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.

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

$$0 = a(2)^2 - 1$$ $$0 = 4a - 1$$ $$4a = 1$$ $$a = \frac{1}{4}$$

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

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

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)

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

$$\frac{1}{4}x^2 - 1 = 0$$ $$\frac{1}{4}x^2 = 1$$ $$x^2 = 4$$ $$x = \pm 2$$

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

\alpha = -2, \beta = 2

$$\frac{1}{4}x^2 - 1 = 2$$ $$\frac{1}{4}x^2 = 3$$ $$x^2 = 12$$ $$x = \pm \sqrt{12} = \pm 2\sqrt{3}$$

x = \pm 2\sqrt{3}