To solve the given problem, we need to compute the difference quotient for the function $f(x) = -x^2 - 6x + 6$. The difference quotient formula is:

$\frac{f(x+h) - f(x)}{h}, \quad h \neq 0$

Step 1: Calculate $f(x + h)$

Substitute $x + h$ into the given function $f(x) = -x^2 - 6x + 6$:

$f(x + h) = -(x + h)^2 - 6(x + h) + 6$

Expand $(x + h)^2$:

$(x + h)^2 = x^2 + 2xh + h^2$

Substitute this back into $f(x + h)$:

$f(x + h) = -(x^2 + 2xh + h^2) - 6(x + h) + 6$

Simplify:

$f(x + h) = -x^2 - 2xh - h^2 - 6x - 6h + 6$

Step 2: Compute $f(x + h) - f(x)$

Now subtract $f(x) = -x^2 - 6x + 6$ from $f(x + h)$:

$f(x + h) - f(x) = \big(-x^2 - 2xh - h^2 - 6x - 6h + 6\big) - \big(-x^2 - 6x + 6\big)$

Distribute the negative sign:

$f(x + h) - f(x) = -x^2 - 2xh - h^2 - 6x - 6h + 6 + x^2 + 6x - 6$

Combine like terms:

$f(x + h) - f(x) = -2xh - h^2 - 6h$

Step 3: Divide by $h$

To find the difference quotient, divide the result by $h$:

$\frac{f(x + h) - f(x)}{h} = \frac{-2xh - h^2 - 6h}{h}$

Simplify by factoring $h$ out of the numerator:

$\frac{-2xh - h^2 - 6h}{h} = -2x - h - 6$

Final Answer:

$\frac{f(x + h) - f(x)}{h} = -2x - h - 6, \quad h \neq 0$

Would you like further clarification on any step or a deeper explanation?

Related Questions:

1. How does the difference quotient relate to the derivative?
2. What happens to the term $-h$ as $h \to 0$?
3. Can you find the derivative of $f(x)$ directly using calculus?
4. How does expanding $(x+h)^2$ affect the calculation?
5. What if $f(x)$ had higher-degree terms or trigonometric components?

Tip:

When simplifying difference quotients, carefully expand and organize terms to avoid missing or combining incorrectly.