Question

How do you find the Vertical, Horizontal, and Oblique Asymptote given `f(x)= (2x+1)/(x-1)`?

Answer 1

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

as `xto+-oo,f(x)to(2+0)/(1-0)`

`rArry=2" is the asymptote"`

Oblique 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 oblique asymptotes.
graph{(2x+1)/(x-1) [-20, 20, -10, 10]}