To solve for the difference quotient $\frac{f(x+h)-f(x)}{h}$ where $f(x) = -x^2 - 5x + 6$, we will follow these steps:

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

Substitute $x+h$ for $x$ in the original function: $f(x+h) = -(x+h)^2 - 5(x+h) + 6$ Expand this: $f(x+h) = -(x^2 + 2xh + h^2) - 5x - 5h + 6$ $= -x^2 - 2xh - h^2 - 5x - 5h + 6$

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

Subtract $f(x)$ from $f(x+h)$: $f(x+h) - f(x) = (-x^2 - 2xh - h^2 - 5x - 5h + 6) - (-x^2 - 5x + 6)$ The terms $-x^2$, $-5x$, and $6$ cancel out: $f(x+h) - f(x) = -2xh - h^2 - 5h$

Step 3: Divide by $h$

Now, divide the expression by $h$: $\frac{f(x+h) - f(x)}{h} = \frac{-2xh - h^2 - 5h}{h}$ Factor out $h$ in the numerator: $\frac{-h(2x + h + 5)}{h}$ Since $h \neq 0$, cancel the $h$: $-2x - h - 5$

Thus, the simplified difference quotient is: $-2x - h - 5$

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Related Questions:

1. How does the difference quotient relate to the derivative of a function?
2. What would happen if we took the limit of the difference quotient as $h \to 0$?
3. How do you find the derivative of $f(x) = -x^2 - 5x + 6$ using the power rule?
4. How can the difference quotient be used to approximate the slope of a tangent line?
5. What is the significance of simplifying the difference quotient as much as possible?

Tip:

When dealing with the difference quotient, it's useful to factor the expression before canceling terms to avoid mistakes.