To find the difference quotient of the function $f(x) = -2x^2 + x$, we need to compute:

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

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

First, substitute $x+h$ into the function $f(x) = -2x^2 + x$:

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

Expand $(x+h)^2$:

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

Simplify the expression:

$f(x+h) = -2x^2 - 4xh - 2h^2 + x + h$

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

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

$f(x+h) - f(x) = \left( -2x^2 - 4xh - 2h^2 + x + h \right) - \left( -2x^2 + x \right)$

Simplify:

$f(x+h) - f(x) = -2x^2 - 4xh - 2h^2 + x + h + 2x^2 - x$

Cancel out the like terms $-2x^2$ and $+2x^2$, $x$ and $-x$:

$f(x+h) - f(x) = -4xh - 2h^2 + h$

Step 3: Divide by $h$

Now divide by $h$:

$\frac{f(x+h) - f(x)}{h} = \frac{-4xh - 2h^2 + h}{h}$

Simplify each term:

$\frac{f(x+h) - f(x)}{h} = -4x - 2h + 1$

Final Answer:

The difference quotient is:

$\frac{f(x+h) - f(x)}{h} = -4x - 2h + 1$

Would you like further details on any of these steps?

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Here are some related questions:

1. What happens when $h \to 0$ in this difference quotient?
2. How would the process change for a different quadratic function?
3. Can we generalize this method for polynomials of higher degrees?
4. What is the significance of the difference quotient in calculus?
5. How can this quotient be used to find the derivative of $f(x)$?

Tip: The difference quotient is an essential concept in calculus as it leads directly to the derivative, which gives the rate of change of a function.