Given the function f(x)=x/(x^2-1) and `beta` =m+n where m is the number of vertical asymptotes and n the number of horizontal asymptotes, then `beta` is: a) 1 b) 2 c) 3 d) 0



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Posted on (Answer #1)

The function `f(x)=x/{x^2-1}` has both a horizontal asymptote and vertical asymptotes.  

Since the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is `y=0` .

To find the vertical asymptotes, we need to factor the denominator.  The asymptotes are given by the zeros of the denominator.

This means that 


has zeros at `x=1` and `x=-1`

so there are two vertical asymptotes.

The sum of asymptotes is 2+1=3, so `beta=3` , which is answer (c).

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