The equation provided is:

$\frac{x^2}{81} - \frac{y^2}{9} = 1$

This is the standard form of a hyperbola, specifically a horizontally-opening hyperbola. Here's how we interpret it:

1. Standard Form: The equation $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ represents a hyperbola centered at $(0, 0)$ with a horizontal transverse axis.

2. Key Features:

   • $a^2 = 81$ → $a = 9$: This means the distance from the center to the vertices along the x-axis is 9. So, the vertices are at $(-9, 0)$ and $(9, 0)$.
   • $b^2 = 9$ → $b = 3$: This determines the shape and asymptotes.

3. Asymptotes: The hyperbola’s asymptotes are given by the equation: $y = \pm \frac{b}{a}x = \pm \frac{3}{9}x = \pm \frac{1}{3}x$

4. Conclusion:

   • The graph corresponds to a hyperbola centered at the origin, opening horizontally with vertices at $(-9, 0)$ and $(9, 0)$, and asymptotes sloping at $\pm \frac{1}{3}$.

Let me know if you need detailed graph sketches or further clarifications. 😊

Would you like additional explanation or assistance?
Here are some related questions for practice:

1. What are the key differences between a hyperbola and an ellipse?
2. How would the graph change if the equation were $\frac{y^2}{81} - \frac{x^2}{9} = 1$?
3. What role do asymptotes play in sketching a hyperbola?
4. How do you find the eccentricity of this hyperbola?
5. How would you rewrite this equation if the hyperbola were centered at $(h, k)$?

Tip: Remember, hyperbolas have two branches, and their asymptotes guide the shape!