Expert: Paul Klarreich - 1/25/2006

Question
I am having trouble finding an equation of the tangent line to the graph of y = e^-2x at the point (1,e^-2).  Can you help me with this please?

Answer
Hi, Steven,

Your question:  I am having trouble finding an equation of the tangent line to the graph of y = e^-2x at the point (1,e^-2). Can you help me with this please?
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You will want to write an equation in the form  y - y0 = m(x - x0), where

  (x0,y0) is a point on the graph, in this case (1,e^-2)
  m is the slope of the tangent, equal to dy/dx at x = 1.

dy/dx = -2e^-2x, which, at x=1, = -2e^-2

So your equation is

y - e^-2 = -2e^-2(x - 1)

Actually, that's the answer, but a little simplification is in order:

y - e^-2 = -2xe^-2 + 2e^-2

y = -2xe^-2 + 3e^-2

That is probably as good as it gets.