# Graphing Inequalities

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10.3Graphing Inequalities

In this Math tutorial, you will learn:

• What is the shape and orientation of the first-order equations with one variable?
• The same for the first-order inequalities with one variable.
• How do we identify the direction of the solution set in linear inequalities with one variable?
• The same for linear inequalities with two variables.
• How to find the standard form of a linear inequality when its graph is given?
• What is the shape of a quadratic equation graph?
• How do we identify the direction of the solution set in quadratic inequalities?

## Introduction

In the previous tutorials on this chapter, we dealt extensively with inequalities - both linear and quadratic. We explained a number of methods to solve them, where it became clear that inequalities have more solutions than equations. As such, it would be appropriate to show inequalities in a graph, to have easier access to the solutions, as we did in the previous chapter with linear equations.

## Graphing First Order Inequalities with One Variable

Let's begin to explain how to graph the inequalities with the simplest type of inequalities - the first order inequalities with one variable. We learned in tutorial 10.1 that we have four possible inequalities of this type:

ax + b>0
ax + b ≥ 0
ax + b < 0

and

ax + b ≤ 0

where a and b are numbers and x is the variable.

Again, we will use the first-order equation with one variable

ax + b > 0

as a boundary line for the corresponding linear inequalities with one variable deriving from it (the four ones shown above).

From the previous tutorials, it is clear that the graph of a first-order equation with one variable represents a vertical line, as the graph concerns only the position of the coordinate x in a number line. Hence, when we represent graphically this type of equation on a XY coordinate system, it shows a vertical line because the y-coordinate does not matter. Look at the graph below. The graph shows the equation 4x - 5 = 0, where a = 4 and b = -5. From theory, we know that such equations have a general solution given by

x = -b/a

In the specific case, this solution is equal to

x = -(-5)/4
= 5/4
= 1.25

as shown in the graph.

Now, let's explain what we obtain when the solutions of the four inequalities deriving from the above equation are required. Thus, if we want to graph the inequality

ax + b > 0

First, we must isolate the variable x. Thus, taking the coefficient a as positive to avoid changes in the inequality sign, yields

ax + b > 0
ax > -b
x > -b/a

In our example, we have

4x - 5 > 0
4x > 5
x > 5/4

This means all solutions of this inequality extend to the right of the above value, as shown in the figure. In this way, it is obvious that the solution set of the inequality

ax + b > 0

extends to the right of the vertical line

x= -b/a

In the figure above, this solution set is shown by the coloured zone that extends on the right of the line x = 1.25.

The dashed line shows that the value x = 1.5 does not belong to the solution set of this inequality, as implied from the sign " > " of the original inequality, which excludes the boundary value as a solution. Hence, we obtain the following rule for this type of inequality:

In linear inequalities of the form ax + b > 0, the solution set represents all values to the right of the boundary value x = -b/a, without including it.

If the original inequality was

4x - 5 ≥ 0

instead, i.e. if the solution set was

x ≥ 1.25

the solution set would also include the vertical line (the boundary line) that shows the equation 4x - 5 = 0 (i.e. x = 1.25). In this case, the boundary line is not dashed but solid instead, as shown below. In this way, we reach the following conclusion about the linear inequality with one variable

ax + b ≥ 0

In linear inequalities of the form ax + b ≥ 0, the solution set represents all values to the right of the boundary value x = -b/a, including it.

Now, let's see what happens if we have to solve graphically the linear inequality with one variable

ax + b < 0

The general solution for this inequality is

ax + b < 0
ax < -b
x < -b/a

which includes all values on the left of the point

x = -b/a

without including it. Therefore, the boundary line x = -b/a is dashed when shown graphically. For example, in the first-order inequality with one variable

2x - 7 < 0

the solution set includes all values that are smaller than 3.5 (x < 3.5), as

2x - 7 < 0
2x < 7
x < 7/2
x < 3.5

When shown graphically, this solution set includes all values on the left of x = 3.5, and the corresponding boundary line is dashed, as the value 3.5 does not belong to the solution set of the original inequality. Look at the figure. In this way, we obtain the following rule for such inequalities:

In linear inequalities of the form ax + b < 0, the solution set represents all values to the left of the boundary value x = -b/a, without including it.

Last, if we have to solve graphically the linear inequality with one variable

ax + b ≤ 0

the solution set contains all values on the left of the boundary value x = -b/a including this one, as there is the combined inequality sign " " involved, which means the original inequality must be less than or equal to zero. For example, the solution set of the inequality

2x - 7 ≤ 0

includes all values from 3.5 to its left. Therefore the boundary value x = 3.5 is shown by a solid vertical line when solving graphically this inequality. Look at the figure. Hence, we obtain the following general rule for this case:

In linear inequalities of the form ax + b ≤ 0, the solution set represents all values to the left of the boundary value x = -b/a, including it.

