SOLUTION: CAN SOMEONE PLEASE HELP ME WITH THIS PROBLEM: FIND THE HORIZONTAL ASYMPTOTE, IF ANY, OF THE GRAPH OF THE RATIONAL FUNCTION f(x)=8x/2x^2 + 1

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: CAN SOMEONE PLEASE HELP ME WITH THIS PROBLEM: FIND THE HORIZONTAL ASYMPTOTE, IF ANY, OF THE GRAPH OF THE RATIONAL FUNCTION f(x)=8x/2x^2 + 1      Log On


   

Question 157930: CAN SOMEONE PLEASE HELP ME WITH THIS PROBLEM:
FIND THE HORIZONTAL ASYMPTOTE, IF ANY, OF THE GRAPH OF THE RATIONAL FUNCTION
f(x)=8x/2x^2 + 1
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Looking at the numerator 8x, we can see that the degree is 1 since the highest exponent of the numerator is 1. For the denominator 2x%5E2%2B1, we can see that the degree is 2 since the highest exponent of the denominator is 2.

Since the degree of the numerator (which is 1) is less than the degree of the denominator (which is 2), the horizontal asymptote is always y=0

So the horizontal asymptote is y=0


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Notice if we graph y=%288x%29%2F%282x%5E2%2B1%29, we can visually verify our answer:

Graph of y=%288x%29%2F%282x%5E2%2B1%29%29 with the horizontal asymptote y=0 (green line)