How do you find the Vertical, Horizontal, and Oblique Asymptote given #(x+3)/(x^2-9)#?

1 Answer
Jul 26, 2017

Answer:

#"vertical asymptote at "x=3#
#"horizontal asymptote at "y=0#

Explanation:

#f(x)=(x+3)/(x^2-9)#

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 denominator is non-zero for this value then it is a vertical asymptote.

#"factorise and simplify f(x)"#

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

#"the removal of the factor "(x+3)" from the numerator"#
#"and denominator indicates a hole at x = - 3"#

#"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"#

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