WAEC 2010 · Paper 2 · Q7

(a)
Copy and complete the following table for the relation $y = \frac12x(x - 6)$ for $-2 \le x \le 8$ .

$x$ $-2$ $-1$ $0$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$

$y$ $8$ $0$ $-4$ $0$

Separate values with commas, e.g. 3, −2
(b)
Using scales of $2\text{ cm}$ to 1 unit on the $x$ -axis, and $2\text{ cm}$ to 2 units on the $y$ -axis, draw the graph of the relation $y = \frac12x(x - 6)$ for $-2 \le x \le 8$ .
Model answer
Plot all eleven points from the table (2 cm to 1 unit across, 2 cm to 2 units up) and join them with one smooth U-shaped curve, symmetrical about $x = 3$ , with its lowest point at $(3, -4.5)$ .

For (c): the curve is below the $x$ -axis for $0 < x < 6$ ; the minimum value is $-4.5$ ; the line $y = 5$ meets it at $x \approx -1.4$ and $x \approx 7.4$ .
(c)
Use the graph to find the: (i) range of values of $x$ for which $y$ is negative; (ii) minimum value of $y$ ; (iii) roots of the equation $\frac12x(x - 6) = 5$ .
Separate values with commas, e.g. 3, −2

The x-axis gives (c)(i); the line y = 5 gives (c)(iii).

Worked solution (try it first)

Plot the eleven points with the scales given and join them with one smooth curve.
It is symmetrical about $x = 3$ .

$y$ is negative where the curve is below the $x$ -axis: between the crossings at $x = 0$ and $x = 6$ , so $0 < x < 6$ .

The lowest point of the curve is $(3, -4.5)$ , so the minimum value of $y$ is $-4.5$ .

Draw the line $y = 5$ and read down from where it meets the curve: $x \approx -1.4$ and $x \approx 7.4$ .
Check: $\frac12x(x - 6) = 5$ gives $x^2 - 6x - 10 = 0$ , so $x = 3 \pm \sqrt{19} = -1.36$ or $7.36$ .