Find the difference quotient of f(x)=2x^2-5x

The **difference quotient** of a function \( f(x) \) is given by the formula:

\[
\frac{f(x + h) - f(x)}{h}
\]

For the function \( f(x) = 2x^2 - 5x \), let's compute the difference quotient step-by-step:

### Step 1: Find \( f(x + h) \)
\[
f(x + h) = 2(x + h)^2 - 5(x + h)
\]
Expanding this, we have:
\[
= 2(x^2 + 2xh + h^2) - 5(x + h)
\]
\[
= 2x^2 + 4xh + 2h^2 - 5x - 5h
\]

### Step 2: Compute \( f(x + h) - f(x) \)
\[
f(x + h) - f(x) = (2x^2 + 4xh + 2h^2 - 5x - 5h) - (2x^2 - 5x)
\]
\[
= 4xh + 2h^2 - 5h
\]

### Step 3: Divide by \( h \)
\[
\frac{f(x + h) - f(x)}{h} = \frac{4xh + 2h^2 - 5h}{h}
\]
\[
= 4x + 2h - 5
\]

### Final Answer
The difference quotient is:
\[
4x + 2h - 5
\]

Would you like more details or have any questions?

### 5 Related Questions
1. What is the geometric interpretation of the difference quotient?
2. How does the difference quotient lead to the derivative as \( h \to 0 \)?
3. How would the difference quotient change if \( f(x) = 3x^3 - 4x + 1 \)?
4. What is the significance of the term \( 2h \) in the difference quotient result?
5. How would you apply the difference quotient to linear functions?

### Tip
The difference quotient represents the average rate of change of the function over the interval from \( x \) to \( x + h \). As \( h \) approaches zero, it becomes the instantaneous rate of change, which is the derivative.

Find the difference quotient of f(x) = 2x^2 - 5x

Explore how to find the difference quotient of the quadratic function f(x) = 2x^2 - 5x with step-by-step solutions. This problem involves key algebraic concepts such as the difference quotient formula, rate of change, and quadratic functions, suitable for Grades 10-12.

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