The table-and-graph question · Quadratics & their graphs

Almost every General Mathematics Paper 2 has a question like this, and it’s worth a lot of marks. Most candidates complete the table, and many draw a reasonable graph. The marks are lost in the last part: reading answers from the graph. This lesson is mostly about that.

The four jobs

Complete the table by putting each $x$ into the equation.
Choose and state your scale, exactly as the question asks, and plot the points.
Draw one smooth curve through all the points.
Read the answers from the graph.

Try it

Table, graph, answersFill in the table

y = 2x² − x − 4

x	−3	−2	−1	0	1	2	3
y	17			−4

Put each x into y = 2x² − x − 4. Every value you get right is plotted at once. Take care with negative x: square it first.

Fill in the table, draw the curve, then use the slider to move the line $y = k$ and watch where it meets the curve.

Job 1: the table

Work out each value carefully, one term at a time. For $y = 2x^2 - x - 4$ at $x = -2$ :

2(-2)^2 - (-2) - 4 = 2(4) + 2 - 4 = 6

Jobs 2 and 3: plotting and drawing

Use the scale in the question (for example, 2 cm to 1 unit on the $x$ -axis). If it asks you to state the scale you used, write it down: that line alone carries a mark.
Join the points with one smooth curve, drawn freehand. Never join them with straight lines with a ruler.
Don’t flatten the bottom. The lowest point is usually between two plotted points, not on one.

Job 4: reading from the graph

This is where most marks are lost, so learn these three moves.

The roots of $ax^2 + bx + c = 0$ . Read the $x$ -values where the curve crosses the $x$ -axis (the line $y = 0$ ). Give them to 1 decimal place.

Solving $ax^2 + bx + c = k$ . Draw the straight line $y = k$ across your graph. Where it cuts the curve, read down to the $x$ -axis. Those $x$ -values are the solutions. On the board above, that’s exactly what the slider does.

Solving a different equation using the same graph. Rearrange the new equation so that one side is exactly your curve’s expression. For example, if the graph is $y = 2x^2 + x - 10$ and you must solve $2x^2 + x - 10 = 2x$ , the other side is $2x$ . So draw the line $y = 2x$ and read the $x$ -values where it crosses the curve.

The turning point. The minimum (or maximum) point is on the line of symmetry, halfway between the roots. Read its coordinates from the graph; you can check the $x$ -value with $x = -\frac{b}{2a}$ .

A past question, step by step

Worked example · WAEC 2023 Paper 2, Q8

WAEC 2023 · Paper 2 · Q8

Copy and complete the table of values for $y = 2x^2 - x - 4$ for $-3 \le x \le 3$ .

$x$	−3	−2	−1	0	1	2	3
$y$	17			−4

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 $y = 2x^2 - x - 4$ for $-3 \le x \le 3$ .

Use the graph to find the: (i) roots of the equation $2x^2 - x - 4 = 0$ ; (ii) values of $x$ for which $y$ increases as $x$ increases; (iii) minimum point of $y$ . Enter the two roots for (i).

Complete the table
Take care with the signs: $2(-2)^2 = 8$ and $-(-2) = +2$ .
$x = -2:\; 8 + 2 - 4 = 6 \qquad x = 1:\; 2 - 1 - 4 = -3$
Think first. Work out y when x = −2 before going on.
Plot the points
The full row is $17, 6, -1, -4, -3, 2, 11$ . Use the scales given, 2 cm to 1 unit on the $x$ -axis and 2 cm to 2 units on the $y$ -axis, and plot every point.
Draw a smooth curve
One smooth U-shape through all seven points. The bottom lies between $x = 0$ and $x = 1$ , not at a plotted point.
Think first. Where does the curve cross the x-axis?
(c)(i) The roots
$2x^2 - x - 4 = 0$ means $y = 0$ , so read where the curve crosses the $x$ -axis:
$x \approx -1.2 \quad\text{and}\quad x \approx 1.7$
Think first. The curve turns round on its line of symmetry. Where is that line?
(c)(ii) Where y increases
Moving to the right, the curve goes down until the lowest point and then up. The lowest point is on the line of symmetry $x = -\frac{b}{2a} = \frac14$ . So $y$ increases for $0.25 < x \le 3$ .
(c)(iii) The minimum point
$y = 2(0.25)^2 - 0.25 - 4 = -4.125$
From the graph, about $(0.3, -4.1)$ ; any reading close to this is accepted.

Your turn

This one also asks you to solve a second equation with the same graph.

WAEC 2018 · Paper 2 · Q11

Quadratics & their graphs

(a)
Copy and complete the table of values for $y = 2x^2 + x - 10$ for $-5 \le x \le 4$ .

$x$ $-5$ $-4$ $-3$ $-2$ $-1$ $0$ $1$ $2$ $3$ $4$

$y$ $5$ $-9$ $-10$ $0$

Model answer

$x$ −5 −4 −3 −2 −1 0 1 2 3 4

$y$ 35 18 5 −4 −9 −10 −7 0 11 26

For example, at $x = -5$ : $y = 2(25) - 5 - 10 = 35$ .
(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 $y = 2x^2 + x - 10$ for $-5 \le x \le 4$ .
Model answer
Plot every point from the table, then join them with one smooth curve (not straight lines between points).

For (c): (i) $2x^2 + x = 10$ is $y = 0$ : the roots are $x = -2.5$ and $x = 2$ . (ii) $2x^2 + x - 10 = 2x$ : draw the line $y = 2x$ ; it meets the curve at $(-2, -4)$ and $(2.5, 5)$ , so $x = -2$ or $x = 2.5$ .
(c)
Use the graph to find the solution of: (i) $2x^2 + x = 10$ ; (ii) $2x^2 + x - 10 = 2x$ .
Show the answer
(i) $x = -2.5$ or $2$ ; (ii) $x = -2$ or $2.5$

$x$	−5	−4	−3	−2	−1	0	1	2	3	4
$y$	35	18	5	−4	−9	−10	−7	0	11	26

Try it on a graph

The curve and the line y = 2x.

Worked solution (try it first)

(a)

Put each $x$ into $y = 2x^2 + x - 10$ .
For $x = -5$ : $50 - 5 - 10 = 35$ .
For $x = -4$ : $32 - 4 - 10 = 18$ .
For $x = -2$ : $8 - 2 - 10 = -4$ .
For $x = 1$ : $2 + 1 - 10 = -7$ .
For $x = 3$ : $18 + 3 - 10 = 11$ .
For $x = 4$ : $32 + 4 - 10 = 26$ .
The row is $35, 18, 5, -4, -9, -10, -7, 0, 11, 26$ .

(b)

With 2 cm to 1 unit across and 2 cm to 5 units up, plot the ten points and join them with a smooth U-shaped curve.

(c)(i)

$2x^2 + x = 10$ is the same as $2x^2 + x - 10 = 0$ , that is $y = 0$ .
The curve crosses the $x$ -axis at $x = -2.5$ and $x = 2$ .

(ii)

$2x^2 + x - 10 = 2x$ means the curve meets the line $y = 2x$ .
Draw the line through $(-3, -6)$ , $(0, 0)$ and $(3, 6)$ .
It meets the curve at $x = -2$ and $x = 2.5$ .
(As a check, the equation is $2x^2 - x - 10 = 0$ , which factorises as $(x + 2)(2x - 5) = 0$ .)

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