To determine which function best models the data provided in the table, we can analyze the given options and see which one fits the pattern of the y-values corresponding to the x-values.
The given table is:
\[ \begin{array}{|c|c|} \hline \text{x-Values} & \text{y-Values} \ \hline 0 & 5 \ 1 & 15 \ 2 & 45 \ 3 & 135 \ 4 & 405 \ \hline \end{array} \]
Upon quickly analyzing the changes in y-values as x increases, we notice that they are not linear. As x increases from 0 to 4, the y-values increase rapidly, suggesting an exponential or polynomial growth.
Let's evaluate the proposed functions one by one:
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f(x) = 2.23 × (1.40)^x; f(−1) = 1.59
- This option looks like an exponential function. We would compute some values for x to check if it matches the data, but based on the rapid growth observed in the original data, it seems too low.
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f(x) = 92x − 63; f(−1) = −155
- This appears to be a linear function. Plugging in x-values like 0, 1, 2, etc., we see it won't yield the correct y-values even before evaluating it at x=-1.
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f(x) = 5 × 3^x; f(−1) = 5/3
- A potential exponential function. The y-values seem to be multiples of each other at successive x-values, and plugging in x=0 gives f(0) = 5, which is correct. Continuing, f(1) = 15, f(2) = 45, f(3) = 135, and f(4) = 405, all match the table. Therefore, this function fits perfectly.
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f(x) = 41.43x² − 73.71x + 19.86; f(−1) = 135
- This appears to be a quadratic function. Evaluating it at x=0 gives f(0)=19.86, drastically different from the first data point, so it doesn't match.
The best model for the data is: f(x) = 5 × 3^x.
Now, we need to calculate f(−1):
\[ f(−1) = 5 × 3^{-1} = 5 × \frac{1}{3} = \frac{5}{3}. \]
So the answer is: f(x) = 5 × 3^x; f(−1) = \frac{5}{3}.