## Calculus 10th Edition

$f(x)$ has a removable discontinuity at $x=-3$, and a nonremovable (vertical asymptote) discontinuity at $x=6.$
$f(x)=\dfrac{x+3}{x^2-3x-18}=\dfrac{(x+3)}{(x+3)(x-6)}=\dfrac{1}{x-6};x\ne-3, x\ne6.$ 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.