SOLUTION:
We are to find if there are holes and at where;
[tex]\begin{gathered} f(x)\text{ = }\frac{x^2-3x\text{ -4}}{x^2-1} \\ \\ \end{gathered}[/tex][tex]\frac{x^2+x-4x\text{ -4}}{(x^{}-1)(x+1)}[/tex][tex]\frac{(x+1)(x-4)}{(x+1)(x-1)}[/tex]A hole exists on the graph of a rational function at any input value that causes both the numerator and denominator of the function to be equal to zero.
Setting x to be "-1" will make both the numerator and the denominator to be zero, so holes exist.
The correct option therefore is, Yes, x = -1 (First Option).