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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:
\[
-2x - h - 5
\]

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

### 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.

Find the difference quotient f(x+h)−f(x)/h, where h ≠ 0, for the function below. f(x) = -x^2 - 5x + 6. Simplify your answer as much as possible.

Learn how to find and simplify the difference quotient for the quadratic function f(x) = -x^2 - 5x + 6. This step-by-step solution walks through expanding, factoring, and simplifying the expression, perfect for high school students in Grades 10-12.

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