Calculus
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 yaxis.

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
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Could someone please help me with these tangent line problems?