What is the average rate of change of the function h(x)=6–3x2 on the interval from x=
–
2 to x=1?


To find the **average rate of change** of a function \( h(x) \) over a given interval \([a, b]\), we use the formula:

\[
\text{Average rate of change} = \frac{h(b) - h(a)}{b - a}
\]

For the function \( h(x) = 6 - 3x^2 \), and the interval from \( x = -2 \) to \( x = 1 \):

1. **Evaluate** \( h(x) \) at the endpoints:
   \[
   h(-2) = 6 - 3(-2)^2 = 6 - 3(4) = 6 - 12 = -6
   \]
   \[
   h(1) = 6 - 3(1)^2 = 6 - 3(1) = 6 - 3 = 3
   \]

2. **Calculate the average rate of change**:
   \[
   \text{Average rate of change} = \frac{h(1) - h(-2)}{1 - (-2)} = \frac{3 - (-6)}{1 + 2} = \frac{3 + 6}{3} = \frac{9}{3} = 3
   \]

Thus, the average rate of change of the function on the interval \([-2, 1]\) is **3**.

Would you like more details or have any questions? Here are some related questions to consider:

1. How does the average rate of change relate to the slope of a secant line?
2. What would the average rate of change be on a different interval, say \([0, 2]\)?
3. How can we interpret the average rate of change for a quadratic function?
4. What is the instantaneous rate of change of \( h(x) \) at \( x = 0 \)?
5. How does the symmetry of a quadratic function affect the average rate of change?

**Tip:** The average rate of change over an interval is equivalent to the slope of the secant line connecting the points on the graph at the interval's endpoints.

What is the average rate of change of the function h(x)=6–3x^2 on the interval from x=–2 to x=1?

Explore how to calculate the average rate of change for the quadratic function h(x) = 6 - 3x^2 on the interval [-2, 1]. This solution demonstrates key algebraic concepts, including evaluating quadratic functions and applying the average rate of change formula. Suitable for students in Grades 9-11.

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