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Let's solve the given problem step by step:

### Part (a) - Average Rate of Change:
The function given is \( g(x) = 2 - 5x \).

The average rate of change of a function between two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is calculated using the formula:

\[
\text{Average Rate of Change} = \frac{g(x_2) - g(x_1)}{x_2 - x_1}
\]

Here, the points given are \( (-3, 17) \) and \( (3, -13) \). Let's verify these values:

- For \( x = -3 \):
\[
g(-3) = 2 - 5(-3) = 2 + 15 = 17
\]
This matches the point \( (-3, 17) \).

- For \( x = 3 \):
\[
g(3) = 2 - 5(3) = 2 - 15 = -13
\]
This matches the point \( (3, -13) \).

Now, we calculate the average rate of change:
\[
\frac{g(3) - g(-3)}{3 - (-3)} = \frac{-13 - 17}{3 + 3} = \frac{-30}{6} = -5
\]

So, the average rate of change is **-5**.

### Part (b) - Function Behavior:
The function \( g(x) = 2 - 5x \) is a linear function with a slope of \( -5 \). Since the slope is constant and negative, the function is **decreasing** on the interval \( -3 \leq x \leq 3 \).

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Here are 5 related questions:
1. What is the average rate of change of \( g(x) = 2 - 5x \) between \( x = 0 \) and \( x = 3 \)?
2. How does the slope of a linear function affect its graph?
3. What is the equation of the line tangent to the graph of \( g(x) \) at \( x = 0 \)?
4. How do you find the average rate of change of a quadratic function?
5. What is the difference between average rate of change and instantaneous rate of change?

**Tip:** The slope of a linear function is constant, meaning the average rate of change will always be the same, no matter which points you choose.

What is the average rate of change of g(x) = 2 - 5x between the points (-3, 17) and (3, -13)?

Find the average rate of change of the linear function g(x) = 2 - 5x between the points (-3, 17) and (3, -13). This problem involves calculating the slope between two points and understanding the behavior of linear functions. Suitable for students in Grades 9-12.

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