ok how do you find the domain of e^(x+1)/(y-2)?

Question

Would it be 2/-2=1?If not why?

Answer 1

E raised to any power will never equal zero, so the domain would be all real numbers.

Answer 2

The domain is the set of values for which the function is defined. From the form of the function I am assuiming it is a function on R^2 (unless you've made a typo and the y should be an x - if so, see later).

e^x is defined for every x, so the numerator is defined everywhere. Similarly the denominator is defined for every (x, y) pair. So the function is defined everywhere that the denominator is not 0, i.e.
{(x, y): x, y ∈ R, y ≠ 2}

If it's supposed to be a function of a single variable, f(x) = e^(x+1) / (x - 2), essentially the same thing applies and the domain will be R \ {2}.