Question 194631
y =1/x,,,,also  called  the  reciprocal  function
,
there  is  a  horizontal  asymptote  at  ,, y=o
,
and  a  vertical  asymptote  at x=0
,
to  rough  sketch  this,  draw  a  standard  x - y  coordinate  system.
sketch  in  asymptotes,  (same  as  x,y axis)
,
starting  at  far  left,  x=- infinity,  y is  on  the  negative  side  of zero( approaches  zero),  
going  to  right  this  function  curves  downward,  to  - infinity  as  x  approaches  zero.
,
skipping  over  the  x=0  asymptote,  The  function  picks  up  at  + infinity
as  we  go right  it  curves  down  to  approach  y=0  as  x = + infinity
,
in  short  in the  3rd  quadrant  it  is  a  "u"  vertex  at  (0,0) pointed  down  and  to  left,
in  the  1st  quadrant,  it  is  a  "u" vertex  at  (0,0) pointing  up  and  to  right. NOTE  that  the  vertex  does  not touch (0,0) , but  only approaches  along  the  45 degree  line,  y=x.
,
hpoefully  this  helps