1. **State the problem:** Find the derivative $F'(x)$ of the function $f(x) = -3x^2 - x$ using the difference quotient method. 2. **Recall the difference quotient formula:** $$F'(x_0) = \lim_{h \to 0} \frac{f(x_0 + h) - f(x_0)}{h}$$ This formula calculates the slope of the tangent line at $x_0$ by finding the limit of the average rate of change as $h$ approaches zero. 3. **Calculate $f(x_0 + h)$:** $$f(x_0 + h) = -3(x_0 + h)^2 - (x_0 + h) = -3(x_0^2 + 2x_0h + h^2) - x_0 - h$$ 4. **Write the difference quotient:** $$\frac{f(x_0 + h) - f(x_0)}{h} = \frac{-3(x_0^2 + 2x_0h + h^2) - x_0 - h - (-3x_0^2 - x_0)}{h}$$ 5. **Simplify the numerator:** $$-3x_0^2 - 6x_0h - 3h^2 - x_0 - h + 3x_0^2 + x_0 = -6x_0h - 3h^2 - h$$ 6. **Substitute back into the difference quotient:** $$\frac{-6x_0h - 3h^2 - h}{h}$$ 7. **Cancel $h$ in numerator and denominator:** $$\frac{\cancel{h}(-6x_0 - 3h - 1)}{\cancel{h}} = -6x_0 - 3h - 1$$ 8. **Take the limit as $h \to 0$:** $$F'(x_0) = \lim_{h \to 0} (-6x_0 - 3h - 1) = -6x_0 - 1$$ 9. **Final answer:** $$\boxed{F'(x) = -6x - 1}$$ This is the derivative of the function $f(x) = -3x^2 - x$.