Question

How do you find the vertical, horizontal or slant asymptotes for `f(x) = x / (3x-1)`?

Answer 1

`"vertical asymptote at " x=1/3`

`"horizontal asymptote at " y=1/3`

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 : `3x-1=0rArrx=1/3" is the asymptote"`

Horizontal asymptotes occur as

`lim_(xto+-oo),f(x)to c " (a constant)"`

divide terms on numerator/denominator by x

`f(x)=(x/x)/((3x)/x-1/x)=1/(3-1/x)`

as `xto+-oo,f(x)to1/(3-0)`

`rArry=1/3" 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{x/(3x-1) [-10, 10, -5, 5]}