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

We now have the following system of equations:

1. $2a + b = -1$
2. $-5a + b = -29$

To eliminate $b$, we can subtract the first equation from the second:

$(-5a + b) - (2a + b) = -29 - (-1) \quad \Rightarrow \quad -7a = -28 \quad \Rightarrow \quad a = 4$

Substituting $a = 4$ back into the first equation:

$2(4) + b = -1 \quad \Rightarrow \quad 8 + b = -1 \quad \Rightarrow \quad b = -9$

The curve crosses the y-axis when $x = 0$. We can find the corresponding $y$ value:

$y = (0)^2 + 4(0) - 9 = -9$

Thus, the curve crosses the y-axis at $(0, -9)$.

To find where the curve crosses the x-axis, we set $y = 0$:

$0 = x^2 + 4x - 9$

Using the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
where $a = 1$, $b = 4$, $c = -9$:

$x = \frac{-4 \pm \sqrt{(4)^2 - 4(1)(-9)}}{2(1)} \quad = \frac{-4 \pm \sqrt{16 + 36}}{2} \quad = \frac{-4 \pm \sqrt{52}}{2} \quad = \frac{-4 \pm 2\sqrt{13}}{2} \quad = -2 \pm \sqrt{13}$

Calculating the two solutions:

1. $x = -2 + \sqrt{13} \approx 1.605$
2. $x = -2 - \sqrt{13} \approx -5.605$

Thus, the points where the curve crosses the x-axis are approximately $1.6$ and $-5.6$.