Question 892661
<pre>
{{{"f(x)"}}}{{{""=""}}}{{{(x^3-1)/(x^2-9)}}}

To find vertical asymptotes, set the denominator = 0

{{{x^2-9=0}}}
{{{(x-3)(x+3)=0}}}
{{{x-3=0}}}; {{{x+3=0}}}
{{{x=3}}};   {{{x=-3}}}

So the two vertical asymptotes are x=-3 and x=3.
We draw those.

{{{drawing(800/3,400,-10,10,-15,15,
green(line(-3,-20,-3,20),line(3,-20,3,20)),
graph(800/3,400,-10,10,-15,15) )}}}

Now we ind what you asked for, the oblique or curvilinear 
asymptotes.  To find that we use long division:

       <u>         x+0</u> 
x²+0x-9)x³+0x²+0x-1
        <u>x³+0x²-9x</u>
           0x²+9x-1
           <u>0x²+0x+0</u>
               9x-1

and we can rewrite the equation: 

{{{"f(x)"}}}{{{""=""}}}{{{x+0+(9x-1)/(x^2-9)}}}

{{{"f(x)"}}}{{{""=""}}}{{{x+(9x-1)/(x^2-9)}}}

When x gets larger and larger in absolute value
the fraction gets closer and closer to 0, so to 
get the asymptote we drop the fraction and set 
y = the quotient only.

So we have an oblique asymptote, which has 
this equation: 

{{{y}}}{{{""=""}}}{{{x}}}

We draw it:

{{{drawing(800/3,400,-10,10,-15,15,
green(line(-3,-20,-3,20),line(3,-20,3,20),line(15,15,-15,-15)),
graph(800/3,400,-10,10,-15,15) )}}}

The we get some points

x|-10| -7 |  -5  |-3.5 |-2.5|
-|---|----|------|-----|-----
y|-11|-8.6|-7.875|-13.5|6.05|


x| 10 | 7  |   5  | 4 | 2.6|
-|----|----|------|---|-----
y|10.9|8.55| 7.75 | 9 |-7.4|

And sketch the graph:

{{{drawing(800/3,400,-10,10,-15,15,
green(line(-3,-20,-3,20),line(3,-20,3,20),line(15,15,-15,-15)),
graph(800/3,400,-10,10,-15,15,(x^3-1)/(x^2-9)) )}}}

Edwin</pre>