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math

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find an equation of the tangent line to the graph of y=e^(2x-3) at point (3/2,1)

thanks!

  • math - ,

    y'(x) = 2e^(2x-3)
    y'(3/2) = 2

    so, you want the line through (3/2,1) with slope 2:

    y-1 = 2(x-3/2)

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