SOLUTION: Find the horizontal and vertical asymptotes {{{y=(5x)/(x^2+1)}}}

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Question 124967: Find the horizontal and vertical asymptotes


y=%285x%29%2F%28x%5E2%2B1%29
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!

y=%285x%29%2F%28x%5E2%2B1%29%29 Start with the given function



Looking at the numerator 5x, we can see that the degree is 1 since the highest exponent of the numerator is 1. For the denominator x%5E2%2B1, we can see that the degree is 2 since the highest exponent of the denominator is 2.


Horizontal Asymptote:

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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Vertical Asymptote:
To find the vertical aysmptote, just set the denominator equal to zero and solve for x

x%5E2%2B1=0 Set the denominator equal to zero


x%5E2=0-1Subtract 1 from both sides


x%5E2=-1 Combine like terms on the right side


x=0%2B-sqrt%28-1%29 Take the square root of both sides

Since you cannot take the square root of a negative number, the answer is not a real number. So in this case, there are no vertical asymptotes.




Notice if we graph y=%285x%29%2F%28x%5E2%2B1%29, we can visually verify our answers:

Graph of y=%285x%29%2F%28x%5E2%2B1%29%29 with the horizontal asymptote y=0 (blue line) . Notice how there are no vertical asymptotes