For the function f(x) = (x²-9)/(x-3), what occurs at x=3?

Factor the numerator as a difference of squares: $x^2 - 9 = (x-3)(x+3)$. Then $f(x) = \frac{(x-3)(x+3)}{x-3} = x + 3$ for $x \neq 3$. The factor $x - 3$ cancels completely, so the function is not defined at $x = 3$ but approaches 6 from both sides. That creates a removable discontinuity, or hole, at $(3, 6)$, choice C. Choice A is the usual reflex when a denominator is zero, but a vertical asymptote needs a factor that remains in the denominator after canceling. Choice B is wrong because the function is a line with a hole. Choice D is wrong because $f(3)$ is undefined.