WAEC 2019 · Paper 2 · Q12

(a)

Copy and complete the table of values for the relation $y = 4x^2 - 8x - 21$ , for $-2.0 \le x \le 4.0$ .

$x$	$-2.0$	$-1.5$	$-1.0$	$-0.5$	$0.0$	$0.5$	$1.0$	$1.5$	$2.0$	$2.5$	$3.0$	$3.5$	$4.0$
$y$	$11$		$-9$		$-21$	$-24$			$-21$		$-9$	$0$

Model answer

$x$	−2.0	−1.5	−1.0	−0.5	0.0	0.5	1.0	1.5	2.0	2.5	3.0	3.5	4.0
$y$	11	0	−9	−16	−21	−24	−25	−24	−21	−16	−9	0	11

For example, at $x = 1.0$ : $y = 4 - 8 - 21 = -25$ .

(b)(i)
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 $y = 4x^2 - 8x - 21$ .
Model answer
Plot every point from the table, then join them with one smooth curve (not straight lines between points). The lowest point is $(1, -25)$ .

For (ii): ( $\alpha$ ) $4x^2 - 8x = 3$ is $y = -18$ , giving $x \approx −0.3 and 2.3$ . ( $\beta$ ) $4x^2 - 7x - 21 = 0$ is $4x^2 - 8x - 21 = -x$ , so draw $y = -x$ : it meets the curve at $x \approx −1.6 and 3.3$ .
(b)(ii)(α)
Use the graph to find the solution set of $4x^2 - 8x = 3$ .
Separate values with commas, e.g. 3, −2
(b)(ii)(β)
Use the graph to find the solution set of $4x^2 - 7x - 21 = 0$ .
Separate values with commas, e.g. 3, −2

The curve with y = −18 (red) and y = −x (green).

Worked solution (try it first)

With 2 cm to 1 unit across and 2 cm to 5 units up, plot the thirteen points and join them with a smooth U-shaped curve.
It is symmetrical about $x = 1$ , where $y = -25$ .

Take $x$ from both sides of $4x^2 - 7x - 21 = 0$ : it becomes $4x^2 - 8x - 21 = -x$ .
So draw the line $y = -x$ through $(-2, 2)$ , $(0, 0)$ and $(4, -4)$ .
It cuts the curve at $x \approx -1.6$ and $x \approx 3.3$ .
The solution set is $\{-1.6, 3.3\}$ .