# Calculus

Find the equation of the tangent to the curve y=e^x at the point where x=-1.

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1. Equation for the tangent :

Delta y/Delta x = y'(-1)------>

[y-exp(-1)]/(x+1) = exp(-1) ------>

y = exp(-1) + (x+1)exp(-1) =

exp(-1)(2 + x)

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2. I really don't understand this. Can you explain it step by step? The answer at the back of the book is y=-ex.

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3. to find the equation of the tangent you will need the point of contact and the slope at that point of contact.

when x= -1, y = e^-1 = 1/e

dy/dx = e^x, so when x = -1 the slope = e^-1 = 1/e

using y = mx + b

1/e = (1/e)(-1) + b
2/e = b

the equation of the tangent is

y = (1/e)x + 2/e

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