find the equation of the tangent line at the given point
y=4x-3x^2 ;(2,-4)
y-(-4)=m(x-2)
what to do from here???
or is like this
3x(4x-1)^1
The slope at any point is y':
y' = 4-6x
y'(2) = -8
Now you have a slope and a point:
y+4 = -8(x-2)