How do you find the vertical, horizontal and slant asymptotes of: #g(x)=2 /(x^2-16)#?

1 Answer
Sep 5, 2016

Answer:

vertical asymptotes at x = ± 4
horizontal asymptote at y = 0

Explanation:

The denominator of g(x) cannot be zero as this would make g(x) 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^2-16=0rArr(x-4)(x+4)=0#

#rArrx=-4" and " x=4" are the asymptotes"#

Horizontal asymptotes occur as

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

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

#g(x)=(2/x^2)/(x^2/x^2-16/x^2)=(2/x^2)/(1-16/x^2)#

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 2) Hence there are no slant asymptotes.
graph{2/(x^2-16) [-10, 10, -5, 5]}