Question

How do you find vertical, horizontal and oblique asymptotes for `(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 set the denominator equal to zero.

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

Horizontal asymptotes occur as `lim_(xto+-oo) , y to 0`

divide terms on numerator/denominator by x

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

as `xto+-oo ,yto(2+0)/(3+0)`

`rArry=2/3" 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+3)/(3x+1) [-10, 10, -5, 5]}