To find the horizontal asymptotes of the function

$f(x) = \frac{16x+1}{\sqrt{4x^2 + 9}},$

we need to analyze the behavior of $f(x)$ as $x \to \infty$ and $x \to -\infty$.

Step 1: Examine the behavior as $x \to \infty$

When $x \to \infty$, the dominant terms in the numerator and denominator will control the behavior of the function. Let's analyze each part:

• The numerator $16x + 1$ behaves like $16x$ for large values of $x$.
• The denominator $\sqrt{4x^2 + 9}$ behaves like $\sqrt{4x^2} = 2x$ for large values of $x$, because the constant term $9$ becomes negligible compared to $4x^2$.

Thus, for large $x$, $f(x) \approx \frac{16x}{2x} = 8.$ Therefore, as $x \to \infty$, $f(x) \to 8$.

Step 2: Examine the behavior as $x \to -\infty$

Similarly, as $x \to -\infty$:

• The numerator $16x + 1$ behaves like $16x$, which approaches a large negative value.
• The denominator $\sqrt{4x^2 + 9}$ behaves like $\sqrt{4x^2} = 2|x|$. For negative values of $x$, this simplifies to $2(-x) = -2x$.

Thus, for large negative $x$, $f(x) \approx \frac{16x}{-2x} = -8.$ Therefore, as $x \to -\infty$, $f(x) \to -8$.

Conclusion:

The function has two horizontal asymptotes:

$y = 8 \quad \text{and} \quad y = -8.$

Final Answer:

The horizontal asymptotes are $y = 8$ and $y = -8$.

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Would you like further details on how asymptotes are derived or have any specific questions?

Here are 5 related questions to deepen your understanding:

1. How do vertical asymptotes differ from horizontal asymptotes?
2. Can a function have more than two horizontal asymptotes?
3. How do slant asymptotes form for rational functions?
4. How do we determine the behavior of a function near an asymptote?
5. How would you calculate the limits of more complex functions with roots?

Tip: The horizontal asymptotes represent the end behavior of the function, showing how it behaves as $x$ grows very large or very small.