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To solve for the average rate of change of the function \( f(x) \) from \( x = 0 \) to \( x = 5 \), we use the formula for the average rate of change between two points:

\[
\text{Average Rate of Change} = \frac{f(b) - f(a)}{b - a}
\]

In this case:
- \( a = 0 \), \( b = 5 \)
- \( f(0) \) and \( f(5) \) are the values of the function at \( x = 0 \) and \( x = 5 \) based on the graph.

From the graph:
- \( f(0) = 5 \) (The value of the function when \( x = 0 \))
- \( f(5) = 0 \) (The value of the function when \( x = 5 \))

Now, apply the values to the formula:

\[
\text{Average Rate of Change} = \frac{f(5) - f(0)}{5 - 0} = \frac{0 - 5}{5} = \frac{-5}{5} = -1
\]

So, the average rate of change of \( f \) from \( x = 0 \) to \( x = 5 \) is \( -1 \).

Would you like more details on this solution?

Here are 5 related questions to explore further:
1. What is the geometric interpretation of the average rate of change?
2. How do you find the average rate of change from a table of values?
3. Can the average rate of change be positive, and what would that signify?
4. How is the average rate of change different from instantaneous rate of change?
5. How does the average rate of change relate to the slope of a secant line?

**Tip:** The average rate of change between two points on a graph can be thought of as the slope of the straight line connecting those two points.

Use the graph of f(x) to evaluate the following: The average rate of change of f from x = 0 to x = 5.

Learn how to calculate the average rate of change of the function f(x) from x = 0 to x = 5 using the graph provided. This example involves key algebraic concepts such as the slope of a secant line and is suitable for students in Grades 8-10.

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