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kow [346]
2 years ago
15

Helpppppppppppppppppppppppp

Mathematics
1 answer:
Vaselesa [24]2 years ago
4 0
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.

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