HomeGraph Quadratic Inequalities: Shade the Solution Region Free original lessons
Graph Quadratic Inequalities: Shade the Solution Region
Learn to graph quadratic inequalities by finding vertex, intercepts, and using test points to shade the correct region on a coordinate plane.
Free lesson
Four exercises
Worked answers
Algebra and Function Practice
Graphing a quadratic inequality means showing all points that satisfy the inequality. Start by graphing the related parabola, then decide whether to shade inside or outside it. This lesson walks through vertex and intercept finding, test-point shading, and common pitfalls. You will practice with four self-contained tasks and a worked example. All numerical values are hypothetical examples for learning.
Practice materials: Printable downloads are not available for this lesson yet. Use the instructions and four exercises below.
Find the Parabola's Key Features
Every quadratic inequality uses a parabola as a boundary. Write the related equation in standard form y = ax^2 + bx + c. The vertex x-coordinate is -b/(2a). Substitute that x back to get the vertex y-coordinate. The y-intercept is c. Find x-intercepts by solving ax^2 + bx + c = 0, using factoring or the quadratic formula. Plot these points and sketch a smooth curve. If the inequality is strict (< or >), draw a dashed parabola because boundary points are not included. If it is non-strict (≤ or ≥), draw a solid parabola because boundary points are included.
- Identify a, b, and c from the related equation.
- Compute vertex x = -b/(2a), then vertex y.
- Find y-intercept (0, c) and x-intercepts if they exist.
- Sketch the parabola with dashed or solid line based on the inequality symbol.
Choose a Test Point to Decide Shading
After drawing the boundary parabola, pick a test point not on the curve, often (0,0) if it is not on the parabola. Substitute its coordinates into the original inequality. If the result is true, shade the region containing that test point. If false, shade the opposite region. For a parabola opening upward with y > ax^2 + bx + c, the solution is usually the region above the parabola; for y < ..., it is below. But always confirm with a test point because the inequality may be written in a different form. Label the shaded region clearly.
- Choose a test point such as (0,0) or (1,1) that is not on the parabola.
- Substitute x and y into the inequality.
- If true, shade the side containing the test point; if false, shade the other side.
- Double-check by testing a second point in the shaded region.
Handle Special Cases and Verify
Sometimes the parabola has no x-intercepts, so the boundary does not cross the x-axis. In that case, use the vertex, the y-intercept, and one or two additional points to sketch the parabola accurately, then use a test point to decide shading. If the inequality is strict, the boundary is dashed and points on the parabola are excluded. If non-strict, the boundary is solid and included. Always verify by testing a point inside and a point outside the shaded region. Keep arithmetic simple and consistent; label any chosen numerical values as hypothetical examples.
- Check whether the parabola opens up, down, left, or right.
- Confirm dashed or solid boundary from the inequality symbol.
- Test one point inside and one outside the shaded region.
- Label axes and shade clearly, noting any excluded boundary.
Worked example
Graph the inequality y > x^2 - 4x + 3. Find the vertex, intercepts, decide dashed or solid, and shade the solution region.
Worked answer
Related equation: y = x^2 - 4x + 3. Vertex x = -(-4)/(2*1) = 2; vertex y = 2^2 - 4*2 + 3 = -1, so vertex (2, -1). y-intercept (0, 3). Solve x^2 - 4x + 3 = 0: (x-1)(x-3)=0, so x-intercepts (1,0) and (3,0). Since inequality is strict (>), draw a dashed parabola opening upward. Test point (0,0): 0 > 0^2 - 4*0 + 3 = 3 is false, so shade the region not containing (0,0), which is above the parabola. The shaded region is above the dashed parabola.
Four practice exercises
Exercise 1
Graph y ≤ x^2 - 2x - 3. Find vertex, intercepts, boundary type, and shade.
Show answer and check
Vertex (1, -4); y-intercept (0, -3); x-intercepts (-1,0) and (3,0); solid parabola opening upward; test (0,0): 0 ≤ -3 false, so shade region not containing (0,0), which is below the parabola (including boundary).
