$$ 1 \& $$ Find and simplify the difference quotient for the following function. $$ f(x)=2 x^{2}-5 x $$

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To find and simplify the difference quotient for the function $$ f(x) = 2x^2 - 5x $$, we will follow these steps:

1. **Define the difference quotient**: The difference quotient is given by the formula:
   $$
   \frac{f(x+h) - f(x)}{h}
   $$
   where $$ h $$ is a small increment.

2. **Calculate $$ f(x+h) $$**: We need to substitute $$ x+h $$ into the function:
   $$
   f(x+h) = 2(x+h)^2 - 5(x+h)
   $$

3. **Expand $$ f(x+h) $$**:
   - First, expand $$ (x+h)^2 $$:
     $$
     (x+h)^2 = x^2 + 2xh + h^2
     $$
   - Now substitute this back into $$ f(x+h) $$:
     $$
     f(x+h) = 2(x^2 + 2xh + h^2) - 5(x + h)
     $$
   - Distributing the terms:
     $$
     f(x+h) = 2x^2 + 4xh + 2h^2 - 5x - 5h
     $$

4. **Calculate $$ f(x+h) - f(x) $$**:
   - Now we need to subtract $$ f(x) $$ from $$ f(x+h) $$:
   $$
   f(x) = 2x^2 - 5x
   $$
   - Therefore:
   $$
   f(x+h) - f(x) = (2x^2 + 4xh + 2h^2 - 5x - 5h) - (2x^2 - 5x)
   $$
   - Simplifying this gives:
   $$
   f(x+h) - f(x) = 4xh + 2h^2 - 5h
   $$

5. **Substitute into the difference quotient**:
   $$
   \frac{f(x+h) - f(x)}{h} = \frac{4xh + 2h^2 - 5h}{h}
   $$

6. **Simplify the expression**:
   - We can factor out $$ h $$ from the numerator:
   $$
   \frac{h(4x + 2h - 5)}{h}
   $$
   - Canceling $$ h $$ (assuming $$ h \neq 0 $$):
   $$
   4x + 2h - 5
   $$

Thus, the simplified difference quotient for the function $$ f(x) = 2x^2 - 5x $$ is:
$$
4x + 2h - 5
$$
The difference quotient simplifies to $$ 4x + 2h - 5 $$.

\( 1 \& \) Find and simplify the difference quotient for the following function. \[ f(x)=2 x^{2}-5 x \]

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