Graphing Quadratic Inequalities in Two Variables

A quadratic inequality in two variables looks almost exactly like a quadratic function, except the equals sign is replaced with an inequality symbol. Instead of a single curve, the solution is an entire region of the coordinate plane. You'll see inequalities written as:

\( y > ax^2 + bx + c \), \( y < ax^2 + bx + c \), \( y \ge ax^2 + bx + c \), or \( y \le ax^2 + bx + c \)

Each one describes every point \((x, y)\) that makes the inequality true. Since a parabola splits the plane into an "inside" region and an "outside" region, graphing the inequality means shading whichever side satisfies the statement. If you haven't reviewed how the boundary curve itself is built, it helps to look back at the quadratic general form first, since that's exactly the curve you'll be sketching here.

Graphing a quadratic inequality in two variables always follows the same routine:

1. Replace the inequality with an equals sign and graph the resulting parabola, \( y = ax^2 + bx + c \). This is called the boundary curve.
2. Choose dashed or solid. If the symbol is strictly \( > \) or \( < \), draw a dashed parabola, since points exactly on the curve are not included. If the symbol is \( \ge \) or \( \le \), draw a solid parabola, since points on the curve are included.
3. Pick a test point that is clearly not on the parabola. The origin \((0, 0)\) is usually the easiest choice, as long as the curve doesn't pass through it.
4. Substitute the test point into the original inequality. If it makes the inequality true, shade the region containing that point. If it makes the inequality false, shade the other region instead.

Graph \( y < x^2 - 2x - 3 \).

First graph the boundary parabola \( y = x^2 - 2x - 3 \). It opens upward, crosses the x-axis at \( x = -1 \) and \( x = 3 \), and has its vertex at \( x = 1 \). Because the inequality is strict (\( < \), not \( \le \)), the parabola itself is drawn dashed.

Now test the origin: substitute \( x = 0, y = 0 \) into \( y < x^2 - 2x - 3 \), giving \( 0 < -3 \). This is false, so the origin is not part of the solution. Shade the region on the opposite side of the parabola from the origin, that is, the region "inside" the upward-opening curve, above the vertex but below the arms, is left unshaded, and the outer region containing points like \((0,0)\)'s opposite side gets shaded. In short: shade below the dashed curve where the inequality holds, and leave the boundary itself unfilled since it's dashed.

Graph \( y \ge -x^2 + 4 \).

The boundary parabola \( y = -x^2 + 4 \) opens downward, crosses the x-axis at \( x = -2 \) and \( x = 2 \), and has its vertex at \( x = 0 \). Since the symbol is \( \ge \), the parabola is drawn solid.

Test a point clearly off the curve, such as \((0, 0)\): substituting gives \( 0 \ge 4 \), which is false. So the region containing the origin is not shaded. Instead, shade the region "inside" the downward-opening parabola, above the curve near the vertex, along with its solid boundary, since points on the curve satisfy the "or equal to" part of the inequality.

A quadratic boundary curve divides the coordinate plane into exactly two regions: one where \( y \) is greater than the quadratic expression, and one where \( y \) is less than it. Every point in the same region behaves the same way with respect to the inequality, so checking just one convenient point tells you how the whole region behaves. This is the same logic used when graphing linear inequalities in two variables, except a parabola replaces a straight line as the dividing boundary.

Students often mix up dashed and solid boundaries, or shade the wrong side after a correct test. Watch out for these:

• Forgetting that \( > \) and \( < \) always mean a dashed curve, never solid.
• Choosing a test point that accidentally lies on the parabola itself, which gives no useful information.
• Shading the region containing the test point when the inequality actually came out false for that point.
• Sketching the parabola inaccurately; getting the vertex and x-intercepts right first makes the shading much easier to place correctly.

If you're comfortable with solving quadratic inequalities in one variable, notice the parallel: there, you found intervals on a number line where a quadratic expression was above or below zero. Here, you're doing the same comparison, but across an entire plane instead of a line, which is why the solution is a shaded region rather than an interval.

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