To determine the vertical and horizontal asymptotes of the given function:

$y = \frac{x - 4}{\sqrt{4x^2 + 3}}$

1. Vertical Asymptotes

A vertical asymptote occurs when the denominator equals zero while the numerator is not zero. In this case, the denominator is:

$\sqrt{4x^2 + 3}.$

• To have a vertical asymptote, the expression inside the square root, $4x^2 + 3$, must be zero.
• However, $4x^2 + 3 > 0$ for all real values of $x$ because $4x^2$ is always positive or zero, and adding 3 makes the expression strictly positive.

Conclusion: There are no vertical asymptotes for this function.

2. Horizontal Asymptotes

To find the horizontal asymptotes, evaluate the behavior of $y$ as $x \to \infty$ and $x \to -\infty$. Analyze the dominant terms in the numerator and denominator for large $|x|$:

• Numerator: $x - 4 \sim x$ (for large $x$)
• Denominator: $\sqrt{4x^2 + 3} \sim \sqrt{4x^2} = 2|x|$.

Thus, for large $|x|$, the function behaves approximately as:

$y \approx \frac{x}{2|x|}.$

Simplify for $x > 0$ and $x < 0$:

1. As $x \to \infty$ (positive $x$):

   $y \approx \frac{x}{2x} = \frac{1}{2}.$

2. As $x \to -\infty$ (negative $x$):

   $y \approx \frac{x}{2(-x)} = -\frac{1}{2}.$

Conclusion: The function has two horizontal asymptotes:

$y = \frac{1}{2} \quad \text{and} \quad y = -\frac{1}{2}.$

Final Answer:

• Vertical Asymptotes: None.
• Horizontal Asymptotes: $y = \frac{1}{2}$ and $y = -\frac{1}{2}$.

Do you need more explanation on any part of this process?

Related Questions:

1. How do you find vertical asymptotes for rational functions?
2. What are oblique asymptotes, and when do they occur?
3. How do you determine horizontal asymptotes for functions with radicals?
4. What happens if the degrees of the numerator and denominator are equal?
5. How do asymptotes relate to the end behavior of a graph?

Tip:

When analyzing a function's behavior at infinity, compare the dominant terms in the numerator and denominator to simplify the limits.