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*girl04*

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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

I AM JESUS LOL!
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y=-e<sup>x</sup>
y`=-e<sup>x</sup>
At the point (1,-e),
m=-e<sup>1</sup>
m=-e
So, the equation of the tangent is
y-y<sub>1</sub>=m(x-x<sub>1</sub>)
y-(-e)=-e(x-1)
y+ex+e=e
y+ex=0
 

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