Graphing Quadratic Inequalities: A Complete Step-by-Step Guide
Graphing quadratic inequalities builds directly on graphing a quadratic equation — you draw the same parabola, then decide which boundary line to use and which region to shade. A quadratic inequality replaces the equals sign in y = ax² + bx + c with <, >, ≤, or ≥, and the solution is no longer a curve but an entire two-dimensional region of the coordinate plane. This guide walks through graphing quadratic inequalities in two variables with three fully worked examples, explains when the parabola itself is dashed versus solid, shows how to pick and test a point to find the correct shaded region, and covers the related skill of solving one-variable quadratic inequalities using interval notation.
Contents
- 01What Is a Quadratic Inequality?
- 02When Is the Parabola Dashed vs Solid?
- 03How Do You Know Which Side to Shade?
- 04Worked Example 1: Graphing y > x² − 4x + 3
- 05Worked Example 2: Graphing y ≤ −x² + 2x + 3
- 06How Do You Solve a One-Variable Quadratic Inequality Like x² − 5x + 6 ≥ 0?
- 07Connecting the Number Line Solution to the Parabola's Graph
- 08Common Mistakes When Graphing Quadratic Inequalities
- 09Practice Problems: Graphing Quadratic Inequalities
- 10FAQ: Graphing Quadratic Inequalities
- 11Practice Graphing Quadratic Inequalities With Instant Feedback
What Is a Quadratic Inequality?
A quadratic inequality in two variables has the form y > ax² + bx + c, y < ax² + bx + c, y ≥ ax² + bx + c, or y ≤ ax² + bx + c. Instead of a single curve, the solution set is a region of the coordinate plane — either everything above the parabola, everything below it, or the parabola itself included as a boundary. Graphing quadratic inequalities always starts the same way: graph the related quadratic equation y = ax² + bx + c as if the inequality symbol were an equals sign, then decide how to draw that boundary and which side to shade. There is also a one-variable version, such as x² − 5x + 6 ≥ 0, where the solution is a set of x-values written in interval notation rather than a shaded region — this guide covers both forms.
Graphing quadratic inequalities in two variables always begins with the parabola y = ax² + bx + c — the inequality symbol only changes how you draw and shade it, not where the curve sits.
When Is the Parabola Dashed vs Solid?
The boundary curve follows the same rule as graphing linear inequalities. A strict inequality (< or >) means points exactly on the parabola do NOT satisfy the inequality, so you draw a dashed (broken) curve to show the boundary is excluded. A non-strict inequality (≤ or ≥) means points on the parabola DO satisfy the inequality, so you draw a solid curve to show the boundary is included. This single check takes two seconds but is worth doing every time, because it is the detail graders check first.
1. Strict inequalities (< or >) → dashed curve
For y > x² − 4x + 3 or y < x² − 4x + 3, the parabola itself is not part of the solution — it marks the edge of the region but is drawn with a dashed line. Any point you pick directly on the curve should fail the inequality when you substitute it in.
2. Non-strict inequalities (≤ or ≥) → solid curve
For y ≥ x² − 4x + 3 or y ≤ x² − 4x + 3, the parabola is part of the solution set, so it is drawn as a solid line. Any point on the curve satisfies the inequality with equality, which is why it belongs to the shaded region.
Strict < or > → dashed curve (boundary excluded). Non-strict ≤ or ≥ → solid curve (boundary included). This is the exact same rule used for graphing linear inequalities.
How Do You Know Which Side to Shade?
Once the boundary is drawn, one test point tells you the entire shaded region. Because the parabola splits the plane into exactly two regions — inside/above and outside/below — a single point that is clearly not on the curve is enough to determine both regions at once. This is faster and more reliable than trying to reason about the inequality symbol alone, especially once the parabola is not a simple upward-opening shape.
1. Step 1: Pick a test point not on the parabola
The origin (0, 0) is the easiest choice whenever the parabola does not pass through it. If (0, 0) happens to sit on the curve, pick another simple point such as (0, 5) or (1, 0) instead.
2. Step 2: Substitute the test point into the inequality
Plug the x and y values of the test point into the original inequality (not just the equation) and simplify both sides.
3. Step 3: Decide which region to shade
If the test point makes the inequality true, shade the region that contains that test point. If the test point makes the inequality false, shade the opposite region — the one that does NOT contain the test point.
