Answer: (2, 8)
Step-by-step explanation:
[tex]f(x)=\dfrac{x^2+4x-12}{x-2}\\\\.\quad=\dfrac{(x+6)(x-2)}{(x-2)}\\\\\text{The denominator cannot equal zero so x - 2}\neq 0\ \rightarrow\ x\neq 2\\\text{Since, (x-2) can be canceled out, there is a hole at x = 2}\\\\f(x)=x+6\\f(2)=2+6\\.\ \ \quad =8\\\\\text{So, the coordinate of the hole is:}\ (2, 8)[/tex]
