How do you find the asymptotes for #f(x) = (x^2 - 8)/(x+3)#?

1 Answer
Dec 28, 2016

Answer:

#"vertical asymptote at " x=-3#
#"oblique asymptote is " y=x-3#

Explanation:

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+3=0rArrx=-3" is the asymptote"#

Horizontal asymptotes occur as

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

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

#f(x)=(x^2/x^2-8/x^2)/(x/x^2+3/x^2)=(1-8/x^2)/(1/x+3/x^2)#

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

This is undefined hence there are no horizontal asymptotes.

Oblique asymptotes occur when the degree of the numerator > degree of the denominator. This is the case here hence there is an oblique asymptote.

Using #color(blue)"polynomial division"#

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

as #xto+-oo,f(x)tox-3#

#rArry=x-3" is the asymptote"#
graph{(x^2-8)/(x+3) [-20, 20, -10, 10]}