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

Question

Elijah Allen · Accepted Answer

#cancel(EE)# vertical asymptotes
#y=1# horizontal asymptote #x rarr +-oo# Given: #f(x)=g(h)/(h(x))# possible asymptotes are #x=x_i# with #x_i# values that zero the denominator #h(x)=0#. Indeed, the existence condition of #f(x) in RR# is #h(x)!=0#. Therefore, #h(x)=0 => x^2+1=0=> x^2=-1=>x_1,x_2 in CC# #h(x)# hasn't any zeros #in RR# #:.f(x): RR rarr RR, AA x in RR# We could looking for possible horizontal asymptotes: #y=L# is a horizontal asymptotes when: #lim_(x rarr +-oo)f(x)=L# #lim_(x rarr +-oo)f(x)=lim_(x rarr +-oo) x^2/(x^2+1)~~cancel(x^2)/cancel(x^2)=1# #:.y=1# horizontal asymptote #x rarr +-oo# graph{x^2/(x^2+1) [-6.244, 6.243, -3.12, 3.123]}

Sadie Alexander · Answer

undefined To find the asymptotes of the function $f(x) = \frac{x^2}{x^2 + 1}$, you need to consider both horizontal and vertical asymptotes.

1. Vertical Asymptotes:
   Vertical asymptotes occur where the denominator of the function becomes zero, but the numerator does not. In this case, $x^2 + 1$ becomes zero when $x^2 = -1$. However, since there are no real solutions to $x^2 = -1$, there are no vertical asymptotes for this function.

2. Horizontal Asymptotes:
   Horizontal asymptotes occur when $x$ approaches positive or negative infinity. To find horizontal asymptotes, compare the degrees of the numerator and denominator polynomials.
   - If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is $y = 0$.
   - If the degree of the numerator is equal to the degree of the denominator, the horizontal asymptote is the ratio of the leading coefficients.
   - If the degree of the numerator is greater than the degree of the denominator, there are no horizontal asymptotes.

In this case, both the numerator and the denominator have the same degree (2). Therefore, the horizontal asymptote is the ratio of the leading coefficients, which is $y = \frac{1}{1} = 1$.

So, the function $f(x) = \frac{x^2}{x^2 + 1}$ has a horizontal asymptote at $y = 1$, and there are no vertical asymptotes.