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

Question

thanks!

Answer 1

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)