Hello!
[tex]\large\boxed{\text{Infinite discontinuity at x = 2}}[/tex]
[tex]f(x) = \frac{x^{2} - 4x - 12}{x - 2}[/tex]
Recall that in a rational function, the denominator contains the vertical asymptote (an infinite discontinuity).
If terms can be canceled from the denominator and the numerator, then the function contains a hole. We can check whether this function contains such by factoring:
[tex]f(x) = \frac{(x - 6)(x + 2)}{x - 2}[/tex]
There is no common factor, so the function only contains a discontinuity at its vertical asymptote.
Find the vertical asymptote by setting the denominator equal to 0:
x - 2 = 0
x = 2.
