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

2 Answers
Apr 19, 2017

Answer:

horizontal asymptote : y=0

vertical asymptote : none

slant asymptote : none

Explanation:

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

Apr 19, 2017

Answer:

#"horizontal asymptote at " y=0#

Explanation:

#"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]}