Question

The horizontal asymptote for \(y = \frac{-8}{x^2-4}\) is given by

• A. \(y = 0\)
• B. \(y = 8\)
• C. \(y = 4\)
• D. \(y = 2\)

Hint:

For any rational function \( y = \frac{P(x)}{Q(x)} \):
- If the degree of the numerator is strictly less than the degree of the denominator, the horizontal asymptote is always the line:
\[ y = 0 \] Here, the degree of the numerator is 0 and the degree of the denominator is 2, so the asymptote is \( y = 0 \).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This problem requires finding the horizontal asymptote of a rational function.
A horizontal asymptote represents the value that the function's output (\( y \)) approaches as the independent variable \( x \) grows infinitely large in either the positive or negative direction (\( x \to \pm\infty \)).

Step 2: Detailed Explanation:
Let us find the limit of the given rational function as \( x \to \infty \):
\[ y = \frac{-8}{x^2 - 4} \] To find the horizontal asymptote, we compute:
\[ L = \lim_{x \to \infty} \frac{-8}{x^2 - 4} \] - Let us divide both the numerator and the denominator by the highest power of \( x \) present in the denominator (which is \( x^2 \)):
\[ L = \lim_{x \to \infty} \frac{\frac{-8}{x^2}}{\frac{x^2}{x^2} - \frac{4}{x^2}} \] \[ L = \lim_{x \to \infty} \frac{-\frac{8}{x^2}}{1 - \frac{4}{x^2}} \] - As \( x \to \infty \), both fractional terms containing \( x \) in their denominators approach zero:
\[ \lim_{x \to \infty} \frac{8}{x^2} = 0 \quad \text{and} \quad \lim_{x \to \infty} \frac{4}{x^2} = 0 \] - Substitute these limits into our expression:
\[ L = \frac{0}{1 - 0} = \frac{0}{1} = 0 \] The same result is obtained as \( x \to -\infty \).
Since the limit of the function as \( x \to \pm\infty \) is \( 0 \), the line \( y = 0 \) (which is the x-axis) is the horizontal asymptote of the rational function.

Step 3: Final Answer:
The horizontal asymptote is \( y = 0 \).
Therefore, the correct choice is Option (A).