Answer
$f(x)$ has a removable discontinuity at $x=-3$, and a nonremovable (vertical asymptote) discontinuity at $x=6.$
Work Step by Step
$f(x)=\dfrac{x+3}{x^2-3x-18}=\dfrac{(x+3)}{(x+3)(x-6)}=\dfrac{1}{x-6};x\ne-3, x\ne6.$
At $x=-3,$ the function has a removable discontinuity since the factor cancels out from both the numerator and denominator.
At $x=6,$ the denominator is $0$ indicating the presence of a vertical asymptote.
The vertical asymptote is irremovable.
