A curve $C$ has equation $\text{f}(x) = \frac{a}{x^2} + bx + c$, where $a$, $b$ and $c$ are constants. It is given that $C$ passes through the points $(-1, 4)$ and $(2, -11)$ and the curve of $y = \frac{1}{\text{f}(x)}$ has a vertical asymptote with equation $x = 1$. Find the equation of $C$.

1. Interpret the vertical asymptote of the reciprocal graph to form the first equation.

Since the curve $y = \frac{1}{\text{f}(x)}$ has a vertical asymptote at $x = 1$, its denominator must equal $0$ when $x = 1$.

$\text{f}(1) = 0$
$\frac{a}{1^2} + b(1) + c = 0$
$a + b + c = 0 \quad \text{--- (1)}$

2. Use the given points on the curve $C$ to form two more equations.

Given that $C$ passes through $(-1, 4)$:
$\text{f}(-1) = 4$
$\frac{a}{(-1)^2} + b(-1) + c = 4$
$a - b + c = 4 \quad \text{--- (2)}$

Given that $C$ passes through $(2, -11)$:
$\text{f}(2) = -11$
$\frac{a}{2^2} + b(2) + c = -11$
$\frac{a}{4} + 2b + c = -11$
$a + 8b + 4c = -44 \quad \text{--- (3)}$

3. Solve the system of linear equations to find $a$, $b$, and $c$.

Subtracting (1) from (2) eliminates $a$ and $c$:
$(a - b + c) - (a + b + c) = 4 - 0$
$-2b = 4 \implies b = -2$

Substitute $b = -2$ into (1):
$a - 2 + c = 0 \implies c = 2 - a$

Substitute $b = -2$ and $c = 2 - a$ into (3):
$a + 8(-2) + 4(2 - a) = -44$
$a - 16 + 8 - 4a = -44$
$-3a - 8 = -44$
$-3a = -36 \implies a = 12$

Substitute $a = 12$ into $c = 2 - a$:
$c = 2 - 12 = -10$

4. State the final equation of the curve $C$.

Substituting $a = 12$, $b = -2$, and $c = -10$ back into the original form for $C$:
$y = \frac{12}{x^2} - 2x - 10$