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 the numerator from the x-intercepts

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

Determine the denominator from the vertical asymptotes

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

Determine the leading coefficient from the 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)