Common Graphs — Question 6

Question 6 - Solution

We are given: $$f(x) = \frac{x^2 - 1}{x^2 - 4} = \frac{(x - 1)(x + 1)}{(x - 2)(x + 2)}$$

Domain: $x \neq \pm 2 \quad \Rightarrow \boxed{(-\infty, -2) \cup (-2, 2) \cup (2, \infty)}$

Vertical Asymptotes: $x = -2$ and $x = 2$

Horizontal Asymptote: Degrees of numerator and denominator are equal: $y = \frac{1}{1} = \boxed{y = 1}$

X-intercepts: Set numerator = 0: $x = \pm 1$

Y-intercept: $f(0) = \frac{-1}{-4} = \frac{1}{4}$