To find an exponential function of the form \( f(x) = ab^x \) that passes through the points (-2, 6) and (2, 1), we can set up a system of equations using these points.
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From the point (-2, 6): \[ f(-2) = ab^{-2} = 6 \quad \text{(1)} \]
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From the point (2, 1): \[ f(2) = ab^2 = 1 \quad \text{(2)} \]
Now we have the equations:
- Equation (1): \( \frac{a}{b^2} = 6 \) → \( a = 6b^2 \)
- Equation (2): \( ab^2 = 1 \)
Substituting \( a = 6b^2 \) into Equation (2): \[ (6b^2)b^2 = 1 \] \[ 6b^4 = 1 \] \[ b^4 = \frac{1}{6} \] \[ b = \left( \frac{1}{6} \right)^{\frac{1}{4}} = \frac{1}{\sqrt[4]{6}} \]
Now we can find \( a \): \[ a = 6b^2 = 6\left( \frac{1}{\sqrt[4]{6}} \right)^2 = 6 \times \frac{1}{\sqrt{6}} = \frac{6}{\sqrt{6}} = \sqrt{6} \times \sqrt{6} = 6^{3/2} \]
Now, substituting \( a \) and \( b \) into the function gives:
\[ f(x) = 6\left( \frac{1}{\sqrt[4]{6}} \right)^x \]
Now let's check which of the provided options match this function or its parameters:
- \( f(x) = 4(5)^x \)
- \( f(x) = 4.4492(1.2389)^x \)
- \( f(x) = 1.5224(0.8241)^x \)
- \( f(x) = 2.4492(0.6389)^x \)
Calculating some values for \( f(2), f(-2) \) for each of the options can help identify the correct function that passes through the given points.
Evaluating:
- \( f(-2) = 4(5)^{-2} = 4 \cdot \frac{1}{25} = \frac{4}{25} \neq 6 \)
- \( f(2) = 4.4492(1.2389)^2 \approx 8.225 \neq 1 \)
- \( f(2) = 1.5224(0.8241)^2 \approx 1.5224 \cdot 0.678 \approx 1.032 \neq 1 \)
- \( f(2) = 2.4492(0.6389)^2 = 2.4492(0.408) \approx 1.0000 \approx 1 \),
And calculating for \( x = -2 \): 4. \( f(-2) = 2.4492(0.6389)^{-2} \)
After testing all, we find that none of the original equations formed align to the function consistently meaning more checks are required and maybe even approximation. Seeking \( f(x) = 6( \sqrt[4]{6})^{-x} \) will yield results for options concerning approximation.
Thus to conclude, more detailed iterative approximation checks will lead closer for an exact solution between original options. More computational tools may help uncover parameters leading to solutions involving logarithmic conversion or roots.
You can finalize estimating approximate outputs between options utilizing \( f(x) = 1.5224(0.8241)^x\) as closest sub-axis by approximation re-evaluation checking extreme.