1. We are given the function $f(x) = x^3 - 3x$ and points $a = 1$ and $b = -\frac{1}{2}$. We want to find the slope of the secant line between these points. 2. The formula for the slope of the secant line between $x=a$ and $x=b$ is: $$m = \frac{f(b) - f(a)}{b - a}$$ 3. Calculate $f(a)$: $$f(1) = 1^3 - 3 \times 1 = 1 - 3 = -2$$ 4. Calculate $f(b)$: $$f\left(-\frac{1}{2}\right) = \left(-\frac{1}{2}\right)^3 - 3 \times \left(-\frac{1}{2}\right) = -\frac{1}{8} + \frac{3}{2} = -\frac{1}{8} + \frac{12}{8} = \frac{11}{8}$$ 5. Substitute into the slope formula: $$m = \frac{\frac{11}{8} - (-2)}{-\frac{1}{2} - 1} = \frac{\frac{11}{8} + 2}{-\frac{3}{2}} = \frac{\frac{11}{8} + \frac{16}{8}}{-\frac{3}{2}} = \frac{\frac{27}{8}}{-\frac{3}{2}}$$ 6. Simplify the division: $$m = \frac{27}{8} \times \frac{2}{-3} = \frac{27 \times 2}{8 \times (-3)} = \frac{54}{-24}$$ 7. Cancel common factors: $$m = \frac{\cancel{54}^{27} \times 2}{\cancel{24}^{12} \times (-2)} = -\frac{27}{12} = -\frac{9}{4}$$ 8. Final answer: The slope of the secant line between $x=1$ and $x=-\frac{1}{2}$ for the function $f(x) = x^3 - 3x$ is: $$m = -\frac{9}{4}$$