### Example 1

Solve graphically the following inequalities and make the proof by taking one value of the variable from the solution set and another value outside the solution set.

1. 1 > -4x
2. -3x ≥ 6

### Solution 1

1. First, we must write this inequality in the standard form. We have
1 ≥ -4x
1 + 4x ≥ -4x + 4x
1 + 4x ≥ 0
1 + 4x - 1 ≥ 0 - 1
4x ≥ - 1
4x + 1 ≥ 0
The graph of this last inequality is shown below. Indeed, if we continued to solve the inequality analytically, we would obtain
4x/4-1/4
x ≤ -1/4
x ≤ -0.25
Now, let's pick one value from each interval - from the solution set and outside it - to confirm the correctness of our solution. Let's take for example x = 3, which belongs to the solution set. Obviously, it must be a value that gives a true inequality. We have
1 ≥ -4x
1 ≥ - 4 ∙ 3
1 ≥ -12 (true)
On the other hand, if we pick a number outside the solution set, for example, x = -2, we obtain
1 ≥ -4x
1 ≥ -4 ∙ (-2)
1 ≥ 8 (false)
Hence, our solution is already confirmed as correct.
2. Again, we write first the given inequality in the standard form. We have
-3x ≥ 6
-3x - 6 ≥ 0
Let's make the coefficient a positive to avoid confusion. Thus, multiplying both sides by -1 yields
(-1)(-3x - 6) ≥ (-1) ∙ 0
3x + 6 ≤ 0
When this inequality is graphed, we obtain Thus, the part of the coordinate plane that lies on the left of the boundary line x = -2, a line that derives from the solution of the inequality -3x ≥ 6, represents the solution set of this inequality. The boundary line x = -2 is also included in this solution set as the original inequality symbol is " ", which is a symbol that combines two other symbols in it: " = " and " > ".

When the variable is not denoted by x but by y, the graph will be horizontal. All the above rules are true except the orientation. Thus, if y > -b/a, the solution set includes the part above the graph without the graph line; if y < -b/a, the solution set includes the part below the graph without the graph line; if y ≥ -b/a, the solution set includes the part above the graph including the graph line as well; and if y ≤ -b/a, the solution set includes the part below the graph as well as the graph line.

For example, the graph of the linear inequality with one variable 4y - 12 ≥ 0 includes the zone above the line y = 3 and the line itself, as a = 4 and b = -12. Thus, since y ≥ -b/a, we have y ≥ - (-12)/4, i.e. y ≥ 3, as shown in the figure below. ## Graphing First Order Inequalities with Two Variables

From the theory explained in the previous tutorials, it is known that linear inequalities in two variables contain two variables at the first power. Their general form is one of the following

ax + by + c > 0
ax + by + c < 0
ax + by + c ≥ 0
ax + by + c ≤ 0

where a and b are coefficients, while c is a constant.

All of them derive from the corresponding linear equation with one variable

ax + by + c = 0

an equation which has a linear graph (hence the name "linear"). The slope of this graph (otherwise known as the "gradient") is obtained by the formula

k = -a/b

As we know, another form of writing a linear equation with one variable is to isolate the variable y and write it in terms of the other variable x in the form

y = mx + n

where m here represents the gradient k, while n is obtained by the formula

n = -c/b

It is better to have the linear inequalities written based on the second form of the corresponding linear equation y = mx + n, as this form allows us to better locate the position of the solution set for that inequality. In this way, we obtain the following four possible linear inequalities with two variables:

y > mx + n
y < mx + n
y ≥ mx + n
y ≤ mx + n

Thus, if we have the first linear inequality y > mx + n, the solution set includes all values (the zone) above the graph without the graph line, while in the second inequality y < mx + n the solution set includes all values (the zone) below the graph without the graph line.

On the other hand, the solution set of the third inequality y ≥ mx + n includes all values above the graph as well as those on the graph line, while the solution set of the fourth inequality y ≤ mx + n includes all values below the graph including those of the graph itself.

Let's consider an example to clarify this point.

