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To solve the problem given in the image, we need to write down the equation of the hyperbola that models the depth of the river over time. The form of the hyperbola is given by:

\[ d = \frac{a}{t - b} + c \]

where \( d \) is the depth in meters, \( t \) is the time in days since the rain started, and \( a \), \( b \), and \( c \) are constants to be determined.

Given data points:
- At \( t = 0 \) (when the rain started), the depth \( d = 3.2 \) meters.
- The maximum depth of 4 meters occurred at \( t = 1 \) day.
- By \( t = 3 \) days, the depth had dropped to 3 meters.

Using these points, we will solve for \( a \), \( b \), and \( c \).

### Step-by-Step Solution:

1. **Initial Condition:**
   At \( t = 0 \), \( d = 3.2 \):
   \[ 3.2 = \frac{a}{0 - b} + c \]

2. **Maximum Depth:**
   At \( t = 1 \), \( d = 4 \):
   \[ 4 = \frac{a}{1 - b} + c \]

3. **Depth After 3 Days:**
   At \( t = 3 \), \( d = 3 \):
   \[ 3 = \frac{a}{3 - b} + c \]

From these equations, we need to solve for \( a \), \( b \), and \( c \).

#### Equation 1:
\[ 3.2 = \frac{a}{-b} + c \]

#### Equation 2:
\[ 4 = \frac{a}{1 - b} + c \]

#### Equation 3:
\[ 3 = \frac{a}{3 - b} + c \]

Subtract Equation 1 from Equation 2 to eliminate \( c \):
\[ 4 - 3.2 = \frac{a}{1 - b} - \frac{a}{-b} \]
\[ 0.8 = a \left( \frac{1}{1 - b} + \frac{1}{b} \right) \]
\[ 0.8 = a \left( \frac{b + (1 - b)}{b(1 - b)} \right) \]
\[ 0.8 = a \left( \frac{1}{b(1 - b)} \right) \]

From here, solve for \( a \):
\[ a = 0.8b(1 - b) \]

Next, subtract Equation 2 from Equation 3 to eliminate \( c \):
\[ 4 - 3 = \frac{a}{1 - b} - \frac{a}{3 - b} \]
\[ 1 = a \left( \frac{1}{1 - b} - \frac{1}{3 - b} \right) \]
\[ 1 = a \left( \frac{(3 - b) - (1 - b)}{(1 - b)(3 - b)} \right) \]
\[ 1 = a \left( \frac{2}{(1 - b)(3 - b)} \right) \]
\[ a = \frac{(1 - b)(3 - b)}{2} \]

Equate the two expressions for \( a \):
\[ 0.8b(1 - b) = \frac{(1 - b)(3 - b)}{2} \]

Divide both sides by \( 1 - b \):
\[ 0.8b = \frac{3 - b}{2} \]
\[ 1.6b = 3 - b \]
\[ 2.6b = 3 \]
\[ b = \frac{3}{2.6} \]
\[ b \approx 1.154 \]

Now substitute \( b \) back into the equation for \( a \):
\[ a = 0.8 \cdot 1.154 \cdot (1 - 1.154) \]
\[ a = 0.8 \cdot 1.154 \cdot (-0.154) \]
\[ a \approx -0.142 \]

Finally, substitute \( a \) and \( b \) back into one of the original equations to solve for \( c \):
\[ 3.2 = \frac{-0.142}{-1.154} + c \]
\[ 3.2 = 0.123 + c \]
\[ c = 3.2 - 0.123 \]
\[ c \approx 3.077 \]

Thus, the equation of the hyperbola is:
\[ d = \frac{-0.142}{t - 1.154} + 3.077 \]

If you have any further questions or need more details, feel free to ask!

Here are 5 numbered questions you might consider asking:

1. Can you explain how hyperbolas are generally used in real-life applications?
2. What are the key differences between hyperbolas and other conic sections like ellipses and parabolas?
3. How do you solve for the intersection points of two hyperbolas?
4. What are some common methods for finding the asymptotes of a hyperbola?
5. Can you show how to derive the standard form of a hyperbola from its geometric definition?

**Tip:** When solving for constants in an equation, always check your work by substituting back into the original equations to ensure consistency and correctness.

Learn how to model depth variation over time using a hyperbola equation. Step-by-step solution provided with specific data points: initial depth, maximum depth, and depth after a certain time period. Suitable for advanced high school students studying advanced algebra and curve fitting.

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