if x=e^t and y=(t-3)^2 , find an equation y=mx+b of the tangent to the curve at (1,9).
so m= that. Now when x=1, (x=e^t), t must be zero, so
now, at point 1,9
9=-6+b,or b=15, so equation for line must be
I just figured it out! thank you!