Question 200397
y= (x^2 +3x -4 )/ (x+1)
.
y= { ( x+4)(x-1) } / (x+1)
.
x  intercepts,  zeros  of  num,,,x=-4,,,x=+1
.
y intercepts  ,  at  x=0,,,,,,,y = -4
.
Vertical  Asymptote,,, zero  of  den ,,,,,x=-1
.
Horizontal  Asymptote,   none,,  x^2 > x  (  Num & den)
.
However  there  is  a  slant  asymptote,,,( x^2 one  power  higher  than  x)
.
,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,x,,,,+2,,,,,,,,,,,,,,,r=-6
,,,,,,,,,,,,,,,,,,,,,,--------------------------
(x+1),,,,,,,,,,,,,!,,,x^2  +3x,,,,,-4
,,,,,,,,,,,,,,,,,,,,,,,,,x^2   +x
,,,,,,,,,,,,,,,,,,,,,,,,,_________
,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,2x  -4
,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,2x  +2
,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,_______
,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,-6
.
Slant  asymptote  is,,,,,,,y= x+2
.
To  get  a  rough  sketch ,  use  x-y  coordinates,  lay  in  intercepts, y=-4,x=-4,x=+1
Show  vertical  asymptote  at  x=-1,  and  slant  asymptote  along  y =x+2
.
Starting  at  left,  curve  is  just  above  slant  asy,,  rises  thru  x=-4,  and  continues  up  at  vert  asy  at  x=-1.  On  the  other  side  of  vert  asy,  it  is  at  = - infinity,  rising  thru (0,-4) and (1,0) 
  to  slant  asy.