A question on geometry about finding the intersection point of a tangent and a curve.?

Question

The curve h(t)=(t^3,t^2-t).show that it's a regular parametrized curve.?
the curve g(t)=(-1+3t,2-3t) is a tangent of h(t). Find where h(t) and g(t) intersect. i.e. find the points such that g(t1)=h(t2) for some t1,t2, in the reals.
which intersection point is the tangent one? give reasons to ur answer.

Answer 1

I think I answered this question
t^3=3t-1 and
t^2-t=2-3t so
t^3-3t+1=0
and t^2+2t-2=0
both equations must have a common root which then is also a root of the remainder of their division which is 3t-3
so if there is a common root it must be 1(Not)
so there are no intersections h(t) and g(t)
Many hours later
Let´s take another look
maybe there are different values of t for h and g that give the same point

t1^3=-1+3t2 and t1^2-t1=2-3t2
This is a system of equations
eliminating t2
t1^3+t1^2-t1-1=0 can be factored as (t1-1)*(t1+1)^2=0
t1=1 gives t2=2/3
t1=-1gives t2=0
The common points are(1,0) and(-1,2)
As t1=-1 is a double root the point (-1,2) is the tangent one

Answer 2

I recon somethings wrong here - it just doesnt work out.

g(t) is a straight line gradient -1

Only points on h(t) with gradient of -1 do not lie on the line g(t) hence there is no tangent.