WAEC 2019 · Paper 2 · Q11✱✱

(a)
Copy and complete the table of values for $y = x^2 + \frac12$ for $-4 \le x \le 4$ .

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

$y$ $9.5$ $0.5$

Model answer

$x$ −4 −3 −2 −1 0 1 2 3 4

$y$ 16.5 9.5 4.5 1.5 0.5 1.5 4.5 9.5 16.5

For example, at $x = -4$ : $y = 16 + 0.5 = 16.5$ .
(b)
Using a scale of 2 cm to 1 unit on the $x$ -axis and 2 cm to 2 units on the $y$ -axis, draw the graph of $y = x^2 + \frac12$ .
Model answer
Plot every point from the table, then join them with one smooth curve (not straight lines between points). The curve is symmetrical about the $y$ -axis, with its lowest point at $(0, 0.5)$ .

For (c): (i) $x^2 = 5$ is $x^2 + \frac12 = 5.5$ , so draw $y = 5.5$ : $x \approx \pm 2.2$ . (ii) $y$ decreases as $x$ increases on the left half, $-4 \le x < 0$ .
(c)
Use the graph to: (i) solve $x^2 = 5$ ; (ii) find the range of values of $x$ for which $y$ decreases as $x$ increases.
Separate values with commas, e.g. 3, −2

$x$	−4	−3	−2	−1	0	1	2	3	4
$y$	16.5	9.5	4.5	1.5	0.5	1.5	4.5	9.5	16.5

y = x² + ½ with the line y = 5.5.

Worked solution (try it first)

With 2 cm to 1 unit across and 2 cm to 2 units up, plot the nine points and join them with a smooth U-shaped curve, symmetrical about the $y$ -axis.

Add $\frac12$ to both sides of $x^2 = 5$ : $x^2 + \frac12 = 5.5$ , that is $y = 5.5$ .
Draw the line $y = 5.5$ .
It meets the curve at $x \approx -2.2$ and $x \approx 2.2$ .
(Exactly, $\pm\sqrt5 \approx \pm 2.24$ .)

Moving from left to right, the curve goes down until its lowest point at $x = 0$ .
So $y$ decreases as $x$ increases for $-4 \le x < 0$ .