How do you find any asymptotes of #f(x)=x/(x-5)#?

Question

Aurora Alvarado · Accepted Answer

VA: #x = 5#
HA: #y=1# (VA) Vertical Asymptote: Set the denominator equal to zero: #x-5 = 0# #x = 5# (HA) Horizontal Asymptote: Divide the coefficients of the x values: #(1x)/(1x)=1# #y=1#

Cameron Alexander · Answer

#"vertical asymptote at "x=5#
#"horizontal asymptote at "y=1# 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"# graph{x/(x-5) [-10, 10, -5, 5]}

Isaiah Allen · Answer

undefined To find the asymptotes of the function f(x) = x/(x-5), we need to consider two types of asymptotes: vertical asymptotes and horizontal asymptotes.

Vertical asymptotes occur when the denominator of the function becomes zero. In this case, the denominator is (x-5), so the vertical asymptote is x = 5.

Horizontal asymptotes can be determined by comparing the degrees of the numerator and denominator. If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is y = 0. If the degree of the numerator is equal to the degree of the denominator, the horizontal asymptote is the ratio of the leading coefficients. In this case, both the numerator and denominator have a degree of 1, so the horizontal asymptote is y = 1.

Therefore, the asymptotes of f(x) = x/(x-5) are x = 5 (vertical asymptote) and y = 1 (horizontal asymptote).