The first thing to is factor the denominator of the rational function:
[tex] x^{2} -2x-3[/tex] to do this we'll need to find two number whose product is -3 and its sum is -2; those numbers are 1 and -3, so:
[tex] x^{2} -2x-3=(x+1)(x-3)[/tex]
Now we can rewrite our rational function as follows:
[tex]f(x)= \frac{x-3}{(x+1)(x-3)} [/tex]
Notice that we have a common factor (x-3) in both numerator and denominator; therefore we can cancel them:
[tex]f(x)= \frac{1}{x+1} [/tex]
Taking all the above into consideration we realize that x=3 is a removable discontinuity; the correct answer is the first one: there is a hole in x=3 and asymptote at x=-1.
