The function described is an exponential function of the form:

$f(x) = ab^x$

You are given the following conditions:

1. $f(1.5) = 17$
2. $f(5) = 11$

Step 1: Set up the exponential function using the given form

We know the general form of the exponential function is:

$f(x) = ab^x$

Substitute the two given values into this equation.

For $x = 1.5$, we have:

$f(1.5) = ab^{1.5} = 17$

For $x = 5$, we have:

$f(5) = ab^5 = 11$

Now, we have two equations:

1. $ab^{1.5} = 17$
2. $ab^5 = 11$

Step 2: Solve the system of equations

We need to solve for $a$ and $b$. Let's divide the second equation by the first:

$\frac{ab^5}{ab^{1.5}} = \frac{11}{17}$

This simplifies to:

$b^{5 - 1.5} = \frac{11}{17}$

$b^{3.5} = \frac{11}{17}$

Now solve for $b$:

$b = \left(\frac{11}{17}\right)^{\frac{1}{3.5}}$

Let’s compute this value:

$b \approx \left(0.647\right)^{\frac{1}{3.5}} \approx 0.925$

Now substitute $b \approx 0.925$ into the first equation $ab^{1.5} = 17$ to find $a$:

$a \times (0.925)^{1.5} = 17$

$a \times 0.887 \approx 17$

$a \approx \frac{17}{0.887} \approx 19.15$

Step 3: Find $f(8)$

Now that we know $a \approx 19.15$ and $b \approx 0.925$, we can calculate $f(8)$:

$f(8) = ab^8$

$f(8) = 19.15 \times (0.925)^8$

Let’s compute this:

$f(8) \approx 19.15 \times 0.515 \approx 9.86$

Final Answer:

The value of $f(8)$ to the nearest hundredth is approximately:

$\boxed{9.86}$

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Would you like further details or have any questions about the solution? Here are some related questions you might find interesting:

1. How can you solve a system of equations involving an exponential function?
2. What would happen if the given points had been different?
3. How do exponential functions behave for large values of $x$?
4. What is the general method for solving for constants in an exponential function?
5. Can we apply this method to other types of functions beyond exponentials?

Tip: When solving for parameters in an exponential function, dividing equations to eliminate one variable is a powerful tool!