1. **State the problem:** We need to find the values of $p$ such that the gradient (slope) of the line joining the points $(3, 3p)$ and $(4, p^2 + 1)$ is $-1$. 2. **Formula for gradient:** The gradient $m$ between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by: $$m = \frac{y_2 - y_1}{x_2 - x_1}$$ 3. **Apply the formula:** Substitute the points and gradient: $$-1 = \frac{(p^2 + 1) - 3p}{4 - 3}$$ 4. **Simplify denominator:** $$-1 = (p^2 + 1) - 3p$$ 5. **Rewrite equation:** $$-1 = p^2 + 1 - 3p$$ 6. **Bring all terms to one side:** $$p^2 - 3p + 1 + 1 = 0$$ $$p^2 - 3p + 2 = 0$$ 7. **Factorize quadratic:** $$(p - 1)(p - 2) = 0$$ 8. **Solve for $p$:** $$p = 1 \quad \text{or} \quad p = 2$$ **Final answer:** The values of $p$ are $1$ and $2$.