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Calculus

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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(x-2)+4e^2 but the program i have to submit it to says its wrong.

  • Calculus -

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

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