WAEC 2009 · Paper 2 · Q11

(a)
Copy and complete the table of values for $y = x^2 - 2$ for $-3 \le x \le 4$ .

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

$y$ $2$ $-2$ $2$

Model answer

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

$y$ $7$ $2$ $-1$ $-2$ $-1$ $2$ $7$ $14$

The row is symmetric about $x = 0$ : $x = 3$ and $x = -3$ both give 7.
(b)
Using a scale of 2 cm to 1 unit on the $x$ -axis and 2 cm to 2 units on the $y$ -axis, draw the graph of $y = x^2 - 2$ .
Model answer
Plot the eight points and join them with a smooth U-shaped curve through $(0, -2)$ . For (c): it cuts the $x$ -axis at $x \approx \pm 1.4$ ; the line $y = 1$ meets it at $x \approx \pm 1.7$ ; the tangent at $x = -1$ has gradient $-2$ .
(c)(i)
Use your graph to find the roots of the equation $x^2 - 2 = 0$ .
Separate values with commas, e.g. 3, −2
(c)(ii)
Use your graph to find the values of $x$ for which $x^2 - 3 = 0$ .
Separate values with commas, e.g. 3, −2
(c)(iii)
Use your graph to find the gradient of the curve at the point where $x = -1$ .

$x$	$-3$	$-2$	$-1$	$0$	$1$	$2$	$3$	$4$
$y$	$7$	$2$	$-1$	$-2$	$-1$	$2$	$7$	$14$

The x-axis gives (c)(i); the line y = 1 gives (c)(ii).

Worked solution (try it first)

Square each $x$ and subtract 2: for example, $x = -3$ gives $9 - 2 = 7$ and $x = 4$ gives $16 - 2 = 14$ .
The row is $7, 2, -1, -2, -1, 2, 7, 14$ .

The roots are where the curve crosses the $x$ -axis: $x \approx -1.4$ and $x \approx 1.4$ ( $\pm\sqrt2 = \pm 1.41$ ).

Write $x^2 - 3 = 0$ as $x^2 - 2 = 1$ .
Draw the line $y = 1$ and read down: $x \approx -1.7$ and $x \approx 1.7$ ( $\pm\sqrt3 = \pm 1.73$ ).

Draw the tangent to the curve at $(-1, -1)$ and pick two points on it, such as $(-2, 1)$ and $(0, -3)$ .
Gradient $= \frac{-3 - 1}{0 - (-2)}$
$= \frac{-4}{2}$
$= -2$ .