Pre-Calc Help?

Question

How do I find the horizontal asymptotes for an equation:

Let's say y=(x^2-9)/(4x^2+1)

Thank you.

Answer 1

If the degree of the numerator equals the degree of the denominator, then it is just the ratio of the coefficients of the term with the highest degree.
The degree of x^2 - 9 is 2.
The degree of 4x^2 + 1 is 2.
The coefficient of x^2 is 1, and the coefficient of 4x^2 is 4
The horizontal asymptote is y = 1/4
Other cases:
1) If the degree of the numerator is greater than the degree of the denominator, then there is no horizontal asymptote.
2) If the degree of the numerator is less than the degree of the denominator, then the horizontal asymptote is y = 0.

Answer 2

If the polynomials in the numerator and denominator have the same degree, as in your example, the horizontal asymptote is at the ratio of the leading coefficients. 1/4 in this case.

Answer 3

y = (x^2 - 9)/(4x^2 + 1)
x = (y^2 - 9)/(4y^2 + 1)

4xy^2 + x = y^2 - 9
4xy^2 - y^2 = -x - 9
y^2(4x - 1) = -x - 9
y^2 = (-x - 9)/(4x - 1)
y = sqrt((-x - 9)/(4x - 1))
y = (sqrt((-x - 9)(4x - 1)))/(4x - 1)
y = (sqrt(-4x^2 + x - 36x + 9))/(4x - 1)
y = (sqrt(-4x^2 - 35x + 9))/(4x - 1)

so the horizontal asympote is (1/4)

Answer 4

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