4. Step 4: Sanity-check with the inequality direction
As a quick cross-check, y > (or y ≥) an expression usually shades above/outside the parabola for an upward-opening curve, and y < (or y ≤) usually shades below/inside it — but always confirm with the test point rather than relying on memory, since a downward-opening parabola reverses this pattern.
One test point off the curve tells you the whole shaded region — substitute it into the original inequality and shade the side where it makes the statement true.
Worked Example 1: Graphing y > x² − 4x + 3
This example uses a strict, greater-than inequality with an upward-opening parabola. The related equation y = x² − 4x + 3 has a = 1, b = −4, c = 3, the same parabola used in earlier graphing guides, so the boundary shape should already look familiar.
1. Step 1: Graph the boundary parabola
Vertex: h = −(−4)/(2×1) = 2, k = (2)² − 4(2) + 3 = 4 − 8 + 3 = −1. Vertex: (2, −1). Y-intercept: (0, 3). X-intercepts: factor x² − 4x + 3 = 0 → (x − 3)(x − 1) = 0 → x = 1 and x = 3, so (1, 0) and (3, 0). Because the inequality is strict (>), draw this parabola as a dashed curve.
2. Step 2: Test a point
Use (0, 0), which is not on the parabola since 0 ≠ (0)² − 4(0) + 3 = 3. Substitute into y > x² − 4x + 3: 0 > (0)² − 4(0) + 3 → 0 > 3. This is false.
3. Step 3: Shade the correct region
Since (0, 0) makes the inequality false, shade the region that does NOT contain (0, 0) — the region above/outside the parabola (the region that includes points like (2, 5), well above the vertex). Check with (2, 5): 5 > (2)² − 4(2) + 3 → 5 > −1, true ✓, confirming (2, 5) belongs to the shaded region.
4. Step 4: Describe the final graph
The solution to y > x² − 4x + 3 is every point strictly above the dashed parabola with vertex (2, −1), x-intercepts (1, 0) and (3, 0), passing through (0, 3). Points exactly on the curve, such as the vertex (2, −1) itself, are not included.
For y > x² − 4x + 3, shading falls on the side away from the failing test point — always let the test point decide, not intuition about the inequality symbol alone.
Worked Example 2: Graphing y ≤ −x² + 2x + 3
This example flips two things at once: the parabola opens downward (a = −1) and the inequality is non-strict (≤), so the boundary is solid rather than dashed. Watch how these two changes affect both the curve and the shaded region compared to Example 1.
1. Step 1: Graph the boundary parabola
For y = −x² + 2x + 3: a = −1, b = 2, c = 3. Vertex: h = −2/(2×(−1)) = −2/(−2) = 1, k = −(1)² + 2(1) + 3 = −1 + 2 + 3 = 4. Vertex: (1, 4) — a maximum since a < 0. Y-intercept: (0, 3). X-intercepts: set −x² + 2x + 3 = 0 → multiply by −1: x² − 2x − 3 = 0 → (x − 3)(x + 1) = 0 → x = 3 or x = −1, giving (3, 0) and (−1, 0). Because the inequality is ≤ (non-strict), draw this downward-opening parabola as a solid curve.
2. Step 2: Test a point
Use (0, 0), which is not on the curve since 0 ≠ 3. Substitute into y ≤ −x² + 2x + 3: 0 ≤ −(0)² + 2(0) + 3 → 0 ≤ 3. This is true.
3. Step 3: Shade the correct region
Since (0, 0) makes the inequality true, shade the region that DOES contain (0, 0) — the region below/outside the downward-opening parabola. Check with (1, 0), directly below the vertex (1, 4): 0 ≤ −(1)² + 2(1) + 3 → 0 ≤ 4, true ✓, confirming this region is correct.
4. Step 4: Describe the final graph
The solution to y ≤ −x² + 2x + 3 is every point on or below the solid parabola with vertex (1, 4), x-intercepts (−1, 0) and (3, 0), passing through (0, 3). Because the boundary is solid, the vertex (1, 4) and every other point on the curve are included in the solution.
A downward-opening parabola with y ≤ shades below the curve — the opposite visual pattern from Example 1, which is why testing a point matters more than memorizing a rule.
How Do You Solve a One-Variable Quadratic Inequality Like x² − 5x + 6 ≥ 0?
