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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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`

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x | `x^2` + x + 1

`x^2`

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x + 1

x

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1

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