My Math Forum Graph of function

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 March 9th, 2009, 11:56 PM #1 Senior Member   Joined: Sep 2008 Posts: 199 Thanks: 0 Graph of function Sketch the graph of : $y=\frac{x^2}{1+x}$ i know that x=-1 is a vertical asymtote and i don think there is a horizontal asymtote .But i can't sketch the graph with only one asymtote and there must be another one . Someone pls help me out with this .. THanks a lot .
 March 10th, 2009, 12:09 AM #2 Senior Member   Joined: Feb 2009 From: Adelaide, Australia Posts: 1,519 Thanks: 3 Re: Graph of function $\frac{x^2}{x+1}= \frac{(x+1)(x-1)+1}{x+1} = \frac{(x+1)(x-1)}{x+1}+\frac{1}{x+1} = x-1+\frac{1}{x+1}$ So y = x-1 is an asymptote.
March 10th, 2009, 04:38 AM   #3
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Quote:
 Originally Posted by mikeportnoy i can't sketch the graph with only one asymtote and there must be another one . Someone pls help me out with this .. THanks a lot .
To find the slant (or "oblique") asymptote, do the polynomial long division. Whatever you get across the top of the division is your other asymptote.

Code:
        x  - 1
-------------
x + 1 )x^2 + 0x + 0
x^2 + 1x
-------------
-1x + 0
-1x - 1
-------------
1

 March 10th, 2009, 06:35 AM #4 Senior Member   Joined: Sep 2008 Posts: 199 Thanks: 0 Re: Graph of function Thanks for both of your help . really appreciate that . .

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