Question

How do you find vertical, horizontal and oblique asymptotes for `(5x-15)/(2x+4)`?

Answer 1

Vertical asymptote: `x=-2`
Horizontal asymptote: `y=5/2`
There is not oblique asymptote.

Explanation:

Given `y=(5x-15)/(2x+4)` (after converting to an equation)

As `(2x+4)rarr 0`
`color(white)("XXX")yrarr +-oo` (whether it is plus or minus depends upon from which side `(2x+4)` approaches zero.
`color(white)("XXX")2x+4=0` implies `x=-2` the vertical asymptote.

The horizontal asymptote is given as the limit value of `y` as `xrarroo`
`color(white)("XXX")lim_(xrarroo)y`
`color(white)("XXX")=lim(xrarroo) (5x-15)/(2x+4)`

`color(white)("XXX")=lim(xrarroo) (5-15/x)/(2+4/x)`

`color(white)("XXX")=5/2`

An oblique asymptote only exists if the degree of the numerator is greater than the degree of the denominator. `(5x-15)` and `(2x+4)` are both of degree `1`; so there is no oblique asymptote.

graph{(5x-15)/(2x+4) [-16.86, 15.2, -7.67, 8.37]}