Question

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Answer 1

The first thing to is factor the denominator of the rational function:
$x^{2} -2x-3$ 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:
$x^{2} -2x-3=(x+1)(x-3)$
Now we can rewrite our rational function as follows:
$f(x)= \frac{x-3}{(x+1)(x-3)}$
Notice that we have a common factor (x-3) in both numerator and denominator; therefore we can cancel them:
$f(x)= \frac{1}{x+1}$

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.