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posted by matthias on Wednesday, October 31, 2012 at 12:15am.
find an equation of the tangent line to the graph of y=e^(2x-3) at point (3/2,1) thanks!
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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