A one-variable quadratic inequality compares a quadratic expression to zero, such as x² − 5x + 6 ≥ 0. Instead of shading a two-dimensional region, the solution is a set of x-values on the number line, written in interval notation. The strategy is to find the roots of the related equation, use them to split the number line into intervals, and test each interval to see whether it satisfies the inequality.
1. Step 1: Solve the related equation to find critical points
Set x² − 5x + 6 = 0 and factor: find two numbers multiplying to 6 and adding to −5 → (−2, −3). So (x − 2)(x − 3) = 0, giving x = 2 and x = 3. These roots are the critical points that divide the number line into three intervals: x < 2, 2 < x < 3, and x > 3.
2. Step 2: Test a value from each interval
Interval x < 2: test x = 0. (0)² − 5(0) + 6 = 6 ≥ 0 ✓ true. Interval 2 < x < 3: test x = 2.5. (2.5)² − 5(2.5) + 6 = 6.25 − 12.5 + 6 = −0.25 ≥ 0 — false. Interval x > 3: test x = 4. (4)² − 5(4) + 6 = 16 − 20 + 6 = 2 ≥ 0 ✓ true.
3. Step 3: Include or exclude the critical points
Since the inequality is ≥ (non-strict), the roots themselves satisfy the inequality: at x = 2, (2)² − 5(2) + 6 = 4 − 10 + 6 = 0 ≥ 0 ✓ true. At x = 3, (3)² − 5(3) + 6 = 9 − 15 + 6 = 0 ≥ 0 ✓ true. Both critical points are included.
4. Step 4: Write the solution in interval notation
Combine the true intervals with their included endpoints: x ≤ 2 or x ≥ 3, written as (−∞, 2] ∪ [3, ∞). This matches the graph of y = x² − 5x + 6: the parabola sits on or above the x-axis (y ≥ 0) everywhere except the open dip between its roots at x = 2 and x = 3.
For a upward-opening parabola, ax² + bx + c ≥ 0 holds outside the roots (or on them), and ax² + bx + c ≤ 0 holds between the roots — the sign always flips at each critical point.
Connecting the Number Line Solution to the Parabola's Graph
The one-variable inequality x² − 5x + 6 ≥ 0 and the two-variable inequality y ≥ x² − 5x + 6 are closely related, and seeing the connection makes both easier to remember. If you graph y = x² − 5x + 6 as a parabola, the one-variable question 'where is x² − 5x + 6 ≥ 0?' is really asking 'where does the parabola sit on or above the x-axis (y ≥ 0)?' Reading the graph directly answers the inequality without any additional algebra once the curve is drawn.
1. Above the x-axis → positive region
For x < 2 or x > 3, the parabola y = x² − 5x + 6 lies above the x-axis, meaning y > 0 there, which matches the true intervals found by testing.
2. Below the x-axis → negative region
For 2 < x < 3, the parabola dips below the x-axis between its two roots, meaning y < 0 there — this is exactly the interval that failed the ≥ 0 test in the worked example above.
3. On the x-axis → the roots themselves
At x = 2 and x = 3, the parabola touches the x-axis exactly, so y = 0 — these are the boundary points that a ≥ or ≤ inequality includes, and a > or < inequality excludes.
A one-variable quadratic inequality ax² + bx + c ≥ 0 or ≤ 0 is just a vertical read of the parabola's graph: check whether the curve sits above, below, or on the x-axis at each x-value.
Common Mistakes When Graphing Quadratic Inequalities
Most errors with graphing quadratic inequalities come from confusing rules that look similar but apply to different situations. Recognizing these patterns in advance prevents the most common point losses on homework and tests.
1. Forgetting to dash the curve for strict inequalities
A solid parabola for y > x² − 4x + 3 is incorrect — the strict > means the boundary itself is excluded, so it must be dashed. Always check the inequality symbol before drawing the curve, not after.
2. Shading the wrong side without testing a point
Guessing 'up' for > and 'down' for < works for upward-opening parabolas but reverses for downward-opening ones, as shown in Example 2. Always substitute a real test point rather than relying on a shortcut that only works half the time.
3. Mixing up the two-variable and one-variable versions
y ≥ x² − 5x + 6 produces a shaded two-dimensional region on the xy-plane; x² − 5x + 6 ≥ 0 produces a one-dimensional set of x-values on the number line, written in interval notation. They use the same parabola but answer different questions — read the problem carefully to see whether y is present.
