WAEC 2016 · Paper 2 · Q10

(a)
Copy and complete the table of values for the relation $y = 2x^2 - 3x - 1$ .

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

$y$ $13$ $-1$

Model answer

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

$y$ $26$ $13$ $4$ $-1$ $-2$ $1$ $8$ $19$

For example $x = -3$ : $18 + 9 - 1 = 26$ , and $x = 1$ : $2 - 3 - 1 = -2$ .
(b)
Using scales of 2 cm to 1 unit on the $x$ -axis and 2 cm to 5 units on the $y$ -axis, draw the graph of $y = 2x^2 - 3x - 1$ for $-3 \le x \le 4$ .
Model answer
Plot every point from the table, then join them with one smooth curve (not straight lines between points).

For (c): (i) $2x^2 - 3x = 1$ is $y = 0$ , so read where the curve crosses the $x$ -axis: $x \approx −0.3 and 1.8$ . (ii) The gradient is 0 at the lowest point, $(0.75, -2.125)$ . (iii) Draw $y = 3x - 1$ (through $(0, -1)$ and $(3, 8)$ ); the curve is below the line between $x = 0$ and $x = 3$ , so shade the area between them there.
(c)
Using the graph: (i) solve the equation $2x^2 - 3x = 1$ ; (ii) find the coordinates of the point where the gradient of the curve is $0$ ; (iii) shade the area for which $y \le 3x - 1$ .
Separate values with commas, e.g. 3, −2

$x$	$-3$	$-2$	$-1$	$0$	$1$	$2$	$3$	$4$
$y$	$26$	$13$	$4$	$-1$	$-2$	$1$	$8$	$19$

The curve, the line y = 3x − 1, and the turning point.

Worked solution (try it first)

Plot the eight points with the given scales and join them with a smooth U-shaped curve (not straight lines).

$2x^2 - 3x = 1$ is the same as $2x^2 - 3x - 1 = 0$ , that is $y = 0$ .
Read where the curve crosses the $x$ -axis: $x \approx -0.3$ and $x \approx 1.8$ .

Draw the line $y = 3x - 1$ through $(0, -1)$ and $(3, 8)$ .
It meets the curve at $x = 0$ and $x = 3$ .
The points of the curve with $y \le 3x - 1$ are where the curve is on or below the line: shade the area between the line (above) and the curve (below), from $x = 0$ to $x = 3$ .