To find the horizontal asymptote for the given rational function $p(x) = \frac{4x^2 + 11x - 6}{2x^2 + 2x - 24}$, let's proceed step-by-step.

Horizontal Asymptote Rules:

1. If the degrees of the numerator and denominator are the same, the horizontal asymptote is $y = \frac{\text{leading coefficient of numerator}}{\text{leading coefficient of denominator}}$.
2. If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is $y = 0$.
3. If the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote (it may have an oblique asymptote instead).

Step-by-Step Solution:

The given function is:

$p(x) = \frac{4x^2 + 11x - 6}{2x^2 + 2x - 24}$

• The degree of the numerator (top) is 2 (since the highest power of $x$ is $x^2$).
• The degree of the denominator (bottom) is also 2.

Since both the numerator and the denominator have the same degree, the horizontal asymptote is given by:

$y = \frac{\text{leading coefficient of the numerator}}{\text{leading coefficient of the denominator}}$

The leading coefficient of the numerator is 4, and the leading coefficient of the denominator is 2. So:

$y = \frac{4}{2} = 2$

Final Answer:

The horizontal asymptote is:

$y = 2$

Would you like more details or have further questions?

Related Questions:

1. How do you find vertical asymptotes for rational functions?
2. What happens if the degree of the numerator is greater than the denominator?
3. Can a function have both horizontal and oblique asymptotes?
4. How do you determine the behavior of the function as $x \to \infty$ beyond the asymptotes?
5. What is the difference between horizontal asymptotes and slant (oblique) asymptotes?

Tip:

To confirm the presence of horizontal asymptotes, always compare the degrees of the numerator and the denominator first—it simplifies the process.