The figure shows the curve y = x³ - 3x, with two points marked: P(1, -2), which lies on the curve…

The key fact. The tangent at $x = t$ is $y = (3t^2 - 3)(x - t) + t^3 - 3t$. Then $x^3 - 3x - \left[(3t^2 - 3)(x - t) + t^3 - 3t\right] = (x - t)^2(x + 2t)$. The double root $x = t$ is the point of contact, and the tangent meets the curve again where $x = -2t$.

(a) For $P$, $t = 1$: $Q$ has $x = -2$. (The tangent at $P$ is the line $y = -2$, and indeed $(-2)^3 - 3(-2) = -2$.)

(b) For $Q$, $t = -2$: $R$ has $x = -2(-2) = 4$. The slope there is $3(4) - 3 = 9$, and the tangent at $Q$ is $y = 9x + 16$, which meets the curve again at $(4, 52)$.

(c) The tangent at $t$ passes through $(2, k)$ when $k = (3t^2 - 3)(2 - t) + t^3 - 3t = - 2 t^{3} + 6 t^{2} - 6$. With $k = -2$: $-2t^3 + 6t^2 - 4 = 0$, that is $- 2 \left(t - 1\right) \left(t^{2} - 2 t - 2\right) = 0$, so $t = 1$ or $t = 1 \pm \sqrt3$. Three real values of $t$ give three different tangents (a line cannot touch a cubic at two points, since that would need four meetings counted with multiplicity). So 3 tangents. One of them is the tangent at $P$ itself, the line $y = -2$, which passes through $A$.

(d) The slopes are $3t^2 - 3$: $0$ for $t = 1$, and $3(1 \pm \sqrt3)^2 - 3 = 9 \pm 6\sqrt3$. The greatest is $9 + 6 \sqrt{3} \approx 19.39$.

(e) Let $\varphi(t) = - 2 t^{3} + 6 t^{2} - 6$. Then $\varphi'(t) = -6t^2 + 12t = -6t(t - 2)$: a local minimum $\varphi(0) = -6$ and a local maximum $\varphi(2) = 2$, with $\varphi \to +\infty$ as $t \to -\infty$ and $\varphi \to -\infty$ as $t \to \infty$. The equation $\varphi(t) = k$ has three real roots exactly when $-6 < k < 2$. So $\alpha = -6$ and $\beta = 2$. At $k = -6$ or $k = 2$ two of the tangents merge, and only two remain.