### Example 2

Find graphically the solution set of the linear inequality

1. 3x - y + 2 > 0
2. -4x + 2y ≥ 5

### Solution 2

1. First, let's express the inequality in the form y (?) 2mx + n for an easier solution. We have
3x - y + 2 > 0
3x + 2 > y
Reading the last inequality from right to left yields
y < 3x + 2
In this way, we can identify the solution set, that consists of the part under the graph of the line y = 3x + 2 without including the line itself, as shown in the figure. 2. Again, let's write the inequality in the form y (?) mx + n to have a better idea of the solution set. We have
-4x + 2y ≥ 5
2y ≥ 4x + 5
2y/24x/2 + 5/2
y ≥ x + 5/2
In this way, we can identify the solution set, which consists of the part above the graph of the line y = x + 5/2 including also the line itself, as shown in the figure. ## Finding the Standard Form of a Linear Inequality from a Given Graph

Sometimes, we have the graph of a linear inequality given but not the inequality shown by that graph. To find the standard form of that inequality, we must first find the corresponding linear equation representing the boundary line of the inequality. For this, we need the coordinates of two known points A and B of the graph to calculate the gradient k by applying the formula

k = ∆y/∆x = yB - yA/xB - xA

Then, using the equation of the line

y = kx + n

we can find the constant n by substituting the coordinates of any from the known points.

Finally, looking at the highlighted region on the graph, you can determine the standard form of the given inequality.

### Example 3

What inequality is shown in the graph below? ### Solution 3

Let's consider two known points A(0, 1) and B(1, 3) as shown in the figure. k = ∆y/∆x
= yB - yA/xB - xA
= 3 - 1/1 - 0
= 2/1
= 2

Hence, in the line

y = 2x + n

We substitute for example the coordinates of point A (x = 0 and y = 1) in the above equation. Thus, we obtain

1 = 2 ∙ 0 + n
n = 1

Therefore, the boundary line (which is included in the solution set of our inequality) has the equation

y = 2x + 1

Since we have the region above the graph highlighted, the inequality shown by this graph is

y ≥ 2x + 1

## Graphing Second-Order Inequalities

As we have explained in the previous tutorials, a second-order equation is an extension of the concept of quadratic equations including a new variable y. This means the quadratic equation

ax2 + bx + c = 0

is a special case of the second-order equation with two variables

y = ax2 + bx + c

where y = 0. The line that represents the graph of second-order equations with two variables is not straight; this line is called a parabola. For example, the graph of the second-order equation with two variables y = x2 + 3x - 4 is shown in the figure below. Following the same reasoning used in the previous paragraphs on graphing the other types of inequalities, it is clear that if a second-order equation is expressed in the standard form as the one shown above, the following rules are true for the four corresponding inequalities (the condition is that a > 0):

1. The solution set of the inequality y > ax2 + bx + c includes the zone above the graph (parabola) but not the graph line.
2. The solution set of the inequality y < ax2 + bx + c includes the zone under the graph (parabola) but not the graph line.
3. The solution set of the inequality y ≥ ax2 + bx + c includes the zone above the graph (parabola) as well as the graph line.
4. The solution set of the inequality y ≤ ax2 + bx + c includes the zone under the graph (parabola) as well as the graph line.

Let's clarify this point through an example.

### Example 4

Solve graphically the following inequalities.

1. 2x2 - y + 3 < 5x
2. 2y - x - 4 ≤ 5x2

### Solution 4

1. First, we must turn the given inequality in the form y (?) ax2 + bx + c, where (?) represents one of the four inequality symbols. We have
2x2 - y + 3 < 5x
2x2 - 5x + 3 < y
Looking at the above inequality from right to left yields
y > 2x2 - 5x + 3

Since the graph is not linear anymore, we need more than two points to plot it. The more points we consider, the more accurate the graph is. We will not dwell too much on this point, as in the upcoming chapters we will explain extensively how to find some special points contained on a parabola that help us plot the graph easier. However, we anticipate that a parabola graph is plotted by joining at least four special points:

1. The two x-intercepts A and B that have the vertical (y) coordinate zero (they are obtained by solving the corresponding quadratic graph ax2 + bx + c = 0, as the x-intercepts represent the x-coordinate of the roots). Thus, we have A(x1, 0) and B(x2, 0);
2. The y-intercept C(0, y) that is obtained by taking x = 0 in the corresponding quadratic equation; and
3. The vertex V of the parabola, represents the minimum or maximum point of the parabola. It has the coordinates V(-b/2a, /4a), where Δ is the discriminant of the corresponding quadratic equation.
Finding all the above points (and some others if possible), we obtain the following graph for the quadratic inequality solved above - a graph that includes only the region above the parabola without the parabola line. 2. Here again, we must turn the given inequality in the form y (?) ax2 + bx + c. We have
2y - x - 4 ≤ 5x2
2y ≤ 5x2 + x + 4
2y/25x2/2 + x/2 + 4/2
y ≤ 5x2/2 + x/2 + 2
or
y ≤ 5/2 x2 + 1/2 x + 2
Applying the procedure described in (a) for the graph plotting, we obtain the following graph, which includes the region below the parabola as well as the parabola itself, because the sign of the inequality is " ". ## Whats next?

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