The problem asks for the average rate of change of the function $y = -x^2 - 8x - 6$ over the interval $[2, 2+h]$.

The formula for the average rate of change of a function $f(x)$ over an interval $[a, b]$ is given by:

$\text{Average rate of change} = \frac{f(b) - f(a)}{b - a}$

Step-by-Step Solution

1. Function given: $f(x) = -x^2 - 8x - 6$

2. Interval: $[2, 2+h]$, so $a = 2$ and $b = 2 + h$.

3. Find $f(2)$: $f(2) = -(2)^2 - 8(2) - 6 = -4 - 16 - 6 = -26$

4. Find $f(2+h)$: $f(2+h) = -(2+h)^2 - 8(2+h) - 6$ Expanding: $-(2+h)^2 = -(4 + 4h + h^2) = -4 - 4h - h^2$ and $-8(2+h) = -16 - 8h$ Therefore: $f(2+h) = -4 - 4h - h^2 - 16 - 8h - 6 = -26 - 12h - h^2$

5. Apply the average rate of change formula: $\frac{f(2+h) - f(2)}{(2+h) - 2} = \frac{(-26 - 12h - h^2) - (-26)}{h}$ Simplifying the numerator: $(-26 - 12h - h^2) - (-26) = -12h - h^2$ Thus, the expression becomes: $\frac{-12h - h^2}{h}$ Simplify by dividing by $h$: $\frac{-12h - h^2}{h} = -12 - h$

Final Answer:

The average rate of change is $\boxed{-12 - h}$.

Would you like further details or have any questions?

Here are some related questions:

1. How do you interpret the meaning of the average rate of change geometrically?
2. Can the average rate of change ever equal zero? If so, under what conditions?
3. What is the difference between average rate of change and instantaneous rate of change?
4. How would the result change if the interval were [3, 3+h] instead?
5. What does the parameter $h$ represent in this context?

Tip: The average rate of change over an interval is the slope of the secant line through the function at the endpoints of the interval.