The domain of a function y = f(x) is the set of all real values that x an take on for which y is real. If the value of x lies in the domain it is possible to plot the point (x, f(x)).

For the function `f(x) = (x+4)/(x^2-9)` , the value of f(x) is real for all real values of x except for the case when the denominator is 0.

If x^-9 = 0

x^2 = 9

x = `+- 3`

**The domain of the function `f(x) = (x+4)/(x^2 - 9)` is R - {-3, 3}**

The graph of this function is:

The domain of a function y=f(x) is a set of values of x for which f(x) is defined.

Now unless otherwise mentioned, we assume that the given function is defined for real values of x i.e. the domain of the function is Real.

The function: `f(x)=(x+4)/(x^2-9)`

is undefined when the denominator is 0.

So assuming a real domain, the values of x for which `(x^2-9)` goes to 0 (and hence f(x) is undefined) are:

`x= +-3 ` .

Hence, the domain of the given function y=f(x) is: `RR - {-3, +3}`

The domain of the function is the defined "x" values that do not result in the denominator equaling 0. In the provided equation:

f(x) = (x+4)/(x^2 + 9) the (x^2 + 9) component cannot have an end result of 0, therefore "x" cannot = 3 or -3.

x < -3 or -3<x<3 or x>3

The domain is all (R) values other than -3 and 3.

The domain is all the x-values. You need to find locations where x cannot be defined in order to find the domain. The main places you need to check are square roots and denominators.

In this problem, there is a denominator, so set it equal to 0 and solve for x.

`(x^(2)-9)=0 `

`x=3, x=-3`

These are the places where x DOES NOT exist. Therefore, it exists in all other places.