How do you find the Vertical, Horizontal, and Oblique Asymptote given #f(x)= x/(x-5)#?

1 Answer
Oct 7, 2016

Answer:

vertical asymptote at x = 5
horizontal asymptote at y = 1

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-5=0rArrx=5" is the asymptote"#

Horizontal asymptotes occur as

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

divide terms on numerator/denominator by x

#f(x)=(x/x)/(x/x-5/x)=1/(1-5/x)#

as #xto+-oo,f(x)to1/(1-0)#

#rArry=1" is the asymptote"#

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