To find the horizontal asymptote of the function $f(x) = \frac{13x^2 + 6x}{12x^2 - 5}$, we compare the degrees of the polynomials in the numerator and the denominator.

Step 1: Compare the degrees of the polynomials

• The degree of the numerator $13x^2 + 6x$ is 2.
• The degree of the denominator $12x^2 - 5$ is 2.

Since both the numerator and denominator have the same degree (degree 2), the horizontal asymptote is determined by the ratio of the leading coefficients.

Step 2: Ratio of the leading coefficients

The leading coefficient of the numerator is 13, and the leading coefficient of the denominator is 12.

Thus, the horizontal asymptote is: $y = \frac{13}{12}$

So, the horizontal asymptote of the function $f(x)$ is $y = \frac{13}{12}$.

---

Do you want more details or have any questions?
Here are 5 related questions to expand your understanding:

1. How do you find the vertical asymptote of a rational function?
2. What happens to a rational function at its horizontal asymptote as $x \to \infty$?
3. What is the horizontal asymptote of a function when the degree of the denominator is higher than the numerator?
4. Can a rational function cross its horizontal asymptote?
5. How do you find the oblique asymptote of a rational function when the degree of the numerator is greater than the denominator?

Tip: When the degrees of the numerator and denominator of a rational function are the same, always divide their leading coefficients