How do you find the vertical, horizontal and slant asymptotes of: #f(x) = 1/(x+3)#?

1 Answer
Aug 31, 2016

Answer:

vertical asymptote at x = - 3
horizontal asymptote at y = 0

Explanation:

The denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the value that x cannot be and if the numerator is non-zero for this value then it is a vertical asymptote.

solve: #x+3=0rArrx=-3" is the asymptote"#

Horizontal asymptotes occur as

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

divide terms on numerator/denominator by x

#f(x)=(1/x)/(x/x+3/x)=(1/x)/(1+3/x)#

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

#rArry=0" is the asymptote"#

Slant asymptotes occur when the degree of the numerator is > degree of the denominator. This is not the case here (numerator-degree 0,denominator-degree 1 ) Hence there are no slant asymptotes.
graph{1/(x+3) [-10, 10, -5, 5]}