To find the function that best models the given data and to determine the value of the function at \(x = -1\), let’s first analyze the data points:
| x | y | |---|----| | 0 | 5 | | 1 | 15 | | 2 | 45 | | 3 | 135| | 4 | 405|
It appears that the y-values are increasing with the x-values in a nonlinear manner. Let's analyze the ratios of \(y\) values to see if there is a pattern:
- \(y(1)/y(0) = 15/5 = 3\)
- \(y(2)/y(1) = 45/15 = 3\)
- \(y(3)/y(2) = 135/45 = 3\)
- \(y(4)/y(3) = 405/135 = 3\)
This suggests that each y-value is obtained by multiplying the previous y-value by 3, which indicates an exponential relation. To further assess this pattern, let’s express the \(y\) values in some defined form. The pattern resembles \(y = 5 \cdot 3^x\):
- At \(x = 0\): \(y = 5 \cdot 3^0 = 5\)
- At \(x = 1\): \(y = 5 \cdot 3^1 = 15\)
- At \(x = 2\): \(y = 5 \cdot 3^2 = 45\)
- At \(x = 3\): \(y = 5 \cdot 3^3 = 135\)
- At \(x = 4\): \(y = 5 \cdot 3^4 = 405\)
Thus, the correct function that models the data points is: \[ f(x) = 5 \cdot 3^x \]
Next, we need to evaluate \( f(-1) \): \[ f(-1) = 5 \cdot 3^{-1} = 5 \cdot \frac{1}{3} = \frac{5}{3} \approx 1.67 \]
The functions listed don't seem to match \(f(x) = 5 \cdot 3^x\) directly, but \(f(x) = 92\) is a constant value, \(f(x) = 41.43x\) is linear, and \(f(x) = 2.23\) is also a constant. None of these represent the exponential model we derived.
Thus, the value of the function at \(x = -1\) is: \[ f(-1) \approx 1.67 \]