Hello @Sean_Olson,
For how the approximation came in, we carry out the taylor expansion of e^{-x}, and take the first-order terms from the expansion result which will be 1 - x.
A brief idea of taylor expansion of f(x) is like to find another function that consists of a series of basis functions (here we have x^0, x^1, x^2, ...). Such function is equal to f(x). The n-th order approximation is that you keep only up to the n-th term of the series and drop the rest.
Your graph is a good illustration of that the first order approximation of f(x=x_0) is nothing but the tangent of f(x) at x=x_0. It is a approximation because, at very close to x=x_0, the two curves are quite like each other. Since it preserves only up to the first order, the approximating curve fails to resemble f(x) in the range far from x=x_0. In other words, you want better approximation in larger range, you keep more terms of the Taylor series. To see this effect, you just add more terms from the page shared by Paul to your graphing tool.
So, the correct way of using your graph here is not just about at exactly the solution point (btw, y=1 is the zeroth order approximation that keeps only the zeroth term, as oppose to 1 - x which keeps both the zeroth and the first term), but looking at the approximating behavior of 1-x to e^{-x} around x=x_0. The more terms you add to your graphing tool, the more obvious the approximating behavior is.
Cheers,
Raymond