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

*print*Print*list*Cite

### 1 Answer

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.

--------------------------------

**Then m=n=1 and p=1.**

-------------------------------

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