NECO 2024 · Paper 2 · Q10

The table is for $y = x^2 + x - 12$ .

$x$	$-4$	$-3$	$-2$	$-1$	0	1	2	3	4
$y$	0				$-12$			0	8

(a)
Copy and complete the table (enter the $y$ -values for $x = -3, -2, -1, 1, 2$ ).
Separate values with commas, e.g. 3, −2
(b)
Using a scale of 2 cm to 1 unit on the $x$ -axis and 2 cm to 5 units on the $y$ -axis, draw the graph of $y = x^2 + x - 12$ for $-4 \le x \le 4$ . On the same axes, draw the graph of $y = 2x + 1$ .
Model answer
Plot every point from the table, then join them with one smooth curve (not straight lines between points). Scale: 2 cm to 1 unit across, 2 cm to 5 units up. The parabola has its lowest point at $(-0.5, -12.25)$ , halfway between the roots $-4$ and $3$ . Draw the straight line $y = 2x + 1$ through two easy points, e.g. $(0, 1)$ and $(4, 9)$ .

For (c): $x^2 - x - 13 = 0$ is the same as $x^2 + x - 12 = 2x + 1$ , so the roots are where the graphs cross: $x \approx −3.1$ and $x \approx 4.1$ . The second crossing is just beyond $x = 4$ (dashed), so extend the curve slightly to read it.
(c)
From your graphs, determine the roots of the equation $x^2 - x - 13 = 0$ .
Separate values with commas, e.g. 3, −2
(d)
Find the minimum value of $y$ from the quadratic graph.

The roots of x² − x − 13 = 0 are where the parabola meets the line y = 2x + 1.

Worked solution (try it first)

Plot the nine points $(-4, 0), (-3, -6), \ldots, (4, 8)$ and join them with a smooth U-shaped curve.
For the line $y = 2x + 1$ , three points are enough: $(-4, -7)$ , $(0, 1)$ and $(4, 9)$ .

Where the line meets the curve, $x^2 + x - 12 = 2x + 1$ .
Taking $2x + 1$ from both sides gives $x^2 - x - 13 = 0$ , so the roots are the $x$ -values of the two crossing points.
Read them from the graph: $x \approx -3.1$ and $x \approx 4.1$ .
(The exact values are $\frac{1 \pm \sqrt{53}}{2}$ , about $-3.14$ and $4.14$ .
The second is just past $x = 4$ , so extend the curve and line slightly to read it.)

The lowest point of the curve is halfway between the roots $-4$ and $3$ of $x^2 + x - 12 = 0$ , at $x = -0.5$ .
There $y = 0.25 - 0.5 - 12 = -12.25$ , so the minimum value of $y$ is $-12.25$ .