Part of the graph of the function $y = x^2 + ax + b$, where $a, b \in \mathbb{Z}$, is shown below - Junior Cycle Mathematics - Question 10 - 2011

To solve for $a$ and $b$, we have:

$2a + b = -1 \quad \text{(1)}$ $-5a + b = -29 \quad \text{(2)}$

Subtracting Equation (1) from Equation (2) gives:

$(-5a + b) - (2a + b) = -29 - (-1)$ $-7a = -28$

Thus: $a = 4$

Substituting $a = 4$ into Equation (1): $2(4) + b = -1$ $8 + b = -1$ $b = -9$

The curve crosses the y-axis when $x = 0$.

Substituting $x = 0$ into the equation gives: $y = (0)^2 + 4(0) - 9 = -9$

Thus, the co-ordinates are $(0, -9)$.

The curve crosses the x-axis when $y = 0$, so we solve: $x^2 + 4x - 9 = 0$

Using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ Here, $a = 1$, $b = 4$, and $c = -9$: $x = \frac{-4 \pm \sqrt{16 + 36}}{2(1)}$ $x = \frac{-4 \pm \sqrt{52}}{2}$ $x = \frac{-4 \pm 2\sqrt{13}}{2}$ $x = -2 \pm \sqrt{13}$

Calculating the approximate values gives: $x \approx 1.6 \quad \text{and} \quad x \approx -5.6$