How do you find the asymptotes for #f(x)=x/(x^2+4)#?

Question

Cora Alexander · Accepted Answer

There is no vertical asymptote, as the denominator will always be unequal to #0# (even #f(x)>=4#, whatever the value of #x#).

For the horizontal asymptote, we look what happens if #x# grows very large (both positive and negative). The #+4# in the denominator will make less and less of a difference, and the function will look more and more like: #f(x)=x/x^2=1/x# In "the language": #lim_(x->oo) f(x)=0# and #lim_(x->-oo) f(x)=0# So there's only #y=0# as horizontal asymptote at the #+-oo# ends of the function. For #x=0->f(x)=0#, so it would seem to be not a proper asymptote (in the meaning "never there but as close as you want"), but it still counts as one, because for the rest of the function it really is "never there". graph{x/(x^2+4) [-10, 10, -5, 5]}

Christian Antunez · Answer

undefined To find the asymptotes of the function $ f(x) = \frac{x}{x^2 + 4} $, we need to determine both the vertical and horizontal asymptotes.

Vertical Asymptotes:
Vertical asymptotes occur where the denominator of the function becomes zero but the numerator does not. Set the denominator equal to zero and solve for x:

$$ x^2 + 4 = 0 $$
$$ x^2 = -4 $$

Since no real number squared can equal a negative number, this equation has no real solutions. Therefore, there are no vertical asymptotes for the function.

Horizontal Asymptotes:
To find horizontal asymptotes, we look at the behavior of the function as x approaches positive or negative infinity. We compare the degrees of the numerator and denominator:

Degree of numerator: 1 (x)
Degree of denominator: 2 (x^2)

Since the degree of the numerator is less than the degree of the denominator, there is a horizontal asymptote at y = 0 (the x-axis).

Therefore, for the function $ f(x) = \frac{x}{x^2 + 4} $, there are no vertical asymptotes, and the horizontal asymptote is y = 0.