Graphing Nonlinear Inequalities and Systems of Nonlinear Inequalities

All of the equations in the systems that we have encountered so far have involved equalities, but we may also encounter systems that involve inequalities. We have already learned to graph linear inequalities by graphing the corresponding equation, and then shading the region represented by the inequality symbol. Now, we will follow similar steps to graph a nonlinear inequality so that we can learn to solve systems of nonlinear inequalities. A nonlinear inequality is an inequality containing a nonlinear expression. Graphing a nonlinear inequality is much like graphing a linear inequality. Recall that when the inequality is greater than, [latex]y>a[/latex], or less than, [latex]y<a,\text{}[/latex] the graph is drawn with a dashed line. When the inequality is greater than or equal to, [latex]y\ge a,\text{}[/latex] or less than or equal to, [latex]y\le a,\text{}[/latex] the graph is drawn with a solid line. The graphs will create regions in the plane, and we will test each region for a solution. If one point in the region works, the whole region works. That is the region we shade.

Figure 8. (a) an example of [latex]y>a[/latex]; (b) an example of [latex]y\ge a[/latex]; (c) an example of [latex]y<a[/latex]; (d) an example of [latex]y\le a[/latex]

Solution

First, graph the corresponding equation [latex]y={x}^{2}+1[/latex]. Since [latex]y>{x}^{2}+1[/latex] has a greater than symbol, we draw the graph with a dashed line. Then we choose points to test both inside and outside the parabola. Let’s test the points [latex]\left(0,2\right)[/latex] and [latex]\left(2,0\right)[/latex]. One point is clearly inside the parabola and the other point is clearly outside.

[latex]\begin{array}{ll}y>{x}^{2}+1\hfill & \hfill \\ 2>{\left(0\right)}^{2}+1\hfill & \hfill \\ 2>1\hfill & \text{True}\hfill \\ \hfill & \hfill \\ \hfill & \hfill \\ \hfill & \hfill \\ 0>{\left(2\right)}^{2}+1\hfill & \hfill \\ 0>5\hfill & \text{False}\hfill \end{array}[/latex]

The graph is shown in Figure 9. We can see that the solution set consists of all points inside the parabola, but not on the graph itself.

Figure 9

Graphing a System of Nonlinear Inequalities

Solution

These two equations are clearly parabolas. We can find the points of intersection by the elimination process: Add both equations and the variable [latex]y[/latex] will be eliminated. Then we solve for [latex]x[/latex].

[latex]\begin{array}\hfill x^{2}−y=0 \\ 2x^{2}+y=12 \\ \text{___________} \\ 3x^{2}=12 \\ x^{2}=4 \\ x=\pm 2\end{array}[/latex]

Substitute the x-values into one of the equations and solve for [latex]y[/latex].

[latex]\begin{array}{r}\hfill {x}^{2}-y=0\\ \hfill {\left(2\right)}^{2}-y=0\\ \hfill 4-y=0\\ \hfill y=4\\ \hfill \\ \hfill {\left(-2\right)}^{2}-y=0\\ \hfill 4-y=0\\ \hfill y=4\end{array}[/latex]

The two points of intersection are [latex]\left(2,4\right)[/latex] and [latex]\left(-2,4\right)[/latex]. Notice that the equations can be rewritten as follows.

[latex]\begin{array}{l}{x}^{2}-y\le 0\hfill \\ {x}^{2}\le y\hfill \\ y\ge {x}^{2}\hfill \\ \hfill \\ 2{x}^{2}+y\le 12\hfill \\ y\le -2{x}^{2}+12\hfill \end{array}[/latex]

Graph each inequality. The feasible region is the region between the two equations bounded by [latex]2{x}^{2}+y\le 12[/latex] on the top and [latex]{x}^{2}-y\le 0[/latex] on the bottom.

Figure 10

Licenses & Attributions