Question

How do you find the vertical, horizontal and slant asymptotes of: `f(x)=(x + 1 )/ ( 2x - 4)`?

Answer 1

vertical asymptote at x = 2
horizontal asymptote at `y=1/2`

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: `2x-4=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)=(x/x+1/x)/((2x)/x-4/x)=(1+1/x)/(2-4/x)`

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

`rArry=1/2" 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{(x+1)/(2x-4) [-10, 10, -5, 5]}