1. Determine the quadratic function:
Given the values of $y$ for specific $x$ values, we can set up a system of equations to find the coefficients of the quadratic function $y = ax^2 + bx + c$.

- For $x = -1$, $y = -1$:
$$a(-1)^2 + b(-1) + c = -1 \implies a - b + c = -1 \quad \text{(Equation 1)}$$
- For $x = 0$, $y = -2$:
$$a(0)^2 + b(0) + c = -2 \implies c = -2 \quad \text{(Equation 2)}$$
- For $x = 1$, $y = 1$:
$$a(1)^2 + b(1) + c = 1 \implies a + b + c = 1 \quad \text{(Equation 3)}$$

2. Solve the system of equations:
Substitute $c = -2$ from Equation 2 into Equations 1 and 3:
- Equation 1:
$$a - b - 2 = -1 \implies a - b = 1 \quad \text{(Equation 4)}$$
- Equation 3:
$$a + b - 2 = 1 \implies a + b = 3 \quad \text{(Equation 5)}$$

Add Equations 4 and 5 to eliminate $b$:
$$(a - b) + (a + b) = 1 + 3 \implies 2a = 4 \implies a = 2$$

Substitute $a = 2$ back into Equation 4:
$$2 - b = 1 \implies b = 1$$

Therefore, the quadratic function is:
$$y = 2x^2 + x - 2$$

3. Find the roots of the quadratic function:
To find where the curve cuts the $x$-axis, set $y = 0$:
$$2x^2 + x - 2 = 0$$

Use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$$a = 2, \quad b = 1, \quad c = -2$$
$$x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 2 \cdot (-2)}}{2 \cdot 2} = \frac{-1 \pm \sqrt{1 + 16}}{4} = \frac{-1 \pm \sqrt{17}}{4}$$

4. Calculate the distance between the roots:
The roots are $x_1 = \frac{-1 + \sqrt{17}}{4}$ and $x_2 = \frac{-1 - \sqrt{17}}{4}$.

The distance between the roots is:
$$\left| \frac{-1 + \sqrt{17}}{4} - \frac{-1 - \sqrt{17}}{4} \right| = \left| \frac{\sqrt{17} + \sqrt{17}}{4} \right| = \left| \frac{2\sqrt{17}}{4} \right| = \frac{\sqrt{17}}{2}$$

The final answer is $\boxed{\frac{\sqrt{17}}{2}}$.