Posted by **Anonymous** on Monday, October 15, 2012 at 11:06am.

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.

- Calculus -
**drwls**, Monday, October 15, 2012 at 11:11am
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.

- Calculus -
**Steve**, Monday, October 15, 2012 at 11:13am
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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