Answer: C

Small Hint:

Factor $x^2-4$, but keep the original restriction $x\ne2$

Big Hint:

Compare the algebraic intersection of the simplified lines with that domain restriction

Solution:

For $x\ne2,$ $$\frac{x^2-4}{x-2}=x+2.$$ The lines $y=x+2$ and $y=2x$ would meet at $x=2,\ y=4.$ But $x=2$ is excluded from the original rational function, so that point is a hole and there is no intersection.

Thus, the correct answer is C.