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

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

Start with the given function

Looking at the numerator , we can see that the degree is since the highest exponent of the numerator is . For the denominator , we can see that the degree is since the highest exponent of the denominator is .

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 , the leading coefficient is

Looking at the denominator , the leading coefficient is

So the horizontal asymptote is the ratio of the leading coefficients. In other words, simply divide by to get

So the horizontal asymptote is

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

Set the denominator equal to zero

Add 16 to both sides

Combine like terms on the right side

Take the square root of both sides

or Break up the "plus/minus" to get two equations

or Take the square root of 16 to get 4

So the vertical asymptotes are the equations and

Notice if we graph , we can visually verify our answers:

Graph of with the horizontal asymptote (blue line) and the vertical asymptotes and (green lines)

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