To analyze the function \( f(t) = 8(2)^t \) for \( t \) ranging from 0 to 6 (representing the first 6 days), we can calculate the values of \( f(t) \) for each day.
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For \( t = 0 \): \[ f(0) = 8(2)^0 = 8(1) = 8 \]
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For \( t = 1 \): \[ f(1) = 8(2)^1 = 8(2) = 16 \]
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For \( t = 2 \): \[ f(2) = 8(2)^2 = 8(4) = 32 \]
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For \( t = 3 \): \[ f(3) = 8(2)^3 = 8(8) = 64 \]
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For \( t = 4 \): \[ f(4) = 8(2)^4 = 8(16) = 128 \]
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For \( t = 5 \): \[ f(5) = 8(2)^5 = 8(32) = 256 \]
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For \( t = 6 \): \[ f(6) = 8(2)^6 = 8(64) = 512 \]
Now, we summarize the results:
- \( f(0) = 8 \)
- \( f(1) = 16 \)
- \( f(2) = 32 \)
- \( f(3) = 64 \)
- \( f(4) = 128 \)
- \( f(5) = 256 \)
- \( f(6) = 512 \)
Next, we observe that this function is an exponential growth function. The number of fruit flies doubles each day, starting from 8 flies on day 0.
When you plot the points \( (0, 8), (1, 16), (2, 32), (3, 64), (4, 128), (5, 256), (6, 512) \), you would expect to see a steep upward curve, illustrating exponential growth.
As I cannot directly provide or visualize graphs here, you should look for a graph that starts at 8, rises quickly, and follows the doubling pattern mentioned, effectively growing from the initial 8 fruit flies to 512 by the end of the 6th day.
If you are provided with multiple graphs to compare, select the one that demonstrates this exponential growth characteristic.