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

1 Answer
Jun 28, 2016

Answer:

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

Explanation:

For this rational function (fraction) the denominator cannot be zero. This would lead to division by zero which is undefined. By setting the denominator equal to zero and solving for #x# we can find the value that #x# cannot be. If the numerator is also non-zero for such a value of #x# then this must be a vertical asymptote.

solve : #x - 5 = 0 rArr x = 5# is the asymptote

Horizontal asymptotes occur as

#lim_(xto+-oo) ytoc" (a constant)"#

divide terms on numerator/denominator by #x#

#((2x)/x)/(x/x-5/x)=2/(1-5/x)#

as #xto+-oo, yto2/(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 degree 1 ) Hence there are no slant asymptotes.
graph{(2x)/(x-5) [-20, 20, -10, 10]}