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

1 Answer
Jul 16, 2016

Answer:

vertical asymptote x = 5
horizontal asymptote y = 2

Explanation:

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]}