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

1 Answer
Nov 28, 2016

Answer:

vertical asymptote at x = 5
horizontal asymptote at y = 2

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)=((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]}