Hence, the function has a zero at x=2, there is a hole in the graph at x=−2, the domain is \( (−∞,−2)∪(−2,4)∪(4,+∞) \nonumber\), and the y-intercept is at (0,12). \[ \lim_{x \to ∞} \frac{x^2−4}{x^2−2x...Hence, the function has a zero at x=2, there is a hole in the graph at x=−2, the domain is \( (−∞,−2)∪(−2,4)∪(4,+∞) \nonumber\), and the y-intercept is at (0,12). \[ \lim_{x \to ∞} \frac{x^2−4}{x^2−2x−8}= \lim_{x \to ∞} \frac{ \frac{x^2}{x^2}−\frac{4}{x^2}} {\frac{x^2}{x^2}−\frac{2x}{x^2}−\frac{8}{x^2}}=\lim_{x \to ∞} \frac{1−\frac{4}{x^2}}{1−\frac{2}{x}−\frac{8}{x^2}}=1 \nonumber\]. The function can be factored \( f(x)=x^3+2x^2−x−2=x^2(x+2)−1(x+2)=(x^2−1)(x+2)=(x−1)(x+1)(x+2) \nonumber\)
