To model the given data using an exponential function of the form:
$f(x) = A \cdot b^x$ we need to determine the parameters $A$ and $b$ based on the two data points:

• $f(1) = -59$
• $f(2) = -87.025$

Step 1: Write two equations

Using the general form $f(x) = A \cdot b^x$ for the given points:

1. For $x = 1$: $-59 = A \cdot b^1 \quad \Rightarrow \quad -59 = A \cdot b$

2. For $x = 2$: $-87.025 = A \cdot b^2$

Step 2: Eliminate $A$

From the first equation, solve for $A$: $A = \frac{-59}{b}$

Step 3: Substitute into the second equation

Plug this value of $A$ into the second equation: $-87.025 = \left(\frac{-59}{b}\right) \cdot b^2$

Simplify: $-87.025 = -59 \cdot b$

Solve for $b$: $b = \frac{-87.025}{-59} = 1.475$

Step 4: Find $A$

Now that we know $b = 1.475$, substitute this value back into the first equation to find $A$: $-59 = A \cdot 1.475 \quad \Rightarrow \quad A = \frac{-59}{1.475} \approx -40$

Step 5: Write the Model

Now that we have $A \approx -40$ and $b \approx 1.475$, the exponential function becomes: $f(x) = -40 \cdot (1.475)^x$

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Do you want further clarification or additional steps?

Here are five related questions:

1. How do we validate if this model fits the given data well?
2. Can exponential functions model negative values reliably?
3. What happens if we shift the base $b$?
4. How do logarithms help solve for unknowns in exponential models?
5. How does changing $A$ affect the graph of the function?

Tip: Exponential models are often sensitive to small changes in data—small errors in parameter estimation can lead to noticeable deviations.