How do you find the horizontal asymptote of the graph of y=3x^6-7x+10/8x^5+9x+10?

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Answer 1 · 2014-09-25T21:53:08

Unfortunately,

$y={3x^6-7x+10}/{8x^5+9x+10}$

does not have any horizontal asymptote; however, it has a slant asymptote $y=3/8x$ (in green).

Its graph looks like this:

![Graph of a slant http://asymptote.](https://useruploads.socratic.org/USNW2LHFS4S5NmA6dwZJ_Capture.JPG)

Let us look at some details.

$lim_{x to pm infty}{3x^6-7x+10}/{8x^5+9x+10}$

by dividing by $x^5$ ,

$=lim_{x to infty}{3x-7/x^4+10/x^5}/{8+9/x^4+10/x^5}$

$={pm infty-0+0}/{8+0+0}=pm infty$

Since the limits at infinity do not exist, there are no horizontal asymptotes.