Given equation x^2-4y^2-6x-8y-31=0. (1) write the equation in standard form. (2) find the vertices and foci. (3) find equations of asymptotes. (4) sketch the graph, using the center, vertices and asymptotes.

Here is the graph of the hyperbola, with the following elements highlighted:

- **Center** at \((3, -1)\), marked in red.
- **Vertices** at \((9, -1)\) and \((-3, -1)\), marked in green.
- **Foci** at \((3 + 3\sqrt{5}, -1)\) and \((3 - 3\sqrt{5}, -1)\), marked in purple.
- **Asymptotes** represented by the dashed red lines, corresponding to the equations \(y + 1 = \frac{1}{2}(x - 3)\) and \(y + 1 = -\frac{1}{2}(x - 3)\).

### Summary of Findings:
1. **Standard Form**: \(\frac{(x - 3)^2}{36} - \frac{(y + 1)^2}{9} = 1\)
2. **Vertices**: \((9, -1)\) and \((-3, -1)\)
3. **Foci**: \((3 + 3\sqrt{5}, -1)\) and \((3 - 3\sqrt{5}, -1)\)
4. **Asymptotes**: \(y + 1 = \frac{1}{2}(x - 3)\) and \(y + 1 = -\frac{1}{2}(x - 3)\)

Would you like more details or have any questions?

Here are 5 related questions to expand your understanding:
1. How does the distance between the foci affect the shape of a hyperbola?
2. How would the equation of the hyperbola change if the transverse axis were vertical instead of horizontal?
3. How do the asymptotes help in sketching the graph of a hyperbola?
4. What are the coordinates of the co-vertices for this hyperbola?
5. How would you convert this hyperbola's equation into parametric form?

**Tip:** When identifying asymptotes of a hyperbola, the slopes are determined by the ratio \(\frac{b}{a}\), giving insight into the orientation and steepness of the asymptotes.

Learn how to graph a hyperbola given its equation in standard form. Understand how to find the vertices, foci, and equations of asymptotes. Includes a detailed solution with graphical representation. Suitable for Grades 11-12 students studying conic sections.

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