Intermediate Algebra: Functions and Graphs

Subsection Graphs of Inequalities in Two Variables

We can gain some insight into the nature of the solutions if we rewrite the inequality as

\begin{gather*} y \ge − x + 10,000 \end{gather*}

This inequality says that for each \(x\)-value, we must choose points with \(y\)-values greater than or equal to \(-x + 10,000\text{.}\) As we saw in the Example above, when \(x = 2000\text{,}\) the solutions have \(y\)-values greater than or equal to 8000. Solutions for \(x = 2000\text{,}\) \(x = 6000\text{,}\) and \(x = 11,000\) are shown in the figure. Do you see a pattern emerging?

The graph of the inequality must show all the points whose coordinates are solutions. We found some solutions to the inequality \(x + y \ge 10,000\) in the Example above by choosing a value for \(x\) and solving for \(y\text{.}\) A more efficient way to find all the solutions of the inequality is to start with the graph of the corresponding equation

\begin{gather*} y = -x + 10,000 \end{gather*}

The graph is a straight line with slope \(-1\text{.}\) Any point above this line has a \(y\)-coordinate greater than \(~-x + 10,000~\) and hence satisfies the inequality. Thus, the graph of the inequality includes all the points on or above the line \(~y = -x + 10,000~\text{,}\) as shown by the shaded region in the figure. Of course, the shaded points are also solutions to the original inequality, \(~x + y \ge 10,000~\text{.}\)

For example, the point \((-1000, ~12,000)\) is a solution, because

On the other hand, the point \((5000, 4000)\) does not lie in the shaded region because its coordinates do not satisfy the inequality.

Subsection Linear Inequalities

A linear inequality can be written in the form

The solutions consist of the line \(ax + by + c = 0\) and a half-plane on one side of that line. We shade the half-plane to show that all its points are included in the solution set. To decide which side of the line to shade, we can solve the inequality for \(y\) in terms of \(x\text{.}\) If we obtain

then we shade the half-plane above the line. If the inequality is equivalent to

then we shade the half-plane below the line.

Example 2.47.

Graph \(4x - 3y \ge 12\)

Solution

We solve the inequality for \(y\text{.}\)

\begin{align*} 4x - 3y \amp\ge 12 \amp \amp \blert{\text{Subtract}~ 4x~ \text{from both sides.}}\\ -3y\amp\ge -4x + 12 \amp \amp \blert{\text{Divide both sides by}~ -3.}\\ y\amp\le \frac{4}{3}x-4 \end{align*}

We graph the corresponding line

\begin{gather*} y = \frac{4}{3}x-4 \end{gather*}

with \(y\)-intercept is \(-4\) and slope \(m=\dfrac{4}{3}\text{.}\) Finally, we shade the half-plane below the line. The completed graph is shown below.

An inequality that uses \(\gt\) or \(\lt\) instead of \(\ge\) or \(\le\) is called strict. The graph of a strict inequality includes only the half-plane and not the line. In that case we use a dashed line for the graph of the equation \(ax + by + c = 0\) to show that it is not part of the solution.

Subsection Using a Test Point

A second method for graphing inequalities does not require us to solve for \(y\text{.}\) Once we have graphed the boundary line, we can decide which half-plane to shade by using a test point. The test point can be any point that is not on the boundary line itself.

Example 2.51.

Graph the solutions of the inequality \(~~3x - 2y\lt 6\)

Solution

First, we graph the line \(3x - 2y = 6\text{,}\) as shown below. We will use the intercept method. The intercepts are \((2, 0)\) and \((0, -3)\text{,}\) so we sketch the boundary line through those points.

Next, we choose a test point. Because \((0, 0)\) does not lie on the line, we choose it as our test point. We substitute the coordinates of the test point into the inequality to obtain

\begin{equation*} 3(\alert{0}) - 2(\alert{0}) \lt 6 \end{equation*}

Because this is a true statement, \((0, 0)\) is a solution of the inequality. Since all the solutions lie on the same side of the boundary line, we shade the half-plane that contains the test point. In this example, the boundary line is a dashed line because the original inequality was strict.

