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Summary

Inequalities on graphs

In a nutshell

Graphical inequalities can be interpreted to identify solutions to inequality problems.

Solutions to inequalities

When being asked to solve an inequality of the form $f(x)<g(x)$ or $f(x)>g(x)$ , it may be helpful to sketch the graphs of $y=f(x)$ and $y=g(x)$ . The inequality can be solved by looking at their relative positions on the graph.

inequality	solution
$f(x)> g(x)$ , $f(x)\geq g(x)$	Points where the graph $y=f(x)$ is above the graph $y=g(x)$ .
$f(x)<g(x)$ , $f(x)\leq g(x)$	Points where the graph $y=f(x)$ is below the graph $y=g(x)$ .

Example

$L_{1}$ has the equation $y= x^2+2x$ .

$L_{2}$ has the equation $y=-3x+6$ .

The diagram shows a sketch of $L_{1}$ and $L_{2}$ .

i) Find the coordinates of the points of intersection, $P_{1}$ and $P_2$ .

ii) Find the solution to the inequality $x^2+2x \lt -3x+6$ .

Maths; Equations and inequalities; KS5 Year 12; Inequalities on graphs

Part i):

The points of intersection can be found where $L_1 = L_2$ :

$x^2+2x = -3x+6$

Rearrange and solve to find $x$ :

$x^2+5x-6 = 0\\ x = \dfrac{-5 \pm \sqrt{5^2 - 4(1)((-6)}}{2(1)}$

$x=1,\ x=-6$ 

Substitute these values for $x$ into one of the equations to find coordinates of $P_1$ and $P_2$ :

$L_1: y=-3x+6$

When $x=1$ :

$y = -3(1)+6 = 3$ 

When $x=-6$ :

$y=-3(-6)+6 = 24$ 

Therefore:

The points of intersection are $\underline{P_1:(6,24)}$ and $\underline{P_2:(1,3)}$ .

Part ii):

The solution to the inequality is when the graph of $y=-3x+6$ is above the graph of $y=x^2+2x$ .

Therefore:

$\underline{\{x:-6\lt x\lt 1\}}$ 

Learn with Basics

Length:

Unit 1

Inequalities: Greater than or less than

Unit 2