The given function is:

\[
f(x) = \frac{8x}{x^2 + 4}
\]

### Simplifying the Function

At first glance, the function is already in its simplest form. The numerator is \(8x\), and the denominator is \(x^2 + 4\). This is a rational function, and you can't simplify it any further.

### Domain of the Function

To find the domain of the function, we look for values of \(x\) that would make the denominator zero. In this case, \(x^2 + 4 = 0\) leads to:

\[
x^2 = -4
\]

Since there are no real numbers whose square is negative, the denominator is never zero for any real \(x\). Therefore, the domain of \(f(x)\) is all real numbers:

\[
\text{Domain: } (-\infty, \infty)
\]

### Asymptotes

- **Vertical Asymptotes**: Since the denominator \(x^2 + 4\) is never zero, there are no vertical asymptotes.
- **Horizontal Asymptote**: As \(x \to \infty\) or \(x \to -\infty\), the function behaves like \( \frac{8x}{x^2} = \frac{8}{x} \), which approaches 0. Therefore, the horizontal asymptote is \( y = 0 \).

### Behavior of the Function

- **At \(x = 0\)**: 

\[
f(0) = \frac{8(0)}{0^2 + 4} = 0
\]

- **As \(x \to \infty\) or \(x \to -\infty\)**: As mentioned, the function tends towards 0, suggesting a horizontal asymptote at \( y = 0 \).

Would you like further details or any specific clarification?

### Related Questions:
1. What are the critical points and extrema of the function \(f(x)\)?
2. How do you find the derivative \(f'(x)\) of the function?
3. Can you find the concavity of \(f(x)\) by calculating the second derivative \(f''(x)\)?
4. What is the range of the function \(f(x)\)?
5. How do you sketch the graph of \(f(x)\) with all key points?

**Tip:** When analyzing a rational function, always check for vertical and horizontal asymptotes to understand its long-term behavior.

Explore the properties of the rational function f(x) = 8x / (x^2 + 4), including its domain, asymptotes, and behavior at infinity. This detailed breakdown covers key mathematical concepts such as the function's domain, horizontal asymptote, and its behavior for Grades 10-12.

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