$$e^{-x}-x+1=0$$
$$\frac{1}{e^x}=x-1$$
$$e^x(x-1) = 1$$
$$\therefore e^x = 1, x-1 = 1$$ Where $$x=0, x=2$$ Or, $$e^x = -1, x-1 = -1$$ Where $$x=nil,x=0 $$
Therefore, there is no solution to the equation. However, plotting the initial curve gives me one x-intercept at x = 1.28 => Why so?
Your method for solving this equation is wrong. The equation $$e^x(x-1)=1$$ is correct, but you can't deduce much from it. In particular, you seem to have factored $1$ in the integers and said that either both factors are $1$ or $-1$. This would work if they work both integers - or, in fact, if either was an integer - and establishes that there are no integer solutions. However, we can have things like $1=\frac{\sqrt{2}}2 \cdot \sqrt{2}$ where $1$ is written as a product in other ways, and your solution fails to account for that. Indeed, for any $x$ other than zero, we have $x\cdot \frac{1}x=1$, so you've missed a lot of solutions when you worked only in the integers. Thus, it shouldn't surprise us too much that there might be a non-integer solution, which is what you're seeing on the graph.
The trouble is that this equation has no solution in terms of elementary functions, as far as I know. One can solve it using the product log function $W$, which has the property that $W(y)e^{W(y)}=y$ - that is, it's the inverse function to $xe^x$. Using this we first divide both sides by $e$: $$e^{x-1}(x-1)=\frac{1}e$$ Then apply the product log: $$x-1 = W\left(\frac{1}e\right)$$ $$x=W\left(\frac{1}e\right)+1$$ which is the root visible on the graph.
Hint you can use approximations here ie $e^x=1+\frac{x}{1!}+\frac{x^2}{2!}...$ till first four -five terms and then get roots. (Though its a very tidious work) .
