WAEC 2012 · Paper 2 · Q9

(a)
Copy and complete the table of values for the relation $y = x^2 + 2x - 2$ .

$x$ $-4$ $-3$ $-2$ $-1$ $0$ $1$ $2$

$y$ $-3$ $-2$

Model answer

$x$ $-4$ $-3$ $-2$ $-1$ $0$ $1$ $2$

$y$ $6$ $1$ $-2$ $-3$ $-2$ $1$ $6$

For example $x = -4$ : $16 - 8 - 2 = 6$ . The values are symmetrical about $x = -1$ , where $y$ is least.
(b)
Using a scale of 2 cm to 1 unit on both axes, draw the graph of $y = x^2 + 2x - 2$ for $-4 \le x \le 2$ .
Model answer
Plot every point from the table, then join them with one smooth curve (not straight lines between points). The lowest point is $(-1, -3)$ .

For (c): (i) the roots are where the curve crosses the $x$ -axis: $x \approx −2.7$ and $0.7$ . (ii) $x^2 + 2x < 0.5$ means $x^2 + 2x - 2 < -1.5$ : draw $y = -1.5$ and take the part of the curve below it, $−2.2 < x < 0.2$ .
(c)
Use your graph to: (i) find the roots of the equation $x^2 + 2x - 2 = 0$ ; (ii) indicate the region where $x^2 + 2x < 0.5$ .
Separate values with commas, e.g. 3, −2

$x$	$-4$	$-3$	$-2$	$-1$	$0$	$1$	$2$
$y$	$6$	$1$	$-2$	$-3$	$-2$	$1$	$6$

The curve and the line y = −1.5.

Worked solution (try it first)

With 2 cm to 1 unit on both axes, plot the seven points and join them with a smooth U-shaped curve.
Its lowest point is at $(-1, -3)$ .

The roots of $x^2 + 2x - 2 = 0$ are where the curve crosses the $x$ -axis ( $y = 0$ ): $x \approx -2.7$ and $x \approx 0.7$ .
(Exactly, $-1 \pm \sqrt3$ .)

Take 2 from both sides: $x^2 + 2x < 0.5$ becomes $x^2 + 2x - 2 < -1.5$ , that is $y < -1.5$ .
Draw the line $y = -1.5$ and shade the part of the curve below it.
The line meets the curve at $x \approx -2.2$ and $x \approx 0.2$ , so the region is $-2.2 < x < 0.2$ .