Question

which equation describes a rational function with \\(x\\)-intercepts at \\(-4\\) and \\(2\\), a vertical asymptote at \\(x = 1\\) and \\(x = -1\\), and a horizontal asymptote at \\(y = -3\\)?

\\(f(x) = \frac{-3(x - 4)(x + 2)}{x^2 - 1}\\)

\\(f(x) = \frac{x^2 - 1}{-3(x + 4)(x - 2)}\\)

\\(f(x) = \frac{x^2 - 1}{-3(x - 4)(x + 2)}\\)

\\(f(x) = \frac{-3(x + 4)(x - 2)}{x^2 - 1}\\)

Explanation:

Determine numerator factors from x-intercepts

$$ \text{x-intercepts at } x = -4 \text{ and } x = 2 \implies \text{Numerator contains } (x + 4)(x - 2) $$

Determine denominator factors from vertical asymptotes

$$ \text{Vertical asymptotes at } x = 1 \text{ and } x = -1 \implies \text{Denominator contains } (x - 1)(x + 1) = x^2 - 1 $$

Determine leading coefficient from horizontal asymptote

$$ \text{Horizontal asymptote at } y = -3 \implies \lim_{x \to \infty} f(x) = -3 \implies f(x) = \frac{-3(x + 4)(x - 2)}{x^2 - 1} $$

Answer:

• (A) \(f(x) = \frac{-3(x - 4)(x + 2)}{x^2 - 1}\)
• (B) \(f(x) = \frac{x^2 - 1}{-3(x + 4)(x - 2)}\)
• (C) \(f(x) = \frac{x^2 - 1}{-3(x - 4)(x + 2)}\)
• (D) \(f(x) = \frac{-3(x + 4)(x - 2)}{x^2 - 1}\) (Correct answer)