How do you find vertical, horizontal and oblique asymptotes for #(x+6 )/ (x^2 - 36)#?

1 Answer
Aug 26, 2016

Answer:

vertical asymptote at x = 6
horizontal asymptote at y = 0

Explanation:

The first step is to factorise and simplify the function.

#f(x)=(cancel(x+6))/(cancel((x+6))(x-6))=1/(x-6)#

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-6=0rArrx=6" 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-6/x)=(1/x)/(1-6/x)#

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

#rArry=0" is the asymptote"#

Oblique asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here ( numerator-degree 0, denominator-degree 1) Hence there are no oblique asymptotes.
graph{1/(x-6) [-10, 10, -5, 5]}