Find the equation of the tangent line and normal line to the curve e^(-x)-y=0 at (-1,0).

Question

Find the equation of the tangent line and normal line to the curve e^(-x)-y=0 at (-1,0).

Tangent line

is y=-e(x+1) correct?

Normal Line

is y=1/e (x+1) correct?

Answer 1

scroll down to near the bottom of this post where I found a possible change in the problem so that it gives the answers you have, but first are two other possible changes to the problem to make it consistent with the point and function given

e^-x - y = 0

is the same as

y = e^-x which is the standard exponential function reflected over the y axis

both which have y intercept at (0,1)

maybe that's the point you meant to use? (-1,0) is on the x axis, but the standard exponential function y=e^x or y=e^-x never touch the x axis. (-1, 0) is not on the graph of y=e^-x as y is always >0

take the derivative to get the slope of the tangent line

y' = -e^-x = the slope of the tangent line at the point (x, e^-x)

it may help to draw a rought sketch of the graph

for the point (0,1) the slope of the tangent line =

y'(0) = -e^-0 = -e^0 = -1 (any number to the zero power = 1)

tangent line at (0,1) is y-1 = (-1)(x-0) in point slope form, or y =-x+1

normal line is a perpendicular line with a negative inverse slope= 1. y-1=(1)(x-0) or y = x+1

or maybe you intended to use the point (-1,e) instead of (-1,0)?? if so, then

y'(-1) = -e^-(-1) = -e^1 = -e = slope of the tangent line when x=-1

tangent line is y-e=-e(x+1)

or y= -ex

normal line is y-e =(1/e)(x+1) = (x+1)/e,

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

Or maybe the problem was intended to read y = e^-x - e, ?? then (-1,0) would be a point on the graph as 0=e^-(-1) - e= e-e = 0

if so, then

f'(x) = -e^-x = slope of the tangent line

at x=-1

f'(-1) = -e^-(-1) = -e^1 = -e

tangent line is y = -e(x+1) =-ex -e

normal line has a negative inverse slope = 1/e

the line is y = (1/e)(x+1) = (x+1)/e