How to determine the equation of tangent? Derivative of e?

Find the equation of the tangent to the curve y = 2 - e^-x at the point where x =1.

I know that you have to find the derivative of that function, which is dy/dx = e^-x

I don't really know how to get the answer beyond that though, because the answer is: x - ey + 2e - 2 = 0

The equation is not in the form of "y=mx+b" and I'm not sure how to find the above answer. If any suggestions or tests can be provided, it would be much appreciated. Please and thank you!

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Re: How to determine the equation of tangent? Derivative of e?

the point of touch is (1, 2-1/e)

the derivative at x=1 is the gradient of the tangent

therefore gradient =1/e

the equation of the tangent is therefore y-(2-1/e)=(1/e)(x-1) ....simplify it to find the formula you mentioned before.

Attachment 28522

Re: How to determine the equation of tangent? Derivative of e?

Note that the description of this topic, "PreCalculus", says "Pre-calculus does not involve calculus".

Re: How to determine the equation of tangent? Derivative of e?

Firstly, you find dy/dx. which equals e^-x. Plug in 1 to find that the slope is e^-1, aka 1/e. Now we have the slope, we need a point. you have 1, so plug 1 back into the original equation to get the point (1, 2- 1/e). You now have the point and the slope, and can make the line from that. I doubt i need to tell you how to do this.