Answer:
See below
Step-by-step explanation:
Factor the numerator and denominator
[tex]\displaystyle f(x)=\frac{x^2-9}{(x^2-x-6)(x^2+6x+9)}\\\\f(x)=\frac{(x+3)(x-3)}{(x-3)(x+2)(x+3)(x+3)}\\\\f(x)=\frac{x-3}{(x-3)(x+2)(x+3)}[/tex]
Because [tex]x-3[/tex] exists in both the numerator and denominator, there will be a hole at [tex]x=3[/tex] because the function is not continuous at that point.
If we check if the function is continuous at [tex]x=2[/tex], we can see that the denominator will not be 0, thus, the function is continuous at [tex]x=2[/tex].