Why: Vertex x = -(-2)/(2*1)=1, y=1-2-3=-4. Factor x^2-2x-3=(x-3)(x+1). Non-strict ≤ means solid. Test point (0,0) gives false, so shade opposite side from (0,0), i.e., below the parabola.
Check: Test a point inside shaded region, e.g., (1,-5): -5 ≤ 1-2-3=-4 true. Test outside, e.g., (1,0): 0 ≤ -4 false.
Exercise 2
Graph y > -x^2 + 4. Find vertex, intercepts, boundary type, and shade.
Show answer and check
Vertex (0,4); y-intercept (0,4); x-intercepts (-2,0) and (2,0); dashed parabola opening downward; test (0,0): 0 > 4 false, so shade region not containing (0,0), which is above the parabola (inside the downward opening).
Why: Vertex at (0,4). Solve -x^2+4=0 gives x=±2. Strict > means dashed. Test (0,0) false, so shade above the parabola.
Check: Test (0,5): 5 > 4 true. Test (0,3): 3 > 4 false.
Exercise 3
Graph y ≥ x^2 + 2x + 1. Find vertex, intercepts, boundary type, and shade.
Show answer and check
Vertex (-1,0); y-intercept (0,1); x-intercept (-1,0) (double root); solid parabola opening upward; test (0,0): 0 ≥ 1 false, so shade region not containing (0,0), which is above the parabola (including boundary).
Why: Vertex x=-2/(2)=-1, y=1-2+1=0. Factor (x+1)^2. Non-strict ≥ means solid. Test (0,0) false, so shade above.
Check: Test (-1,1): 1 ≥ 0 true. Test (0,0): 0 ≥ 1 false.
Exercise 4
Graph y < -x^2 + 6x - 5. Find vertex, intercepts, boundary type, and shade.
Show answer and check
Vertex (3,4); y-intercept (0,-5); x-intercepts (1,0) and (5,0); dashed parabola opening downward; test (0,0): 0 < -5 false, so shade region not containing (0,0), which is below the parabola (outside the downward opening).
Why: Vertex x=-6/(2*-1)=3, y=-9+18-5=4. Factor -(x-1)(x-5). Strict < means dashed. Test (0,0) false, so shade below.
Check: Test (3,5): 5 < 4 false. Test (3,3): 3 < 4 true.
Common mistakes
- Shading the wrong side without testing a point. Always substitute a test point not on the parabola and shade according to the truth value.
- Using a solid line for a strict inequality (< or >). Use a dashed parabola for strict inequalities to show boundary points are excluded.
- Forgetting to find the vertex and intercepts, leading to an inaccurate sketch. Compute vertex x = -b/(2a), y-intercept c, and x-intercepts before sketching.
- Assuming the shaded region is always above the parabola. Test a point; for y < quadratic, the solution is usually below, but confirm with substitution.
Self-check rubric
- Correctly identifies vertex and at least one intercept.
- Uses dashed boundary for strict inequality or solid for non-strict.
- Chooses a valid test point and substitutes correctly.
- Shades the correct region based on test-point result.
- Labels axes and boundary clearly; valid alternative sketches accepted if key features and shading are correct.
Questions and answers
How do I know whether to shade inside or outside the parabola?
Pick a test point not on the parabola. If substituting its coordinates makes the inequality true, shade the region containing that point. Otherwise shade the opposite region.
When should the parabola be dashed versus solid?
Use a dashed parabola for strict inequalities (< or >) because boundary points are not included. Use a solid parabola for ≤ or ≥ because boundary points are included.
What if the quadratic has no x-intercepts?
The parabola still has a vertex and y-intercept. Sketch it using those points and a few additional points, then use a test point to decide shading.
Can I use (0,0) as a test point every time?
Only if (0,0) is not on the parabola. If it is, choose another point such as (1,1) or (0,1).