Question

How do you find the vertical, horizontal or slant asymptotes for `(3x-2) /(2x+5)`?

Answer 1

vertical asymptote `x=-5/2`
horizontal asymptote `y=3/2`

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 : 2x + 5 = 0 `rArrx=-5/2`

`rArrx=-5/2" is the asymptote"`

Horizontal asymptotes occur as

`lim_(xto+-oo),ytoc" (a constant)"`

divide terms on numerator/denominator by x

`((3x)/x-2/x)/((2x)/x+5/x)=(3-2/x)/(2+5/x)`

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

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