curve C defined by these parametric equations
$x = t^3 - 3t^2$
$y = t^3 - 3t$
I need to find the equations of two tangents at the point $(-4,2)$
I have attempted to solve this problem myself but I got stuck.
Here is what I've done so far:
1) Found out that t came out to be $t = 2$ and $t = -1$
2) When I plugged $t = 2,-1$ into $dy/dx, t=2$ comes out to be undefined
3) $dy/dx = (3t^2-3)/(3t^2-6t) = (t^2-1)/(t^2-2) $
Re-submission of my question, I hope I did the right format this time Thank you
The curve is self intersecting. So at one point ( 2.-1) there are two slopes.
For t= 2, slope is 3/2
For t= -1 ,slope is 0
So equation of first tangent is $$ (y+1)/(x-2) = 3/2 $$
and equation of second tangent is $$ (y+1)/(x-2) = 0 , y=-1. $$