How do you find the vertical, horizontal or slant asymptotes for # g(x) = (X+6) /( x^2 - 36)#?

1 Answer
Dec 27, 2016

Answer:

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

Explanation:

Factorise and simplify.

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

Since (x + 6) has been removed, x = -6 is an excluded value.

The denominator of g(x) cannot be zero as this would make g(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),g(x)toc" ( a constant)"#

divide terms on numerator/denominator by x

#g(x)=(1/x)/(x/x-6/x)=(1/x)/(1-6/x)#

as #xto+-oo,g(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 (numerator-degree 0, denominator-degree 1 ) Hence there are no slant asymptotes.
graph{1/(x-6) [-10, 10, -5, 5]}