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]}
