SOLUTION: Please help me out with this question: How to find the oblique or curvilinear asymptotes for this function . f(X)= (X^3-1)/(x^2-9) Thanks :)

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Question 892661: Please help me out with this question:
How to find the oblique or curvilinear asymptotes for this function
. f(X)= (X^3-1)/(x^2-9)
Thanks :)
Found 2 solutions by josgarithmetic, Edwin McCravy:
Answer by josgarithmetic(39616) About Me  (Show Source):
You can put this solution on YOUR website!
Perform the polynomial division. You should likely find a remainder, which may become increasingly small as x tends to plus or minus infinity, meaning that the quotient without the remainder will be the slant asymptote.

You have a rational expression function. Divide x%5E3-1 by x%5E2-9.
The quotient found is highlight_green%28x%2B%289x-1%29%2F%28x%5E2-9%29%29.
Notice the degree of the numerator of the remainder is smaller than the degree of the denominator of the remainder. As x goes to either extreme, the quotient gets closer to highlight%28y=x%29, which is the slant asymptote.

Compare with the graph:

graph%28300%2C300%2C-15%2C15%2C-15%2C15%2C%28x%5E3-1%29%2F%28x%5E2-9%29%29

,and then showing the included line for the asymptote,
graph%28300%2C300%2C-15%2C15%2C-15%2C15%2C%28x%5E3-1%29%2F%28x%5E2-9%29%2Cx%29

Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!
%22f%28x%29%22%22%22=%22%22%28x%5E3-1%29%2F%28x%5E2-9%29

To find vertical asymptotes, set the denominator = 0

x%5E2-9=0
%28x-3%29%28x%2B3%29=0
x-3=0; x%2B3=0
x=3;   x=-3

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



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

                x+0 
x²+0x-9)x³+0x²+0x-1
        x³+0x²-9x
           0x²+9x-1
           0x²+0x+0
               9x-1

and we can rewrite the equation: 

%22f%28x%29%22%22%22=%22%22x%2B0%2B%289x-1%29%2F%28x%5E2-9%29

%22f%28x%29%22%22%22=%22%22x%2B%289x-1%29%2F%28x%5E2-9%29

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%22%22=%22%22x

We draw it:



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:



Edwin