What is the range of (f(x)=\frac{x^2-16}{x-4})?

Correct answer: ( \mathbb{R}\setminus{8} )

First factor the numerator: \(x^2-16=(x-4)(x+4)\). Since the original denominator is \(x-4\), the original function is defined only when \(x\ne4\). For every allowed input, cancellation gives \(f(x)=x+4\). Thus its graph is the line \(y=x+4\), but with the point corresponding to \(x=4\) removed.

If the missing input \(x=4\) were allowed, the missing output would be \(4+4=8\). Could another input produce output 8? The equation \(x+4=8\) gives only \(x=4\), which is forbidden. Every other real output is produced by \(x=y-4\), and that input is not 4 when \(y\ne8\). Therefore the range is \(\mathbb R\setminus\{8\}\), option A.