WAEC 2010 · Paper 2 · Q10✱✱

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

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

$y$

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 a graph of the relation $y = -x^2 + x + 2$ .
Model answer
Plot the seven points from the table and join them with one smooth curve (not straight lines). Scale: 2 cm to 1 unit on $x$ , 2 cm to 2 units on $y$ . The curve opens downwards, crosses the $x$ -axis at $x = -1$ and $x = 2$ , and is highest at $(0.5, 2.25)$ .
(c)
From the graph, find the: (i) minimum value of $y$ ; (ii) roots of the equation $x^2 - x - 2 = 0$ ; (iii) gradient of the curve at $x = -0.5$ .
Separate values with commas, e.g. 3, −2

The x-axis gives (c)(ii); the tangent at x = −0.5 gives (c)(iii).

Worked solution (try it first)

Plot the seven points with the scales given and join them with one smooth curve.
It opens downwards, with its top at $x = 0.5$ , $y = 2.25$ .

For $-3 \le x \le 3$ the lowest point of the curve is at the left end, $x = -3$ : the minimum value of $y$ is $-10$ .

$x^2 - x - 2 = 0$ is the same as $-x^2 + x + 2 = 0$ , so read where the curve crosses the $x$ -axis: $x = -1$ and $x = 2$ .

Draw the tangent to the curve at $(-0.5, 1.25)$ .
It passes through about $(-1.5, -0.75)$ and $(0.5, 3.25)$ : a rise of 4 over a run of 2, so the gradient is $2$ .