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We are tasked with finding the difference quotient for the given function \( f(x) = x^2 - x + 1 \), specifically for:

\[
\frac{f(2+h) - f(2)}{h}, \quad h \neq 0.
\]

### Step 1: Calculate \( f(2+h) \)
Substitute \( x = 2+h \) into \( f(x) = x^2 - x + 1 \):

\[
f(2+h) = (2+h)^2 - (2+h) + 1.
\]

Simplify \( (2+h)^2 \):
\[
(2+h)^2 = 4 + 4h + h^2.
\]

Now substitute back:
\[
f(2+h) = 4 + 4h + h^2 - (2+h) + 1.
\]

Simplify further:
\[
f(2+h) = 4 + 4h + h^2 - 2 - h + 1 = h^2 + 3h + 3.
\]

### Step 2: Calculate \( f(2) \)
Substitute \( x = 2 \) into \( f(x) = x^2 - x + 1 \):
\[
f(2) = 2^2 - 2 + 1 = 4 - 2 + 1 = 3.
\]

### Step 3: Difference Quotient
Substitute \( f(2+h) \) and \( f(2) \) into the difference quotient formula:
\[
\frac{f(2+h) - f(2)}{h} = \frac{(h^2 + 3h + 3) - 3}{h}.
\]

Simplify the numerator:
\[
(h^2 + 3h + 3) - 3 = h^2 + 3h.
\]

So the difference quotient becomes:
\[
\frac{f(2+h) - f(2)}{h} = \frac{h^2 + 3h}{h}.
\]

### Step 4: Simplify
Factor \( h \) out of the numerator:
\[
\frac{h^2 + 3h}{h} = \frac{h(h+3)}{h}.
\]

Since \( h \neq 0 \), cancel \( h \) from the numerator and denominator:
\[
\frac{h(h+3)}{h} = h+3.
\]

### Final Answer:
\[
\frac{f(2+h) - f(2)}{h} = h + 3, \quad h \neq 0.
\]

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Let me know if you have any questions or want a deeper explanation!  

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**Related Questions:**
1. What is the difference quotient and why is it useful in calculus?
2. How would the result change if \( f(x) \) was a cubic function instead?
3. How do you generalize the process of finding \( \frac{f(a+h)-f(a)}{h} \)?
4. Can the difference quotient be used to approximate derivatives?
5. How do you interpret the limit of the difference quotient as \( h \to 0 \)?

**Tip:** Always simplify the numerator completely before canceling terms in the difference quotient!

Find the difference quotient and simplify your answer: f(x) = x^2 - x + 1, (f(2+h) - f(2))/h, h ≠ 0.

Find and simplify the difference quotient (f(2+h) - f(2))/h for the quadratic function f(x) = x^2 - x + 1. This step-by-step solution explores algebraic simplification, quadratic expansions, and difference quotient concepts, suitable for high school students (Grades 9-12).

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