How do you find the asymptotes for #f(x)=(3x^2+2) / (x^2 -1)#?

Question

Benjamin Achtenberg · Accepted Answer

The vertical asymptotes are #x=1# and # x=-1#

The horizontal asymptote is #y=3# The asymptotes occur where the denominator approaches zero, and where #x# becomes very large, either positively or negatively. #f(x) = (3x^2+2)/(x^2-1)# In this case the denominator is the difference of two squares so the function can be rewritten as #f(x)= (3x^2+2)/((x-1)(x+1))# The denominator is zero when either #x=1# or # x=-1# and these are therefore the vertical asymptotes. AS #x# becomes very large, either positively or negatively, #lim_(x->oo) =(3cancel(x^2) )/cancel(x^2)# The horizontal asymptote is therefore #y=3#

Ariana Alvarez · Answer

undefined To find the asymptotes of $ f(x) = \frac{{3x^2 + 2}}{{x^2 - 1}} $, we need to identify vertical asymptotes, horizontal asymptotes, and possibly oblique asymptotes.

1. Vertical Asymptotes: Set the denominator equal to zero and solve for $ x $.
   $ x^2 - 1 = 0 $
   $ (x - 1)(x + 1) = 0 $
   $ x = 1 $ and $ x = -1 $

Therefore, the vertical asymptotes are $ x = 1 $ and $ x = -1 $.

2. Horizontal Asymptotes: Compare the degrees of the numerator and the denominator.
   Degree of numerator = 2, Degree of denominator = 2
   Since the degrees are equal, divide the leading coefficients.
   $ \frac{{3}}{{1}} = 3 $

Therefore, the horizontal asymptote is $ y = 3 $.

There are no oblique asymptotes for this function. Therefore, the vertical asymptotes are $ x = 1 $ and $ x = -1 $, and the horizontal asymptote is $ y = 3 $.