Inequalities on a graph · Inequalities

An inequality in two letters, such as $y \le \frac12x + 1$ , is satisfied by a whole region of the graph: every point on one side of a straight line.

y ≤ ½x + 1Solid line: points on the line count. Shade below.

y > ½x + 1Dashed line: points on the line don't count. Shade above.

The method

Draw the boundary line: replace the inequality sign with $=$ and plot two or three points.
Solid or dashed? Solid for $\le$ or $\ge$ (the line is included); dashed for $<$ or $>$ (it isn’t).
Test a point not on the line, usually the origin $(0, 0)$ . Put its coordinates into the inequality.
Shade: if the test point satisfies the inequality, its side is the region; if not, it’s the other side.

Try it

Regions on a graphSwitch inequalities on and off, and drag the test point

3 ≤ 6 ✓x + y ≤ 62 ≥ 1 ✓x ≥ 11 > 0 ✓y > 0

The test point is (2, 1). Put its coordinates into each inequality: it satisfies all of them, so it is inside the shaded region. A solid boundary line (≤, ≥) is part of the region; a dashed one (<, >) is not.

The region that satisfies all the inequalities is where every one of them is true at once. Drag the test point in and out of the shaded region and watch which inequality fails.

More: regions on a graph

Worked example · WAEC 2017

WAEC 2017 · Paper 2 · Q12 (b)

Using a scale of 2 cm to 1 unit on both axes, draw on a graph sheet the region which satisfies the following inequalities simultaneously: $y < x + 1$ ; $2y \ge -2x + 3$ ; $2x < 3$ ; $y + 1 > 0$ .

The boundary lines
$y = x + 1$ (dashed); $2y = -2x + 3$ , that is $y = -x + 1\frac12$ (solid); $2x = 3$ , that is $x = 1\frac12$ (dashed, a vertical line); $y + 1 = 0$ , that is $y = -1$ (dashed, a horizontal line).
Think first. Replace each sign with = and tidy up.
Test the origin
$0 < 0 + 1$ ✓ so shade the origin’s side of $y = x + 1$ (below it). $0 \ge 0 + 3$ ✗ so the region is the side of $2y = -2x + 3$ away from the origin (above it). $0 < 3$ ✓, to the left of $x = 1\frac12$ . $0 + 1 > 0$ ✓, above $y = -1$ .
Think first. Put (0, 0) into each inequality.
The region
The region is the triangle between $y = x + 1$ , $y = -x + 1\frac12$ and $x = 1\frac12$ , with corners at $\left(\frac14, 1\frac14\right)$ , $\left(1\frac12, 2\frac12\right)$ and $\left(1\frac12, 0\right)$ . The line $y = -1$ lies below the triangle, so it doesn’t cut any of it off.
Think first. Where do the lines meet?

Reading an inequality from a graph

To go the other way, find the equation of the boundary line, then choose the sign: $\ge$ or $>$ for the region above the line, $\le$ or $<$ for below, and solid or dashed to decide whether the equals sign is included.

The same idea works with a curve. On the graph of $y = x^2 - 2x - 3$ , " $x^2 - 2x - 3 < 0$ " means " $y < 0$ ": the part of the curve below the $x$ -axis, between the roots. Read the range of $x$ from the graph.

More: reading an inequality from a graph

Quadratic inequalities without a graph

You don’t need to draw the curve, only picture it. Get 0 on one side, then:

solve the equation to find the roots $a$ and $b$ (where the curve crosses the axis);
for a U-shaped curve (positive $x^2$ term), it is below the axis between the roots and above it outside them.

Between or outside< 0 between the roots; > 0 outside them

So $x^2 - 3x - 10 > 0$ becomes $(x - 5)(x + 2) > 0$ , with roots $-2$ and $5$ . “Greater than 0” is outside: $x < -2$ or $x > 5$ . If the $x^2$ term is negative, multiply through by $-1$ first and turn the sign round.

More: quadratic inequalities

Your turn

WAEC 2012 · Paper 2 · Q7 (a)

(a)
(i) Using a scale of 2 cm to 1 unit on both axes, draw on the same graph sheet the graphs of $y - \frac{3x}{4} = 3$ and $y + 2x = 6$ . (ii) From your graph, find the coordinates of the point of intersection of the two graphs. (iii) Show, on the graph sheet, the region satisfied by the inequality $y - \frac34x \ge 3$ .
Separate values with commas, e.g. 3, −2

Try it on a graph

The two lines; the shaded region is y − ¾x ≥ 3.

Worked solution (try it first)

(a)(i)

Rearrange each equation for $y$ : $y = \frac34x + 3$ and $y = 6 - 2x$ .
Plot points for each.
For $y = \frac34x + 3$ : $(-4, 0)$ , $(0, 3)$ , $(4, 6)$ .
For $y = 6 - 2x$ : $(0, 6)$ , $(1, 4)$ , $(3, 0)$ .
Join each set with a straight line.

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