How do you find the vertical, horizontal and slant asymptotes of: y=(2x )/ (x-5)y=2xx5?

1 Answer
Jun 28, 2016

vertical asymptote x = 5x=5
horizontal asymptote y = 2y=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 xx we can find the value that xx cannot be. If the numerator is also non-zero for such a value of xx then this must be a vertical asymptote.

solve : x - 5 = 0 rArr x = 5x5=0x=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]}