1. **State the problem:** We need to find the equation of the secant line passing through the points $(-4, h(-4))$ and $(2, h(2))$ where $h(x) = x^3 + 6$. 2. **Calculate the function values at the given points:** $$h(-4) = (-4)^3 + 6 = -64 + 6 = -58$$ $$h(2) = 2^3 + 6 = 8 + 6 = 14$$ 3. **Find the slope $m$ of the secant line:** The slope formula is $$m = \frac{y_2 - y_1}{x_2 - x_1}$$ Substitute the points: $$m = \frac{14 - (-58)}{2 - (-4)} = \frac{14 + 58}{2 + 4} = \frac{72}{6}$$ Show cancellation: $$m = \frac{\cancel{72}}{\cancel{6}} = 12$$ 4. **Use point-slope form to find the equation:** Point-slope form is $$y - y_1 = m(x - x_1)$$ Using point $(-4, -58)$: $$y - (-58) = 12(x - (-4))$$ Simplify: $$y + 58 = 12(x + 4)$$ 5. **Expand and solve for $y$:** $$y + 58 = 12x + 48$$ Subtract 58 from both sides: $$y = 12x + 48 - 58$$ $$y = 12x - 10$$ **Final answer:** $$\boxed{y = 12x - 10}$$