How do you find the vertical, horizontal or slant asymptotes for #1/(x^2+4)#?

Question

Liam Brennan · Accepted Answer

horizontal asymptote : y=0

vertical asymptote : none

slant asymptote : none

because the numerator is of lower degree than the denominator then there are no slant asymptotes and the horizontal asymptote is : y=0 to find the vertical asymptote you put the denominator= 0 #x^2+4=0# #x^2=-4# #x=sqrt(-4)# you can see that it has no real solutions so there are no vertical asymptotes of this rational function

Elias Ackerman · Answer

#"horizontal asymptote at " y=0#

#"let " f(x)=1/(x^2+4)#

The denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the values that x cannot be and if the numerator is non-zero for there values then they are vertical asymptotes.

#"solve " x^2+4=0rArrx^2=-4#

This has no real solutions, hence there are no vertical asymptotes.

Horizontal asymptotes occur as

#lim_(xto+-oo),f(x)toc" ( a constant)"#

divide terms on numerator/denominator by the highest power of x, that is #x^2#

#f(x)=(1/x^2)/(x^2/x^2+4/x^2)=(1/x^2)/(1+4/x^2)#

as #xto+-oo,f(x)to0/(1+0)#

#rArry=0" is the asymptote"#

Slant asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here, hence there are no slant asymptotes. graph{1/(x^2+4) [-10, 10, -5, 5]}

Greyson Brown · Answer

undefined For the function $ \frac{1}{x^2 + 4} $, there are no vertical asymptotes. There are horizontal and slant (or oblique) asymptotes.

To find the horizontal asymptote:
- As $ x $ approaches positive or negative infinity, the function $ \frac{1}{x^2 + 4} $ approaches 0. Hence, the horizontal asymptote is $ y = 0 $.

To find the slant asymptote (if it exists):
- Divide the numerator by the denominator using long division or polynomial division. Here, we have:
$$ \frac{1}{x^2 + 4} = \frac{0}{x^2} + \frac{1}{x^2 + 4} $$
- The quotient is $ 0 $ and the remainder is $ \frac{1}{x^2 + 4} $.
- Since the degree of the numerator is less than the degree of the denominator, the slant asymptote does not exist.

Therefore, for $ \frac{1}{x^2 + 4} $:
- There is a horizontal asymptote at $ y = 0 $.
- There is no vertical asymptote.
- There is no slant (or oblique) asymptote.