Question

Find the vertical asymptotes of \[ f(x)=\frac{-8}{x^2-4}. \] 

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

Hint:

To find vertical asymptotes, simplify the fraction completely, and then solve for where the denominator equals 0.
For \(x^2 - 4 = 0 \implies x^2 = 4 \implies x = \pm 2\).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
A vertical asymptote of a rational function \(f(x) = \frac{P(x)}{Q(x)}\) represents a vertical line \(x = c\) where the function values grow without bound (approaching \(\infty\) or \(-\infty\)) as \(x\) approaches \(c\).

Step 2: Key Formula or Approach:
For a rational function in its lowest terms, the vertical asymptotes occur at the real roots of the denominator, where:
\[ Q(x) = 0 \quad \text{and} \quad P(x) \neq 0 \]

Step 3: Detailed Explanation:
We are given the rational function:
\[ f(x) = \frac{-8}{x^2 - 4} \]
Here, the numerator is a constant polynomial:
\[ P(x) = -8 \]
Since \(P(x)\) is never zero, there is no possibility of a hole (removable discontinuity) in this function.
To find the vertical asymptotes, we set the denominator polynomial equal to zero:
\[ Q(x) = 0 \]
\[ x^2 - 4 = 0 \]
Factoring the difference of squares:
\[ (x - 2)(x + 2) = 0 \]
This quadratic equation has two real roots:
\[ x = 2 \quad \text{and} \quad x = -2 \]
As \(x\) approaches either 2 or \(-2\), the denominator approaches 0 while the numerator remains \(-8\).
Consequently, the absolute value of the function approaches infinity:
\[ \lim_{x \to 2} |f(x)| = \infty \]
\[ \lim_{x \to -2} |f(x)| = \infty \]
Thus, both \(x = 2\) and \(x = -2\) (or \(x = \pm 2\)) are vertical asymptotes of the function.
This matches Option (C).

Step 4: Final Answer:
Therefore, the correct option is (C).