Question

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

Answer 1

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

Explanation:

The denominator of f(x) cannot be zero as this is 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- 3 = 0 `rArrx=3/2" is the asymptote"`

Horizontal asymptotes occur as

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

divide terms on numerator/denominator by x

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

as `xto+-oo,f(x)to(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+5)/(2x-3) [-20, 20, -10, 10]}