Question

How do you find the horizontal and vertical asymptote of this function `(3x-2) / (x+1)`?

Answer 1

The vertical asymptote is the line passing through the `x` coordenate that makes the denominator equal to zero, i.e., the line of equation `x=-1`.
The horizontal asymptote can be found by observing the behaviour of the functions for `x` very large doing:
`lim_(x->oo)(3x-2)/(x+1)=lim_(x->oo)(x(3-2/x))/(x(1+1/x))=lim_(x->oo)(cancel(x)(3-2/x))/(cancel(x)(1+1/x))=`
doing the limit:
`lim_(x->oo)(3-2/x)/(1+1/x)=(3-cancel(2/x))/(1+cancel(1/x))=3`
The function approximates the value `3` when `x` becomes large.
So, the horizontal line of equation `y=3` will be a horizontal asymptote:

Graphically:
graph{(3x-2)/(x+1) [-43.4, 49.07, -17.12, 29.13]}