The graph shows the relation y = ax2 + bx + c and the line y-x=1. Calculate a, b, c, the greatest value of the relation, the truth set of the simultaneous equations, and the values of x for which y > 0.

Step 1: Find the equation of the parabola $y = ax^2 + bx + c$.

From the graph, the parabola intersects the x-axis at roots $\alpha = -2$ and $\beta = 2$.

Sum of roots: $\alpha + \beta = -2 + 2 = 0 = -\frac{b}{a} \implies b = 0$.

Thus, $y = ax^2 + c$.

Vertex at $(0, -1)$, so at $x=0$, $y = c = -1$.

Substitute root $x=2$, $y=0$:

Equation: $y = \frac{1}{4}x^2 - 1$.

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

Step 2: Find the coordinates of vertex $V$.

From the graph, $V$ is at $(0, -1)$.

For $y = \frac{1}{4}x^2 - 1$, vertex at $x = -\frac{b}{2a} = 0$, $y = -1$.

(0, -1)

Step 3: Find $\alpha$ and $\beta$.

From graph and equation, roots where $y=0$:

Thus, $\alpha = -2$, $\beta = 2$ (or vice versa).

\alpha = -2, \beta = 2

Step 4: Find $x$ when $y=2$.

x = \pm 2\sqrt{3}