The given expression is:

$\frac{4x^2 + 3x}{12x^2 + 4}$

To find the horizontal asymptote, we need to analyze the behavior of the rational function as $x$ approaches infinity.

Steps:

1. Degree of numerator and denominator: The degree of the numerator $4x^2 + 3x$ is 2 (since the highest power of $x$ is $x^2$), and the degree of the denominator $12x^2 + 4$ is also 2.

2. Horizontal asymptote rule: For rational functions, if the degrees of the numerator and denominator are the same, the horizontal asymptote is given by the ratio of the leading coefficients of the numerator and the denominator.

   • The leading coefficient of the numerator is 4.
   • The leading coefficient of the denominator is 12.

   So, the horizontal asymptote is:

   $y = \frac{4}{12} = \frac{1}{3}$

Conclusion:

The horizontal asymptote is $y = \frac{1}{3}$.

Would you like further details on any part of this solution or have any other questions?

Here are 5 related questions for further practice:

1. How do you find vertical asymptotes in rational functions?
2. What happens to the asymptote if the degree of the numerator is greater than the degree of the denominator?
3. What are the conditions for having no horizontal asymptote?
4. Can a rational function have both a horizontal and oblique asymptote?
5. How does the presence of a constant in the numerator or denominator affect the asymptotes?

Tip: Always focus on the leading terms when finding horizontal asymptotes. The behavior of lower degree terms becomes negligible as $x \to \infty$.