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To find the horizontal asymptote of the function \( f(x) = \frac{(3x+8)(5-4x)}{(2x+1)^2} \), we need to examine the behavior of \( f(x) \) as \( x \to \infty \) or \( x \to -\infty \).

### Steps to find the horizontal asymptote:

1. **Degree of the numerator**:  
   Expand the numerator, \( (3x+8)(5-4x) \):
   \[
   (3x+8)(5-4x) = 3x(5) + 3x(-4x) + 8(5) + 8(-4x)
   = 15x - 12x^2 + 40 - 32x.
   \]
   Simplifying, we get:
   \[
   -12x^2 - 17x + 40.
   \]
   The degree of the numerator is 2, and its leading term is \( -12x^2 \).

2. **Degree of the denominator**:  
   Expand the denominator, \( (2x+1)^2 \):
   \[
   (2x+1)^2

Let f be the function defined by f(x) = (3x+8)(5-4x) / (2x+1)^2. Which of the following is a horizontal asymptote to the graph of f?

Determine the horizontal asymptote of the function f(x) = (3x+8)(5-4x) / (2x+1)^2 by analyzing the degrees of the numerator and denominator. Suitable for students in Grades 10-12.

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