For $f(x)=\dfrac{x^2-2x-3}{x^2-4}$, the $x$-intercepts and the $y$-intercept are:

1. $x$-intercepts need $f(x)=0$, so set the numerator to zero. 2. $x^2-2x-3=(x-3)(x+1)=0$ gives $x=3$ and $x=-1$. 3. The $y$-intercept is $f(0)=\dfrac{0-0-3}{0-4}=\dfrac{-3}{-4}=\dfrac34$. 4. Both negatives cancel, so the $y$-intercept is positive. 5. $-2$ and $2$ are the asymptotes, not intercepts; the sign-flipped pair mis-factorises the numerator. _Source: NECTA ACSEE 2023 Advanced Mathematics 142/1, Question 6: Functions_