A curve is such that $\mathrm{f}''(x) = 6x - 10$. The tangent to the curve $y = \mathrm{f}(x)$ at the point $(1, -2)$ passes through the point $(0, 13)$. Find the equation of the curve.

1. Find the gradient of the tangent to the curve at $x = 1$.

The tangent passes through $(1, -2)$ and $(0, 13)$.
Gradient of tangent, $m = \frac{13 - (-2)}{0 - 1} = \frac{15}{-1} = -15$
Therefore, $\mathrm{f}'(1) = -15$.

2. Integrate $\mathrm{f}''(x)$ to find the general expression for $\mathrm{f}'(x)$.

$\mathrm{f}'(x) = \int (6x - 10) \, dx$
$\mathrm{f}'(x) = 3x^2 - 10x + C$

3. Find the constant of integration $C$.

Substitute $x = 1$ and $\mathrm{f}'(1) = -15$:
$3(1)^2 - 10(1) + C = -15$
$3 - 10 + C = -15$
$-7 + C = -15$
$C = -8$
So, $\mathrm{f}'(x) = 3x^2 - 10x - 8$.

4. Integrate $\mathrm{f}'(x)$ to find the general expression for $\mathrm{f}(x)$.

$\mathrm{f}(x) = \int (3x^2 - 10x - 8) \, dx$
$\mathrm{f}(x) = x^3 - 5x^2 - 8x + D$

5. Find the constant of integration $D$ to determine the equation of the curve.

The curve passes through $(1, -2)$. Substitute $x = 1$ and $\mathrm{f}(1) = -2$:
$(1)^3 - 5(1)^2 - 8(1) + D = -2$
$1 - 5 - 8 + D = -2$
$-12 + D = -2$
$D = 10$
Therefore, the equation of the curve is $y = x^3 - 5x^2 - 8x + 10$.