For the function f(x) = ((x-2)(x+3))/((x-2)(x-5)), which statement correctly describes its behavior at x=2 and x=5?

The factor $(x - 2)$ appears in both the numerator and the denominator, so it cancels, leaving $\frac{x+3}{x-5}$ for $x \neq 2$. At $x = 2$, the original function is undefined, but the simplified form gives a finite value of $-5$, so there is a removable discontinuity (a hole). At $x = 5$, the factor $(x - 5)$ remains in the denominator, so the function blows up and there is a vertical asymptote. That is choice C. Choice A treats both as asymptotes. Choice B swaps the roles. Choice D ignores that both $x = 2$ and $x = 5$ make the original denominator zero.