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Given the function f(x)=x/(x^2-1) and `beta` =m+n where m is the number of vertical...

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yapayapa | (Level 2) Honors

Posted July 20, 2013 at 1:48 PM via web

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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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lfryerda | High School Teacher | (Level 2) Educator

Posted July 20, 2013 at 2:00 PM (Answer #1)

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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 

`x^2-1=(x-1)(x+1)`

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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