Question

the amount of water, \\(g(x)\\), in liters, in a large tank after \\(x\\) days is modeled in the table.

part a: is the amount of water in the tank increasing or decreasing, and what are the values of \\(a\\) and \\(b\\), if \\(g(x) = ab^x\\)? show all necessary steps. (6 points)

part b: what is the percent rate of change? explain. (6 points)

part c: write an equation to represent \\(g(x)\\). identify the parts of the equation. (6 points)

Explanation:

Determine if the amount of water is increasing or decreasing

Using the Exponential Decay Models knowledge point
The values of \(g(x)\) represent the amount of water in liters:

• At \(x = 0\), \(g(0) = 15,000\)
• At \(x = 1\), \(g(1) = 12,300\)
• At \(x = 2\), \(g(2) = 10,086\)
• At \(x = 3\), \(g(3) = 8,270.52\)

Since \(15,000 > 12,300 > 10,086 > 8,270.52\), the amount of water is decreasing.

Find the values of a and b

Using the Exponential Decay Models knowledge point
The model is given by \(g(x) = ab^x\).

• For \(x = 0\):

$$ g(0) = a \cdot b^0 = a \implies a = 15,000 $$

• For \(x = 1\):

$$ g(1) = a \cdot b^1 = 15,000 \cdot b = 12,300 \implies b = \frac{12,300}{15,000} = 0.82 $$

Calculate the percent rate of change

Using the Percent Rate of Change knowledge point
The decay factor is \(b = 0.82\).
The rate of decay \(r\) is calculated as:

$$ r = 1 - b = 1 - 0.82 = 0.18 $$

Converting to a percentage:

$$ 0.18 \times 100\% = 18\% $$

This means the amount of water decreases by \(18\%\) each day.

Write and identify parts of the equation

Using the Exponential Decay Models knowledge point
The equation representing \(g(x)\) is:

$$ g(x) = 15,000(0.82)^x $$

Identifying the parts of the equation:

• \(g(x)\): The remaining amount of water in liters after \(x\) days.
• \(15,000\): The initial amount of water in liters at day \(0\).
• \(0.82\): The decay factor, representing the fraction of water remaining each day.
• \(x\): The number of days elapsed.

Answer:

Part A

• Behavior: The amount of water is decreasing because the values of \(g(x)\) decrease as \(x\) increases.
• Value of \(a\): \(a = 15,000\) (the initial value when \(x = 0\)).
• Value of \(b\): \(b = 0.82\) (calculated from \(\frac{g(1)}{g(0)} = \frac{12,300}{15,000}\)).

Part B

• Percent Rate of Change: \(18\%\) decrease per day.
• Explanation: Since the decay factor \(b = 0.82\), the rate of decay is \(r = 1 - 0.82 = 0.18\), which corresponds to an \(18\%\) daily reduction.

Part C

• Equation: \(g(x) = 15,000(0.82)^x\)
• Parts of the Equation:
• \(g(x)\): The amount of water remaining in the tank after \(x\) days.
• \(15,000\): The initial amount of water in liters (when \(x = 0\)).
• \(0.82\): The decay factor (representing that \(82\%\) of the water remains each day).
• \(x\): The time in days.