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If y = mx+n is oblique asymptote of function y = (x^2+x+1)/x, what is p=mXn(m times n)?

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lixalixa | Honors

Posted July 9, 2013 at 5:30 PM via web

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If y = mx+n is oblique asymptote of function y = (x^2+x+1)/x, what is p=mXn(m times n)?

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embizze | High School Teacher | (Level 1) Educator Emeritus

Posted July 9, 2013 at 5:45 PM (Answer #1)

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Given `y=(x^2+x+1)/x` ; if the oblique asymptote is y=mx+n find p=mn:

The oblique asymptote is the line that the graph of the function approaches as x tends to plus/minus infinity.

Perform long division:

`y=(x^2+x+1)/x=x+1+1/x` As x tends to infinity the function approaches the line y=x+1.

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Then m=n=1 and p=1.

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The graph of the function and the line y=x+1(in red):

** Long division:

 

                    x  +  1  +  `1/x`
                 -----------------
             x  |  `x^2` + x +  1
                    `x^2`
                   -----
                             x  +  1
                             x
                            --------
                                     1
 

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