Part of the graph of the function $y = x^2 + ax + b$ where $a, b ext{ are } extbf{Z}$ is shown below - Junior Cycle Mathematics - Question 11 - 2014

We now have the two equations from above:

2a + b = -1 \tag{1} -5a + b = -29 \tag{2}

Subtracting the first equation from the second:

-7a = -28 \\ a = 4$$ Now substituting $a = 4$ into equation (1): $$2(4) + b = -1 \\ 8 + b = -1 \\ b = -1 - 8 \\ b = -9$$ Thus, we find that $a = 4$ and $b = -9$.

The curve crosses the $y$-axis when $x = 0$. Substituting this in:

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

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

The curve crosses the $x$-axis when $y = 0$. So we use:

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

This is a quadratic equation of the form $ax^2 + bx + c = 0$. We will use the quadratic formula:

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

Substituting $a = 1$, $b = 4$, and $c = -9$:

= \frac{-4 \pm \sqrt{52}}{2} \\ = \frac{-4 \pm 2\sqrt{13}}{2} \\ = -2 \pm \sqrt{13}$$ Calculating the roots gives: $$x_1 = -2 + \sqrt{13} \approx 0.6 \\ x_2 = -2 - \sqrt{13} \approx -4.6$$ Thus, the points where the curve crosses the $x$-axis are approximately $(0.6, 0)$ and $(-4.6, 0)$.