Graph y=3/x-2

Question

Vertical asymptotes and horizontal asymptotes and holes

Answer 1

y=(3/x)-2

first, yu kant divide bi zero so as gotu x=0, kerv gotu infinity

on rite side, gotu +infinity=up off the paper

on left side, down off paper

as x get big, (3/x) gotu 0, so y merj intu horizontal line...y=-2

on rite side, it come down from hi up, tern tu get neer asymptote

left side, come up from down belo