Question 222548
To find oblique (or horizontal) asymptotes for rational functions:<ol><li>If the degree of the numerator is equal to or larger than the degree of the denominator, then divide the numerator by the denominator (using long or synthetic division).</li><li>At this point any fraction that remains will have a numerator whose degree is less than the degree of the denominator. As x approaches positive or negative infinity, this fraction will approach zero in value.</li><li>If all you have is a fraction then you have a horizontal asymptote of y = 0.</li><li>If you have more than a fraction, then your asymptote is the graph of the non-fractional part of your function.</li></ol>
Since the degree of the numerator is 2 and the degree of the denominator is 1 we need to divide: {{{T(x) = (x^2-8x+4)/(x+5)}}}. I'll use dymthetic division:
<pre>
-5 |    1   -8   4
----        -5  65
       ------------
        1  -13  69
</pre>
{{{T(x) = (x^2-8x+4)/(x+5) = x - 13 + 69/(x+5)}}}
As x approaches positive or negative infinity the fraction at the end approaches zero in value. So T(x) approaches x-13 in value for these values of x. Our oblique asymptote is y = x - 13. (If there was no "x" in the non-fractional part of the divided T(x), then we'd have a horizontal asymptote.)<br>
If you going to graph T(x), then it may be helpful to look a little more at the divided T(x). As x approaches infinity, the fraction part is a very small <b>positive</b> number. So T(x) is always just a little but more than x-13. So the graph of T(x) will approach the asymptote from above. And as x approaches negative infinity the fraction will be a very small <b>negative</b> number. So T(x) will always be a little bit less than x-13. So the graph of T(x) will approach the asymptote from below.
Here's a graph of T(x) (which shows the oblique asymptote, too):
{{{graph(600, 600, -45, 75, -60, 60, x - 13 + 69/(x+5), x-13)}}}