How do you find the Vertical, Horizontal, and Oblique Asymptote given #(2x-4)/(x^2-4)#?

1 Answer
Aug 4, 2018

Answer:

#"vertical asymptote at "x=-2#
#"horizontal asymptote at "y=0#

Explanation:

#"let "f(x)=(2x-4)/(x^2-4)#

#"factor numerator/denominator"#

#f(x)=(2cancel((x-2)))/(cancel((x-2))(x+2))=2/(x+2)#

#"the removal of the factor "(x-2)" indicates a removable"#
#"discontinuity (hole) at "x=2#

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+2=0rArrx=-2" is the asymptote"#

#"Horizontal asymptotes occur as"#

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

#"divide terms on numerator/denominator by "x#

#f(x)=(2/x)/(x/x+2/x)=(2/x)/(1+2/x)#

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

#y=0" is the asymptote"#

Oblique asymptotes occur when the degree of the numerator is greater than the degree of the denominator. This is not the case here hence there are no oblique asymptotes.
graph{(2x-4)/(x^2-4) [-10, 10, -5, 5]}