To find the intercepts of the equation:

$y = \frac{x^2 - 9}{x^2 - 4}$

1. x-intercept (Set $y = 0$):

• For $y = 0$, the numerator must be 0 (since a fraction is 0 only when the numerator is 0). $x^2 - 9 = 0$ This is a difference of squares: $(x - 3)(x + 3) = 0$ So, the solutions are: $x = 3 \quad \text{and} \quad x = -3$ Thus, the x-intercepts are $(3, 0)$ and $(-3, 0)$.

2. y-intercept (Set $x = 0$):

$y = \frac{0^2 - 9}{0^2 - 4} = \frac{-9}{-4} = \frac{9}{4}$ So, the y-intercept is: $\left(0, \frac{9}{4}\right)$

3. Exclusions from the Domain (Undefined Points):

The function is undefined where the denominator equals 0: $x^2 - 4 = 0 \quad \Rightarrow \quad (x - 2)(x + 2) = 0$ This gives: $x = 2 \quad \text{and} \quad x = -2$ Thus, the function has vertical asymptotes (not intercepts) at $x = 2$ and $x = -2$.

Answer:

The intercepts are: $(3, 0), (-3, 0), \left(0, \frac{9}{4}\right)$

Would you like a further breakdown of these calculations?

Here are 5 related questions for deeper exploration:

1. How do you determine if a point is a vertical asymptote or a hole in the graph?
2. How does a difference of squares factorization work?
3. What does the graph of this rational function look like near the asymptotes?
4. What are horizontal asymptotes, and does this function have one?
5. How do vertical and horizontal asymptotes affect the shape of the graph?

Tip: Always check for domain restrictions when solving rational equations to ensure you don't miss any undefined points!