Is the denominator of the function (x2-8), or is it x2?
As written, with parenthesis for clarification, the function is:
f(x) = (1/x2) - 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/x2), 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/(x2-8), we will again consider where the denominator may equal 0, and exclude these values from the domain.
x2 - 8 = 0
x2 = 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.