SOLUTION: determine the horizontal asymptotes of the function f(x)=(3x^2+4x-12/x^2-16)

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Question 172180: determine the horizontal asymptotes of the function
f(x)=(3x^2+4x-12/x^2-16)
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!

y=%283x%5E2%2B4x-12%29%2F%28x%5E2-16%29%29 Start with the given function



Looking at the numerator 3x%5E2%2B4x-12, we can see that the degree is 2 since the highest exponent of the numerator is 2. For the denominator x%5E2-16, 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 and the denominator are the same, we can find the horizontal asymptote using this procedure:

To find the horizontal asymptote, first we need to find the leading coefficients of the numerator and the denominator.

Looking at the numerator 3x%5E2%2B4x-12, the leading coefficient is 3

Looking at the denominator x%5E2-16, the leading coefficient is 1

So the horizontal asymptote is the ratio of the leading coefficients. In other words, simply divide 3 by 1 to get %283%29%2F%281%29=3


So the horizontal asymptote is y=3





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

x%5E2-16=0 Set the denominator equal to zero


x%5E2=0%2B16Add 16 to both sides


x%5E2=16 Combine like terms on the right side


x=0%2B-sqrt%2816%29 Take the square root of both sides


x=sqrt%2816%29 or x=-sqrt%2816%29 Break up the "plus/minus" to get two equations


x=4 or x=-4 Take the square root of 16 to get 4



So the vertical asymptotes are the equations x=4 and x=-4

Notice if we graph y=%283x%5E2%2B4x-12%29%2F%28x%5E2-16%29, we can visually verify our answers:

Graph of y=%283x%5E2%2B4x-12%29%2F%28x%5E2-16%29%29 with the horizontal asymptote y=3 (blue line) and the vertical asymptotes x=4 and x=-4 (green lines)