For a particular curve $\frac{\text{d}^2 y}{\text{d}x^2} = 3x - 2$. The tangent to the curve at the point $A(4, -40)$ is parallel to the line $3x - y = 2$. Find the equation of the curve.

1. Integrate the second derivative to find the first derivative

$\frac{\text{d}y}{\text{d}x} = \int (3x - 2) \, \text{d}x$
$\frac{\text{d}y}{\text{d}x} = \frac{3}{2}x^2 - 2x + c$

2. Determine the gradient of the tangent at $x = 4$

The line is $3x - y = 2 \implies y = 3x - 2$.
The gradient of this line is $3$.
Since the tangent is parallel to this line, $\frac{\text{d}y}{\text{d}x} = 3$ at $x = 4$.

3. Find the constant of integration $c$

Substitute $x = 4$ and $\frac{\text{d}y}{\text{d}x} = 3$:
$3 = \frac{3}{2}(4)^2 - 2(4) + c$
$3 = 24 - 8 + c$
$3 = 16 + c \implies c = -13$
So, $\frac{\text{d}y}{\text{d}x} = \frac{3}{2}x^2 - 2x - 13$

4. Integrate the first derivative to find the equation of the curve

$y = \int \left(\frac{3}{2}x^2 - 2x - 13\right) \, \text{d}x$
$y = \frac{1}{2}x^3 - x^2 - 13x + d$

5. Find the constant of integration $d$ using the point $A(4, -40)$

Substitute $x = 4$ and $y = -40$:
$-40 = \frac{1}{2}(4)^3 - (4)^2 - 13(4) + d$
$-40 = 32 - 16 - 52 + d$
$-40 = -36 + d \implies d = -4$

6. State the final equation of the curve

$y = \frac{1}{2}x^3 - x^2 - 13x - 4$