Question

How do you find all the asymptotes for `(4+3x)/(6-12x)`?

Answer 1

Vertical Asymptote: `x=1/2`
Horizontal asymtote: `y=-1/4`

Explanation:

The first one is the vertical asymptote found by setting the condition:
`6-12x!=0` this is done to avoid a zero in the denominator (that cannot be evaluated) and gives you that must be `x!=1/2`

So the vertical line with equation `x=1/2` will be an asymptote of your function (cannot be crossed by it).

The horizontal asymptote can be found observing the behaviour of your function for very big values of `x` by doing:

`lim_(x->oo)((4+3x)/(6-12x))` this will tell you whether or not your function tends to get near to some value.

`lim_(x->oo)((4+3x)/(6-12x))=lim_(x->oo)(x(4/x+3))/(6x(1/x-2))=`
`=lim_(x->oo)(cancel(x)(4/x+3))/(6cancel(x)(1/x-2))=-1/4`

So basically your function when `x` becomes very large tends to become equal to `-1/4` so you have a horizontal asymptote (representing the line near which your functions tends) given as:
`y=-1/4`

Graphically:
graph{(4+3x)/(6-12x) [-10, 10, -5, 5]}