SOLUTION: give the equation of the oblique asymptote, if any, of the function. T(x) x^2 - 8x + 4 / x + 5 this being a fraction, and / being the fraction line

Algebra ->  Rational-functions -> SOLUTION: give the equation of the oblique asymptote, if any, of the function. T(x) x^2 - 8x + 4 / x + 5 this being a fraction, and / being the fraction line      Log On


   

Question 222548: give the equation of the oblique asymptote, if any, of the function.
T(x) x^2 - 8x + 4 / x + 5
this being a fraction, and / being the fraction line
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
To find oblique (or horizontal) asymptotes for rational functions:
  1. 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).
  2. 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.
  3. If all you have is a fraction then you have a horizontal asymptote of y = 0.
  4. If you have more than a fraction, then your asymptote is the graph of the non-fractional part of your function.

Since the degree of the numerator is 2 and the degree of the denominator is 1 we need to divide: T%28x%29+=+%28x%5E2-8x%2B4%29%2F%28x%2B5%29. I'll use dymthetic division:
-5 |    1   -8   4
----        -5  65
       ------------
        1  -13  69

T%28x%29+=+%28x%5E2-8x%2B4%29%2F%28x%2B5%29+=+x+-+13+%2B+69%2F%28x%2B5%29
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.)

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 positive 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 negative 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%28600%2C+600%2C+-45%2C+75%2C+-60%2C+60%2C+x+-+13+%2B+69%2F%28x%2B5%29%2C+x-13%29