Is the denominator of the function (x^{2}-8), or is it x^{2}?

As written, with parenthesis for clarification, the function is:

f(x) = (1/x^{2}) - 8

The domain of the function is the set of real number values that we may input into the function and get a real number value as an output. Since division by 0 is undefined, the domain of the function (as written) is all real numbers except 0.

If we compare this function to (1/x^{2}), which has a range of (0, infinity). Now the function that you wrote has been vertically shifted down 8 units, and the range would be (-8,infinity)

If, instead, you meant:

f(x) = 1/(x^{2}-8), we will again consider where the denominator may equal 0, and exclude these values from the domain.

x^{2} - 8 = 0

x^{2} = 8 (taking the square root of both side and evaluating the result)

x = ±2√(2)

Then the function has two vertical asymptotes at these values (x=2√(2) and x= - 2√(2)), and the domain is all real numbers except these two values.

Since the highest power of x in the numerator is less than the highest power of x in the denominator, there is a horizontal asymptote at y=0.

The range of this function is all real numbers not equal to 0.