Calculus
posted by Alex .
Find the equation of the line tangent to the graph of y=4 e^x at x=2.
I always get confused with tangent lines
I worked this out to be
4e^x(x2)+4e^2 but the program i have to submit it to says its wrong.

The equation of the tangent to the graph
y=f(x) at x=a:
y=f(a)+f'(a)(xa)
(4e^x)'=4e^x
y=4e^2+4e^2(x2)
y=4e^2*x4e^2
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