Please help me graph this:

Set the denominator = 0 to find the vertical asymptotes:



 and 

So the vertical asymptotes are these two lines:





Since the degree of the numerator is greater
than the degree of the denominator, there are
no horizontal asymptotes.  However, since the
degree of the numerator is 1 more than the 
degree of the denominator, there is a slant
or oblique asymptote.  To find it, divide the
denominator into the numerator, being sure
to put in the necessary zero place holders 
for the missing terms. The slant asymptote
is 

            y = quotient

So we divide

                        x + 0
           ——————————————————
x² + 0x - 1)x³ + 0x² + 0x + 0
            x³ + 0x² -  x
            —————————————
                 0x² +  x + 0
                 0x² + 0x - 0
                 ————————————
                        x

So y = quotient is the line y = x + 0

or just the line y = x.  So draw this

line:

 

Now get some points, plot them, and draw a
smooth curve through them:
(-5,-5.2), (-3,-3.8), (-2,-2,7), (-1.5,-2.7), 
(-.9,3.8), (-.5,.17), (0,0), (.5,.-.17), (.9,-3.8)
(1.5,2.7), (2,2.7), (3,3.8), (5,5.2)

 


Edwin