To find the exponential function that fits the data provided, we will look for a function of the form:
\[ P(x) = a \cdot b^x \]
We have the following data points from the table:
- \( P(-2) = 2500 \)
- \( P(-1) = 500 \)
- \( P(0) = 100 \)
- \( P(1) = 20 \)
- \( P(2) = 4 \)
We'll use two of the points to create equations to solve for \( a \) and \( b \).
From \( P(0) = 100 \): \[ P(0) = a \cdot b^0 = a = 100 \]
Now substituting \( a = 100 \) into the other points:
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For \( P(-1) = 500 \): \[ 100 \cdot b^{-1} = 500 \implies b^{-1} = \frac{500}{100} = 5 \implies b = \frac{1}{5} \]
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For \( P(-2) = 2500 \): \[ 100 \cdot b^{-2} = 2500 \implies b^{-2} = \frac{2500}{100} = 25 \implies b^2 = \frac{1}{25} \implies b = \frac{1}{5} \]
Thus, we have \( a = 100 \) and \( b = \frac{1}{5} \).
Now, we can write the function: \[ P(x) = 100 \left( \frac{1}{5} \right)^x \]
To verify, we can calculate \( P(1) \) and \( P(2) \):
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For \( P(1) \): \[ P(1) = 100 \left( \frac{1}{5} \right)^1 = 100 \cdot \frac{1}{5} = 20 \]
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For \( P(2) \): \[ P(2) = 100 \left( \frac{1}{5} \right)^2 = 100 \cdot \frac{1}{25} = 4 \]
Both values are correct according to the data from the table. Thus, the exponential function that represents the relation shown in the table is:
\[ P(x) = 100 \left( \frac{1}{5} \right)^x \]
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