Answer:
C) [tex]\{x|x\neq-4 \:or \:4\}[/tex]
Step-by-step explanation:
[tex]\frac{x^2+2x-8}{x^2-16}\\ \\\frac{(x+4)(x-2)}{(x+4)(x-4)}\\ \\x\neq-4 \:or \:4[/tex]
Here, [tex]x\neq-4[/tex] is a hole on the graph since [tex]x+4[/tex] exists in both the numerator and denominator.
