t+6 ..find dy/dx and then where is the tangent to the curve vertical (give the cartesian coordiantes of the points. and find the tang. to the curve horizontal
Now, if dy/dx= inf, then
(t-5)(t+1)=0 check that.
solve for t. Put that value of t in the original equations to get x,y for the point where the tangent is vertical.