Both come from the denominator being zero, so I can't tell which one I'm supposed to get.
Accepted answer
Factor the numerator and denominator completely first — the answer depends on whether the factor **cancels**.
- Factor appears in the denominator **only** → **vertical asymptote**. Nothing offsets it, so the function blows up.
- Factor appears in **both** and cancels → **hole** (removable discontinuity) at that x-value.
For `(x² − 4)/(x − 2) = (x−2)(x+2)/(x−2)`, the `(x − 2)` cancels, so there's a hole at `x = 2`, not an asymptote. The simplified function is `x + 2`, but `x = 2` is still excluded from the domain — cancelling doesn't put the point back.
To give the hole's coordinates, plug the x-value into the *simplified* expression: here `(2, 4)`.