How do you find the vertical, horizontal and slant asymptotes of: #f(x) = (2x-1) / (x - 2)#?

1 Answer
Aug 25, 2016

Answer:

vertical asymptote at x = 2
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-2=0rArrx=2" 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-1/x)/(x/x-2/x)=(2-1/x)/(1-2/x)#

as #xto+-oo,f(x)to(2-0)/(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-1)/(x-2) [-10, 10, -5, 5]}