That's correct and that's all you have to do. You have found the gradient of the tangent line in the point x=1 at the curve.
hi,
i don't understand how i would find the gradient at the following points at the given points:
y=3x^2-4x+1 at x=1
i know i would start it:
dy/dx=6x-4
then substitute x to get:
6-4=2
would 2 be the gradient and that is all i do or is there more to do?
thanks!