To determine the average rate of change of the function $h(x) = x^2 + 7x + 9$ over the interval $[-9, 0]$, we can use the formula for the average rate of change:

$\text{Average Rate of Change} = \frac{h(b) - h(a)}{b - a}$

where $a$ and $b$ are the endpoints of the interval. For this problem:

• $a = -9$
• $b = 0$

Step 1: Calculate $h(a) = h(-9)$

$h(-9) = (-9)^2 + 7(-9) + 9$ $h(-9) = 81 - 63 + 9 = 27$

Step 2: Calculate $h(b) = h(0)$

$h(0) = (0)^2 + 7(0) + 9$ $h(0) = 9$

Step 3: Compute the Average Rate of Change

$\text{Average Rate of Change} = \frac{h(0) - h(-9)}{0 - (-9)} = \frac{9 - 27}{9} = \frac{-18}{9} = -2$

Thus, the average rate of change of the function over the interval $[-9, 0]$ is $-2$.

Would you like more details, or do you have any questions?

Here are 5 related questions for further practice:

1. What is the average rate of change of the function $h(x) = x^2 + 7x + 9$ over the interval $[1, 4]$?
2. Determine the slope of the secant line between $x = -3$ and $x = 2$ for the function $h(x) = x^2 + 7x + 9$.
3. What is the average rate of change of the function $h(x) = 2x^2 - 3x + 5$ over the interval $[0, 3]$?
4. Find the instantaneous rate of change of $h(x) = x^2 + 7x + 9$ at $x = -9$.
5. How does the average rate of change of a linear function differ from that of a quadratic function?

Tip: The average rate of change for a quadratic function over an interval can be thought of as the slope of the secant line connecting the endpoints on the graph.