WAEC 2008 · Paper 2 · Q8✱✱

(a)
Copy and complete the table of values for the relation $y = -3x^2 + 12x$ for $-2 \le x \le 6$ .

$x$ $-2$ $-1$ $0$ $1$ $2$ $3$ $4$ $5$ $6$

$y$ $-36$ $0$ $0$

Model answer

$x$ $-2$ $-1$ $0$ $1$ $2$ $3$ $4$ $5$ $6$

$y$ $-36$ $-15$ $0$ $9$ $12$ $9$ $0$ $-15$ $-36$

The row is symmetric about $x = 2$ , where $y$ is greatest (12).
(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 $y = -3x^2 + 12x$ .
Model answer
Plot the nine points from the table and join them with a smooth, upside-down U-shaped curve. It crosses the $x$ -axis at $0$ and $4$ and has its highest point, $(2, 12)$ , on the line $x = 2$ .
(c)
On the same axes, draw the line $y = 5x - 10$ .
Model answer
Plot, for example, $(-2, -20)$ , $(0, -10)$ and $(4, 10)$ and join them with a straight line. It meets the curve at $(-1, -15)$ and at about $(3.3, 6.7)$ .
(d)
On your graph, shade the region for which $7x + 10 > 3x^2$ .
Model answer
Rearranged, $7x + 10 > 3x^2$ is $-3x^2 + 12x > 5x - 10$ : the curve is above the line. Shade the region enclosed between the curve (above) and the line (below), from $x = -1$ to $x = 3\frac13$ .

$x$	$-2$	$-1$	$0$	$1$	$2$	$3$	$4$	$5$	$6$
$y$	$-36$	$-15$	$0$	$9$	$12$	$9$	$0$	$-15$	$-36$

The curve and the line; the region for (d) lies between them.

Worked solution (try it first)

Take $5x$ from both sides of $7x + 10 > 3x^2$ and rearrange: $-3x^2 + 12x > 5x - 10$ .
So the region is where the curve lies above the line.
The curve and line meet where $3x^2 - 7x - 10 = 0$ , i.e. $(3x - 10)(x + 1) = 0$ , so $x = -1$ and $x = 3\frac13$ .
Shade the region between the line and the curve from $x = -1$ to $x = 3\frac13$ .