曲线 $C$ 的方程为 $y = \frac{a}{x} + bx^2 + c$，其中 $a$，$b$ 和 $c$ 是常数。已知 $C$ 经过坐标为 $(1, 2)$ 的点，并在 $x = -1$ 处与 $x$ 轴相交。在 $x = 0.5$ 的点处， $C$ 的斜率为 $-1$。求 $a$，$b$ 和 $c$ 的值，并写出 $C$ 的方程。

1. Form two equations using the known points $(1, 2)$ and $(-1, 0)$ on the curve.

The equation of the curve is $y = \frac{a}{x} + bx^2 + c$.

Substitute $(1, 2)$:
$2 = \frac{a}{1} + b(1)^2 + c$
$a + b + c = 2 \quad \text{--- (1)}$

Since the curve intersects the $x$-axis at $x = -1$, it passes through $(-1, 0)$. Substitute $(-1, 0)$:
$0 = \frac{a}{-1} + b(-1)^2 + c$
$-a + b + c = 0 \quad \text{--- (2)}$

2. Differentiate the equation of the curve to find an expression for the gradient.

Rewrite the equation as $y = ax^{-1} + bx^2 + c$.

Differentiating with respect to $x$:
$\frac{dy}{dx} = -ax^{-2} + 2bx$

3. Form a third equation using the given gradient condition.

Given that the gradient is $-1$ when $x = 0.5$:
$-a(0.5)^{-2} + 2b(0.5) = -1$
$-a\left(\frac{1}{0.25}\right) + b = -1$
$-4a + b = -1 \quad \text{--- (3)}$

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

Subtract equation (2) from equation (1):
$(a + b + c) - (-a + b + c) = 2 - 0$
$2a = 2 \implies a = 1$

Substitute $a = 1$ into equation (3):
$-4(1) + b = -1$
$b = 3$

Substitute $a = 1$ and $b = 3$ into equation (1):
$1 + 3 + c = 2$
$c = -2$

5. State the values of the constants and the final equation of the curve.

The values are $a = 1$, $b = 3$, and $c = -2$.

Substituting these into the original equation, the equation of $C$ is:
$y = \frac{1}{x} + 3x^2 - 2$