logs question (1 Viewer)

[*girl04*]

Find the equation of the tangent to the curve to the curve y = -e^x

at the point ( 1, -e)

Answer: ex+y=0

I keep getting an answer with 3 in it and 2x. If anyone could tell me how to approach/ do this sort of question properly it would be appreciated!

thankyou

 

[thunderdax]

y=-e^x
y`=-e^x
At the point (1,-e),
m=-e¹
m=-e
So, the equation of the tangent is
y-y₁=m(x-x₁)
y-(-e)=-e(x-1)
y+ex+e=e
y+ex=0