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 the equations:

1. Add the two equations: $$$$

2a + b = -1 \ 5a + b = 29 \

$$Subtract the first from the second:$$

(5a + b) - (2a + b) = 29 - (-1) \ 3a = 30 \ \Rightarrow a = 10

2. Substitute $a = 10$ back into one of the equations:

2(10) + b = -1 \ 20 + b = -1 \ \Rightarrow b = -21

Thus, $a = 10$ and $b = -21$.

The curve crosses the $y$-axis when $x = 0$. Substituting $x = 0$ into the equation gives:

$$y = (0)^2 + 10(0) - 21 = -21$$

Thus, the coordinates are (0, -21).

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

$$0 = x^2 + 10x - 21 \\$$

Using the quadratic formula:

$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\$$

Here, $a = 1, b = 10, c = -21$:

$$x = \frac{-10 \pm \sqrt{(10)^2 - 4(1)(-21)}}{2(1)} \\$$

Calculating gives:

$$x = \frac{-10 \pm \sqrt{100 + 84}}{2} \\ = \frac{-10 \pm \sqrt{184}}{2} \\ = \frac{-10 \pm 2\sqrt{46}}{2} \\ = -5 \pm \sqrt{46}$$

The x-intercepts are approximately:

1. $x \approx -5 + 6.782 = 1.8$ (to one decimal place)
2. $x \approx -5 - 6.782 = -11.8$ (to one decimal place) Thus, the curve crosses the x-axis at approximately (1.8, 0) and (-11.8, 0).