To model the given data using an exponential function of the form $f(x) = Ab^x$, we have the following points:

$(1, -59) \quad \text{and} \quad (2, -87.025)$

Step 1: Set up the equations

Using the given points, we can create two equations based on the general form of the exponential function:

1. For $x = 1$, $f(1) = -59$:

$f(1) = A b^1 = A b$ $A b = -59 \quad \text{(Equation 1)}$

2. For $x = 2$, $f(2) = -87.025$:

$f(2) = A b^2 = -87.025$ $A b^2 = -87.025 \quad \text{(Equation 2)}$

Step 2: Solve the system of equations

We have the following system of equations:

1. $A b = -59$
2. $A b^2 = -87.025$

We can solve this system step by step.

Divide Equation 2 by Equation 1 to eliminate $A$:

$\frac{A b^2}{A b} = \frac{-87.025}{-59}$ $b = \frac{-87.025}{-59} = 1.474$

Now substitute $b = 1.474$ back into Equation 1:

$A \cdot 1.474 = -59$ $A = \frac{-59}{1.474} \approx -40.03$

Step 3: Write the final model

Now that we have $A \approx -40.03$ and $b \approx 1.474$, the exponential function is:

$f(x) = -40.03 \cdot (1.474)^x$

Conclusion

The model for the given data is:

$f(x) = -40.03 \cdot (1.474)^x$

Would you like further details or explanations on any part of this solution?

Here are five related questions:

1. How do you interpret the constants $A$ and $b$ in the exponential model?
2. How can we estimate the value of $f(x)$ for other $x$-values using this model?
3. How would the model change if the data showed a decay instead of growth?
4. Can this model be used to predict negative values for $x$? Why or why not?
5. What is the significance of the base $b$ being greater than 1?

Tip: Exponential growth or decay can be easily modeled by the form $f(x) = A b^x$, but be cautious when the base $b$ is negative—it typically doesn't represent standard growth or decay.