Check what happens at the limits of the function as x goes to infinity.
Well... f(x) = e^x + 2. As x goes to positive infinity, e^x + 2 grows infinitely big, so theres obviously no horizontal asymptote there.
But when x goes to negative infinity, this happens:
[input]e^(-x) + 2 = 1/(e^x) + 2[/input]
As x gets big, e^x gets huge, so 1/(e^x) goes to zero and the function goes to 0+2.
Your horizontal asymptote is f(x)=2
Hope this is clear. Btw, I have no idea how to solve this using algebra, this is a typical calculus question.