4. Including or excluding critical points incorrectly in interval notation
For x² − 5x + 6 > 0 (strict), the roots x = 2 and x = 3 are excluded, so the answer uses parentheses: (−∞, 2) ∪ (3, ∞). For x² − 5x + 6 ≥ 0 (non-strict), the roots are included, so the answer uses brackets: (−∞, 2] ∪ [3, ∞). Matching bracket type to inequality type is the same rule used throughout interval notation.
The strict vs non-strict rule controls both the dashed/solid boundary in two-variable graphs and the parenthesis/bracket choice in interval notation — it is the same logic applied twice.
Practice Problems: Graphing Quadratic Inequalities
Work through each problem before checking the solution. For the two-variable problems, identify the boundary parabola, whether it is dashed or solid, and which region is shaded. For the one-variable problem, find the critical points and write the answer in interval notation.
1. Problem 1 — y < x² + 2x − 8
Boundary: y = x² + 2x − 8. Vertex: h = −2/2 = −1, k = 1 − 2 − 8 = −9, vertex (−1, −9). X-intercepts: (x + 4)(x − 2) = 0 → x = −4, x = 2. Strict <, so dashed curve. Test (0, 0): 0 < (0)² + 2(0) − 8 → 0 < −8, false, so shade the region NOT containing (0, 0) — below/inside the parabola. Check (−1, −20): −20 < −9, true ✓.
2. Problem 2 — y ≥ −x² + 4
Boundary: y = −x² + 4, vertex (0, 4), opens downward. X-intercepts: −x² + 4 = 0 → x² = 4 → x = ±2. Non-strict ≥, so solid curve. Test (0, 0): 0 ≥ −(0)² + 4 → 0 ≥ 4, false, so shade the region NOT containing (0, 0) — above/outside the parabola, including the curve itself.
3. Problem 3 — Solve x² − x − 6 < 0
Factor: (x − 3)(x + 2) = 0 → critical points x = −2 and x = 3. Test x = 0 (between the roots): (0)² − 0 − 6 = −6 < 0 ✓ true. Test x = −5: 25 + 5 − 6 = 24 < 0, false. Test x = 5: 25 − 5 − 6 = 14 < 0, false. Since only the interval between the roots works and the inequality is strict, answer: (−2, 3).
FAQ: Graphing Quadratic Inequalities
These are the questions students ask most often when first learning to graph quadratic inequalities.
1. How is graphing a quadratic inequality different from graphing a quadratic equation?
A quadratic equation y = ax² + bx + c graphs as a single curve — the parabola itself. A quadratic inequality replaces the equals sign with <, >, ≤, or ≥, so the solution becomes an entire shaded region of the plane on one side of that same parabola, with the curve drawn dashed or solid depending on whether the boundary is included.
2. Do I always need to test a point, or can I just look at the inequality symbol?
Always test a point. For an upward-opening parabola, > and ≥ typically shade above/outside while < and ≤ shade below/inside — but for a downward-opening parabola, as in Example 2, this pattern reverses. A single substituted test point avoids relying on a rule that only works some of the time.
3. What is the difference between x² − 5x + 6 ≥ 0 and y ≥ x² − 5x + 6?
x² − 5x + 6 ≥ 0 is a one-variable inequality; its solution is a set of x-values written in interval notation, such as (−∞, 2] ∪ [3, ∞). y ≥ x² − 5x + 6 is a two-variable inequality; its solution is a shaded region of the coordinate plane above or on the same parabola. Both use the same parabola but answer different questions.
4. Can the solution region of a quadratic inequality be empty?
Yes, in special cases. For example, y < x² + 1 has no solution where y is also required to be less than 0 for all x, since the parabola's minimum value is already 1. More commonly this shows up in the one-variable case: x² + 1 < 0 has no real solution because x² + 1 is always positive, so the interval notation answer is the empty set ∅.
5. How do I check my final shaded graph is correct?
Pick two points: one clearly inside your shaded region and one clearly outside it. Substitute both into the original inequality. The inside point should make the inequality true, and the outside point should make it false. If either check fails, re-examine the test point step and confirm the boundary parabola's vertex and intercepts are calculated correctly.
Practice Graphing Quadratic Inequalities With Instant Feedback
Graphing quadratic inequalities gets easier with repetition, especially once you have worked through several examples with both upward- and downward-opening parabolas. If you want to check a boundary curve, verify a shaded region, or see a one-variable inequality solved step by step, Solvify's AI tools can walk through the same process shown in this guide on any problem you enter.
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