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Graphing simultaneous quadratic inequalities

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Graphing Systems of Quadratic Inequalities

This lesson shows how to graph a system of two quadratic inequalities: drawing each boundary parabola, deciding solid or dashed and which side to shade, then identifying the overlap as the final solution region, with two fully worked examples.

What Is a System of Quadratic Inequalities?

A system of quadratic inequalities is simply two (or more) quadratic inequalities that must be true at the same time. Each inequality on its own describes a shaded region of the coordinate plane bounded by a parabola, similar to what you see when graphing a single quadratic inequality in two variables. When you combine two of these inequalities into a system, the solution is no longer one shaded region, it is the overlap of both regions.

Before graphing, it helps if each quadratic is written in general form, \(y = ax^2 + bx + c\), so you can quickly find the vertex and direction of opening. If you need a refresher, see quadratic function in general form.

Steps to Graph a System of Quadratic Inequalities

  1. Treat each inequality as an equation and graph its boundary parabola.
  2. Use a dashed curve for \(<\) or \(>\), and a solid curve for \(\le\) or \(\ge\).
  3. Pick a test point not on the curve and check whether it satisfies the inequality; shade the region containing all points that work.
  4. Repeat steps 1 to 3 for the second inequality on the same graph.
  5. The solution to the system is the region where both shaded areas overlap.

If you also need the exact points where the two parabolas cross, that is the same algebra used in solving systems of quadratic equations, since the boundary curves of the inequalities are quadratic equations in disguise.

Example 1: Parabolas Opening Toward Each Other

Graph the solution to the system:

\(y > x^2 - 4\)

\(y < -x^2 + 4\)

The first boundary, \(y = x^2 - 4\), opens upward with vertex \((0, -4)\). Since the inequality is strict (\(>\)), draw it dashed.

Graph of y equals x squared minus 4 Plot of y = x**2 - 4 for x in [-4, 4] -4 -2 0 2 4 -5 0 5 10 x y Intersection point Intersection point
Boundary curve of \(y = x^2 - 4\), dashed because the inequality is strict.

Test the origin: \(0 > 0 - 4\) is true, so shade above this parabola.

The second boundary, \(y = -x^2 + 4\), opens downward with vertex \((0, 4)\), also dashed.

Graph of y equals negative x squared plus 4 Plot of y = -x**2 + 4 for x in [-4, 4] -4 -2 0 2 4 -10 -5 0 5 x y Intersection point Intersection point
Boundary curve of \(y = -x^2 + 4\), dashed because the inequality is strict.

Test the origin again: \(0 < 0 + 4\) is true, so shade below this parabola.

To find exactly where the two curves cross, set them equal: \(x^2 - 4 = -x^2 + 4\), which gives \(2x^2 = 8\), so \(x = \pm 2\) and \(y = 0\). The two parabolas meet at \((-2, 0)\) and \((2, 0)\).

Since the origin satisfied both inequalities, the solution is the lens-shaped region between the two curves, sitting between \(x = -2\) and \(x = 2\), where the region shaded above the upward parabola overlaps the region shaded below the downward parabola.

Example 2: Parabolas Opening Away From Each Other

Graph the solution to the system:

\(y < x^2 - 1\)

\(y > -x^2 + 1\)

The first boundary, \(y = x^2 - 1\), opens upward with vertex \((0, -1)\), drawn dashed since the inequality is strict.

Graph of y equals x squared minus 1 Plot of y = x**2 - 1 for x in [-3, 3] -3 -2 -1 0 1 2 3 -2 0 2 4 6 8 x y Intersection point Intersection point
Boundary curve of \(y = x^2 - 1\), dashed because the inequality is strict.

Testing the origin: \(0 < 0 - 1\) is false, so this time shade below the parabola instead.

The second boundary, \(y = -x^2 + 1\), opens downward with vertex \((0, 1)\), also dashed.

Graph of y equals negative x squared plus 1 Plot of y = -x**2 + 1 for x in [-3, 3] -3 -2 -1 0 1 2 3 -8 -6 -4 -2 0 2 x y Intersection point Intersection point
Boundary curve of \(y = -x^2 + 1\), dashed because the inequality is strict.

Testing the origin: \(0 > 0 + 1\) is false, so shade above this parabola.

Setting the boundaries equal, \(x^2 - 1 = -x^2 + 1\) gives \(2x^2 = 2\), so \(x = \pm 1\) and \(y = 0\). The parabolas meet at \((-1, 0)\) and \((1, 0)\).

Because the origin failed both tests, the solution here is not one central region. It is the two outer regions where \(|x| > 1\): the space below the upward parabola and above the downward parabola on either side of the intersection points. This is a good reminder that a system of quadratic inequalities does not always give a single, bounded region.

Common Mistakes to Avoid

  • Forgetting to switch between dashed and solid curves based on whether the inequality is strict or includes equality.
  • Shading the wrong side of a parabola because the test point was accidentally chosen on the boundary curve itself.
  • Assuming the overlap is always a single, closed region, when opening the parabolas in opposite directions can split the solution into separate pieces.
  • Mixing up which parabola opens up and which opens down before deciding the shading direction.

Quick Recap

To sketch the solution to a system of quadratic inequalities, graph each boundary parabola with the correct line style, use a test point to shade each inequality separately, and then identify where the shaded regions overlap. Finding the intersection points of the two parabolas algebraically confirms exactly where that overlapping region begins and ends.

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