Here is a summary of our test point method for graphing inequalities.

We can choose any point for the test point as long as it does not lie on the boundary line. We chose \((0, 0)\) in Example 2.51 because the coordinates are easy to substitute into the inequality. If the test point is a solution, we shade the half-plane including that point. If the test point is not a solution, we shade the other half-plane. For example, suppose we had chosen \((5, 0)\) as the test point in Example 2.51. When we substitute its coordinates into the inequality, we find

which is a false statement. Thus, \((5, 0)\) is not a solution to the inequality, so the solutions must lie on the other side of the boundary line. Note that using \((5, 0)\) as the test point gives us the same solutions we found in Example 2.51.

Recall that the equation of a vertical line has the form

where \(k\) is a constant, and a horizontal line has an equation of the form

Similarly, the inequality \(x\ge k\) may represent the inequality in two variables

Its graph is then a region in the plane.

Example 2.54.

Graph \(~~x\ge 2~~\) in the plane.

Solution

First, we graph the equation \(x = 2\text{;}\) its graph is a vertical line. Because the origin does not lie on this line, we can use it as a test point. Substitute \(0\) for \(x\) (there is no \(y\)) into the inequality to obtain

\begin{gather*} 0\ge 2 \end{gather*}

Since this is a false statement, we shade the half-plane that does not contain the origin. We see in the figure below that the graph of the inequality contains all points whose \(x\)-coordinates are greater than or equal to \(2\text{.}\)

Subsection Systems of Inequalities

Some applications are best described by a system of two or more inequalities. The solutions to a system of inequalities include all points that are solutions to each inequality in the system. The graph of the system is the intersection of the shaded regions for each inequality in the system. For example, the figure at right shows the solutions of the system

\begin{gather*} y \gt x \qquad\text{ and }\qquad y \gt 2 \end{gather*}

To describe the solutions of a system of inequalities, it is useful to locate the vertices, or corner points, of the boundary.

Example 2.58.

Graph the solution set of the system below and find the coordinates of its vertices.

\begin{align*} x - y - 2 \amp\le 0\\ x + 2y - 6 \amp\le 0\\ x \ge 0, ~~~~~y \amp\le 0 \end{align*}

Solution

The last two inequalities, \(x\ge 0\) and \(y\ge 0\text{,}\) restrict the solutions to the first quadrant. We graph the line \(x-y-2 = 0\text{,}\) and use the test point \((0, 0)\) to shade the half-plane including the origin. Finally we graph the line \(x - 2y - 6 = 0\) and again use the test point \((0, 0)\) to shade the half-plane below the line. The intersection of the shaded regions is shown at right.

To find the coordinates of the vertices \(A\text{,}\) \(B\text{,}\) \(C\text{,}\) and \(D\text{,}\) we solve simultaneously the equations of the two lines that intersect at each vertex. Thus,

For \(A\text{,}\) we solve the system

\begin{align*} x \amp= 0\\ y\amp =0 \end{align*}

to find \((0, 0)\)

For \(B\text{,}\) we solve the system

\begin{align*} x \amp= 0\\ x + 2y \amp= 6 \end{align*}

to find \((0, 3)\)

For \(C\text{,}\) we solve the system

\begin{align*} x + 2y \amp= 6\\ x -y \amp= 2 \end{align*}

to find \(\left(\dfrac{10}{3},\dfrac{4}{3} \right)\)

For \(D\text{,}\) we solve the system

\begin{align*} y \amp= 0\\ x -y \amp= 2 \end{align*}

to find \(\left(2,0 \right)\)

The vertices are the points \((0, 0)\text{,}\) \((0, 3)\text{,}\) \(\left(\dfrac{10}{3},\dfrac{4}{3} \right)\text{,}\) and \((2, 0)\text{.}\)