Posted by
**Sammy** on
.

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

dx/dt= 6t^2-24t-30

dy/dt= 2t-4

dy/dx= (t-2)/(3t^2-12t-15)

Now, if dy/dx= inf, then

3t^2-12t-15=0 or

(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.