1. **State the problem:** We need to find the difference quotient $$\frac{f(x+h)-f(x)}{h}$$ for the function $$f(x) = -x^2 + 6x + 3$$ where $$h \neq 0$$. 2. **Write the formula:** The difference quotient is given by $$\frac{f(x+h)-f(x)}{h}$$ This represents the average rate of change of the function over the interval from $$x$$ to $$x+h$$. 3. **Calculate $$f(x+h)$$:** Substitute $$x+h$$ into the function: $$f(x+h) = - (x+h)^2 + 6(x+h) + 3$$ Expand the square: $$= - (x^2 + 2xh + h^2) + 6x + 6h + 3$$ Distribute the minus sign: $$= -x^2 - 2xh - h^2 + 6x + 6h + 3$$ 4. **Calculate $$f(x+h) - f(x)$$:** $$f(x+h) - f(x) = (-x^2 - 2xh - h^2 + 6x + 6h + 3) - (-x^2 + 6x + 3)$$ Simplify by removing parentheses: $$= -x^2 - 2xh - h^2 + 6x + 6h + 3 + x^2 - 6x - 3$$ Combine like terms: $$= \cancel{-x^2} - 2xh - h^2 + \cancel{6x} + 6h + \cancel{3} + \cancel{x^2} - \cancel{6x} - \cancel{3}$$ $$= -2xh - h^2 + 6h$$ 5. **Divide by $$h$$:** $$\frac{f(x+h) - f(x)}{h} = \frac{-2xh - h^2 + 6h}{h}$$ Factor $$h$$ from numerator: $$= \frac{h(-2x - h + 6)}{h}$$ Cancel $$h$$: $$= \cancel{\frac{h}{h}}(-2x - h + 6)$$ $$= -2x - h + 6$$ 6. **Final answer:** $$\boxed{-2x - h + 6}$$ This expression gives the average rate of change of the function $$f(x)$$ over the interval $$[x, x+h]$$.