1. **State the problem:** We need to find the gradient of the tangent line $T$ to the curve $y = 5x^3 - 8x$ at the point $(-2, -24)$. 2. **Recall the formula:** The gradient of the tangent line to a curve at a point is given by the derivative of the function evaluated at that point. 3. **Find the derivative:** Differentiate $y = 5x^3 - 8x$ with respect to $x$: $$\frac{dy}{dx} = 15x^2 - 8$$ 4. **Evaluate the derivative at $x = -2$:** $$\frac{dy}{dx}\bigg|_{x=-2} = 15(-2)^2 - 8 = 15 \times 4 - 8 = 60 - 8 = 52$$ 5. **Interpretation:** The gradient of the tangent line $T$ at the point $(-2, -24)$ is $52$. 6. **Final answer:** The gradient of $T$ is $52.0$ (to 1 decimal place).