How do you find the vertical, horizontal and slant asymptotes of: #y = (x)/(x^2+4)#?

1 Answer
Nov 30, 2016

Answer:

horizontal asymptote at y = 0

Explanation:

The denominator of y cannot be zero as this would make y undefined. Equating the denominator to zero and solving gives the values that x cannot be and if the numerator is non-zero for these 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),ytoc" ( a constant)"#

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

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

as #xto+-oo,yto0/(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 ( numerator-degree 1, denominator-degree 2 ) Hence there are no slant asymptotes.
graph{(x)/(x^2+4) [-10, 10, -5, 5]}