The function you provided is:

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

To analyze this function, let's break it down:

1. Domain:
   The function is defined for all real values of $x$ because the denominator $x^2 + 4$ is never zero (since $x^2 + 4 \geq 4$ for all real $x$).

2. Simplification:
   The expression is already in its simplest form. There is no factor that cancels between the numerator and denominator.

3. Asymptotes:

   • Horizontal Asymptote: As $x \to \infty$, the highest degree term in the denominator $x^2$ dominates the function. So, $\lim_{x \to \infty} \frac{8x}{x^2 + 4} = \lim_{x \to \infty} \frac{8}{x} = 0$ Thus, the horizontal asymptote is $y = 0$.

   • Vertical Asymptote:
     Since the denominator $x^2 + 4$ is never zero, there are no vertical asymptotes.

4. Intercepts:

   • x-intercept: Set $f(x) = 0$ and solve for $x$: $\frac{8x}{x^2 + 4} = 0$ The numerator must be zero, so $8x = 0$ implies $x = 0$. Thus, the x-intercept is $(0, 0)$.

   • y-intercept: This is the same as the x-intercept since $f(0) = 0$.

5. Behavior at Infinity: As $x \to \infty$ or $x \to -\infty$, the function approaches the horizontal asymptote $y = 0$.

Summary:

• Domain: All real numbers $\mathbb{R}$
• Horizontal Asymptote: $y = 0$
• No Vertical Asymptote
• Intercept: $(0, 0)$

Would you like to go deeper into any part of this explanation? Or plot the graph?

Here are 5 related questions to extend this analysis:

1. What is the derivative of the function $f(x) = \frac{8x}{x^2 + 4}$?
2. How do we find the critical points of the function $f(x)$?
3. What is the second derivative of $f(x)$ and what does it tell us about concavity?
4. Can we analyze the behavior of this function as $x \to -\infty$?
5. How would we integrate $f(x) = \frac{8x}{x^2 + 4}$?

Tip: Always check the domain of a rational function by ensuring the denominator is never zero.