posted by Anonymous .
Determine an equation of the tangent line to the curve of f (x) = 3e^4x − 3 at the point where the curve crosses the y-axis.
The curve crosses the y axis when x = 0.
f(0) = 3*e^0 - 3 = 0 @ x=0
dy/dx = 12 e^4x = 12 @ x=0
The straight tangent line at the origin (the tanglent point) is
y = 12 x.
y=0 when x=0, so we are looking at the line through (0,0)
f'(x) = 12e^4x
f'(0) = 12
the line is thus y=12x