How do you find the vertical, horizontal and slant asymptotes of: #f(x)=(2x) /( x-5)#?

Question

How do you find the vertical, horizontal and slant asymptotes of:  #f(x)=(2x) /( x-5)#?

Elijah Allen · Accepted Answer

vertical asymptote x = 5
horizontal asymptote y = 2

The denominator of f(x) cannot be zero as this is 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 of x then it is a vertical asymptote. solve: x - 5 = 0 → x = 5 is the asymptote Horizontal asymptotes occur as #lim_(xto+-oo),f(x)toc" ( a constant)"# divide terms on numerator/denominator by x #((2x)/x)/(x/x-5/x)=2/(1-5/x)# as #xto+-oo,f(x)to2/(1-0)# #rArry=2" is the asymptote"# Slant 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 slant asymptotes. graph{(2x)/(x-5) [-20, 20, -10, 10]}

Cora Alexander · Answer

undefined To find the vertical asymptote, set the denominator equal to zero and solve for $ x $. Vertical asymptote: $ x = 5 $. 
There are no horizontal asymptotes. 
To find the slant asymptote, perform polynomial long division. $ f(x) = 2 + \frac{10}{x-5} $. The slant asymptote is the quotient, $ y = 2 $.