What is the horizontal asymptote of the equation f(x) = x^2 -100/(x^2 - 64)?

Question

And can anyone help on what is the process to find it?

Answer 1

We could find t he limit as x goes to infinity.

f(x) = (1 - 100/x^2) / (1 - 64/x^2)
I just divided every term by x^2.
So as x goes to inf, f(x) goes to 1/1 = 1

Horizontal aymptote is at y = 1.
---
Alternatively, you can make x the subject of the formula
y*(x^2 - 64) = x^2 - 100
x^2 * (y - 1) = 64*y - 100
x^2 = (64*y - 100) / (y - 1)

Now you see that as y approaches 1, x^2 approaches infinity.

Answer 2

If it's really x² - 100/(x²-64), it doesn't have one. It's just the parabola f(x) = x² interrupted by vertical asymptotes at x = ±8.

If it's [x² - 100] / [x² - 64], then the horizontal asymptote is y = 1, since x²/x² = 1. You just read off the coefficient of term of highest degree in numerator over term of highest degree in denominator. If degrees are the same, you have a horizontal asymptote: lim (n→∞) ax^n / bx^n = a/b. If top degree is 1 more than bottom degree, you have a slant (oblique) asymptote. If bottom degree is higher, horizontal asymptote is y=0.