一条三次曲线经过点 (2, 3) 和 (-3, -22)。如果它在 (1, -6) 处有一个驻点，求该曲线的方程。

1. Set up the general equation for the cubic curve and find its derivative.

Let the equation of the cubic curve be $y = ax^3 + bx^2 + cx + d$.
Differentiating with respect to $x$:
$\frac{dy}{dx} = 3ax^2 + 2bx + c$

2. Form equations using the coordinates of the given points on the curve.

Substitute the coordinates of the points $(1, -6)$, $(2, 3)$, and $(-3, -22)$ into the curve's equation:
At $(1, -6)$:
$a(1)^3 + b(1)^2 + c(1) + d = -6$
$a + b + c + d = -6 \quad \text{--- (1)}$

At $(2, 3)$:
$a(2)^3 + b(2)^2 + c(2) + d = 3$
$8a + 4b + 2c + d = 3 \quad \text{--- (2)}$

At $(-3, -22)$:
$a(-3)^3 + b(-3)^2 + c(-3) + d = -22$
$-27a + 9b - 3c + d = -22 \quad \text{--- (3)}$

3. Form an equation using the stationary point condition.

The curve has a stationary point at $x = 1$, which means $\frac{dy}{dx} = 0$ when $x = 1$.
$3a(1)^2 + 2b(1) + c = 0$
$3a + 2b + c = 0 \quad \text{--- (4)}$

4. Solve the simultaneous equations to find $a$, $b$, $c$, and $d$.

Subtract (1) from (2) to eliminate $d$:
$(8a + 4b + 2c + d) - (a + b + c + d) = 3 - (-6)$
$7a + 3b + c = 9 \quad \text{--- (5)}$

Subtract (1) from (3) to eliminate $d$:
$(-27a + 9b - 3c + d) - (a + b + c + d) = -22 - (-6)$
$-28a + 8b - 4c = -16$
Divide by $-4$:
$7a - 2b + c = 4 \quad \text{--- (6)}$

Subtract (6) from (5) to eliminate $a$ and $c$:
$(7a + 3b + c) - (7a - 2b + c) = 9 - 4$
$5b = 5 \implies b = 1$

Substitute $b = 1$ into (5) and (4):
$7a + 3(1) + c = 9 \implies 7a + c = 6 \quad \text{--- (7)}$
$3a + 2(1) + c = 0 \implies 3a + c = -2 \quad \text{--- (8)}$

Subtract (8) from (7):
$4a = 8 \implies a = 2$

Substitute $a = 2$ into (8):
$3(2) + c = -2 \implies c = -8$

Substitute $a = 2, b = 1, c = -8$ into (1):
$2 + 1 - 8 + d = -6 \implies d = -1$

5. State the final equation of the curve.

Substituting the values of $a, b, c$, and $d$ back into the general equation:
$y = 2x^3 + x^2 - 8x - 1$