Answer:
-3 (removable), +5
Step-by-step explanation:
Maybe you have ...
[tex]f(x)=\dfrac{x+3}{x^2-2x-15}=\dfrac{x+3}{(x+3)(x-5)}=\dfrac{1}{x-5}\quad x\ne-3[/tex]
This will have discontinuities (points where the function is undefined) at ...
The discontinuity as x = -3 is removable by defining f(-3) = -1/8.