To find the exponential regression function of the form $f(x) = ab^x$ through the points $(3, 3)$, $(4, 4)$, and $(5, 12)$, we'll follow these steps:

1. Transform the exponential model into a linear one: $f(x) = ab^x$ Taking the natural logarithm (ln) of both sides: $\ln(f(x)) = \ln(a) + x \ln(b)$ This transforms the model into a linear equation of the form $y = mx + c$, where: $\text{ln}(f(x)) = \ln(a) + x \ln(b)$ Here, $\ln(f(x))$ plays the role of $y$, $x$ remains the same, $\ln(b)$ is the slope, and $\ln(a)$ is the intercept.

2. Prepare the data for regression: Take the natural log of the $y$-values of each point:

   • For $(3, 3)$: $\ln(3) \approx 1.0986$
   • For $(4, 4)$: $\ln(4) \approx 1.3863$
   • For $(5, 12)$: $\ln(12) \approx 2.4849$

   Now, we have the transformed points: $(3, 1.0986), (4, 1.3863), (5, 2.4849)$

3. Find the linear regression line: Using technology (such as a calculator, Excel, or software), we can find the equation of the line $\ln(f(x)) = mx + c$ through the points $(3, 1.0986), (4, 1.3863), (5, 2.4849)$.

   The result is: $\ln(f(x)) = 0.6931x - 1.9006$

4. Convert back to the exponential form: Now, exponentiate both sides to return to the exponential model: $f(x) = e^{0.6931x - 1.9006}$ This simplifies to: $f(x) = e^{-1.9006} \cdot e^{0.6931x}$ Approximating the constants: $f(x) \approx 0.1499 \cdot 2^x$

Thus, the exponential regression function is: $f(x) = 0.1499 \cdot 2^x$

Would you like more details, or do you have any questions?

Here are 5 related questions to explore further:

1. How does the method change if you have more than three points?
2. How would you approach this if the function was logarithmic instead of exponential?
3. Can you solve this regression problem by hand without technology?
4. What is the difference between exponential growth and decay in such models?
5. How do different values of $a$ and $b$ affect the shape of an exponential function?

Tip: When dealing with exponential models, taking the natural log is a powerful tool to simplify calculations by transforming them into linear form.