https://file.aiquickdraw.com/mathbot/file/1725605399311i0fxc1o2.jpg

The image shows two hyperbolas with equations:

1) \(\frac{(x - 1)^2}{16} - \frac{y^2}{25} = 1\)
2) \(\frac{x^2}{16} - \frac{(y + 1)^2}{25} = 1\)

We are tasked with identifying the **center**, **vertices**, **co-vertices**, and **foci** for each hyperbola, and sketching their graphs.

### 1) Equation: \(\frac{(x - 1)^2}{16} - \frac{y^2}{25} = 1\)
This is the standard form of a horizontal hyperbola:
\[
\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1
\]
- **Center**: \((h, k)\) is \((1, 0)\)
- **Vertices**: Since \(a^2 = 16\), \(a = 4\). The vertices are located at \((h \pm a, k)\), which are \((1 - 4, 0) = (-3, 0)\) and \((1 + 4, 0) = (5, 0)\).
- **Co-vertices**: Since \(b^2 = 25\), \(b = 5\). The co-vertices are at \((h, k \pm b)\), or \((1, 0 \pm 5)\), which gives \((1, 5)\) and \((1, -5)\).
- **Foci**: The foci are found using \(c^2 = a^2 + b^2\), so \(c^2 = 16 + 25 = 41\), and \(c = \sqrt{41} \approx 6.4\). The foci are at \((h \pm c, k)\), or approximately \((1 - 6.4, 0) = (-5.4, 0)\) and \((1 + 6.4, 0) = (7.4, 0)\).

### 2) Equation: \(\frac{x^2}{16} - \frac{(y + 1)^2}{25} = 1\)
This is the standard form of a vertical hyperbola:
\[
\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1
\]
- **Center**: \((h, k)\) is \((0, -1)\)
- **Vertices**: Since \(b^2 = 25\), \(b = 5\). The vertices are located at \((h, k \pm b)\), which are \((0, -1 - 5) = (0, -6)\) and \((0, -1 + 5) = (0, 4)\).
- **Co-vertices**: Since \(a^2 = 16\), \(a = 4\). The co-vertices are at \((h \pm a, k)\), or \((0 \pm 4, -1)\), giving \((4, -1)\) and \((-4, -1)\).
- **Foci**: Using \(c^2 = a^2 + b^2\), we get \(c^2 = 16 + 25 = 41\), and \(c = \sqrt{41} \approx 6.4\). The foci are at \((h, k \pm c)\), or approximately \((0, -1 - 6.4) = (0, -7.4)\) and \((0, -1 + 6.4) = (0, 5.4)\).

These are the characteristics for both hyperbolas. Would you like a more detailed explanation or clarification on any of these points?

Here are 5 related questions you might find helpful:
1. How do you graph hyperbolas step by step?
2. What is the difference between horizontal and vertical hyperbolas?
3. How do you find the asymptotes of a hyperbola?
4. How do foci relate to the shape of a hyperbola?
5. How does changing \(a^2\) and \(b^2\) affect the graph of the hyperbola?

**Tip**: For hyperbolas, the relationship \(c^2 = a^2 + b^2\) helps to find the foci, which are crucial in understanding the hyperbola's spread and orientation.

Identify the center, vertices, co-vertices, and foci of the hyperbolas. Then sketch the graph.
1) (x - 1)^2 / 16 - y^2 / 25 = 1
2) x^2 / 16 - (y + 1)^2 / 25 = 1

Learn how to identify the center, vertices, co-vertices, and foci of two hyperbolas: (x - 1)^2 / 16 - y^2 / 25 = 1 and x^2 / 16 - (y + 1)^2 / 25 = 1. This solution walks you through using the standard form of hyperbolas and finding key points.

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