Question

How do you find the vertical, horizontal or slant asymptotes for `f(x) = (2x+3)/(3x+1 )`?

Answer 1

vertical asymptote `x = -1/3`
horizontal asymptote `y = 2/3`

Explanation:

Vertical asymptotes occur as the denominator of a rational function tends to zero. To find the equation let the denominator equal zero.

solve : `3x + 1= 0 rArr x = -1/3`

`rArr x = -1/3" is the asymptote "`

Horizontal asymptotes occur as `lim_(xto+-oo) f(x) to 0`

divide terms on numerator/denominator by x

`((2x)/x + 3/x)/((3x)/x +1/x) = (2 + 3/x)/(3 + 1/x)`

as `x to+-oo , 3/x" and " 1/x to 0`

`rArr y = 2/3 " is the asymptote "`

This is the graph of f(x).
graph{(2x+3)/(3x+1) [-10, 10, -5, 5]}