For $f(x)=\dfrac{x^2-2x-3}{x^2-4}$, the vertical and horizontal asymptotes are:

1. Vertical asymptotes occur where the denominator vanishes: $x^2-4=0$, so $x=\pm 2$. 2. Check the numerator does not vanish there — $x^2-2x-3=(x-3)(x+1)$ is non-zero at $x=\pm2$, so both are genuine asymptotes. 3. Numerator and denominator have equal degree, so the horizontal asymptote is the ratio of leading coefficients. 4. Both leading coefficients are $1$, giving $y=1$. 5. $x=\pm 4$ takes the constant $-4$ as the root instead of solving $x^2=4$; $y=0$ would need the numerator degree to be lower. _Source: NECTA ACSEE 2023 Advanced Mathematics 142/1, Question 6: Functions_