Regions between linear and quadratic boundaries
| English | 中文 | Pinyin |
|---|---|---|
| boundary/ˈbaʊndəri/ | 边界 | biān jiè |
A point can lie above a parabola but still fail a second condition. Which points satisfy both boundaries?
- A point can lie above a parabola but still fail a second condition. Which points satisfy both boundaries?
- This lesson studies boundary 边界: The line or curve separating points that satisfy an inequality from those that do not.
Choose the mathematical structure
- Draw each equality boundary first. Use a solid line or curve for ≤ or ≥ and a dashed one for < or >. For y greater than a function, shade above its graph; for y less than it, shade below. For AND, keep only the overlap. Test a point away from every boundary to confirm the chosen side.
- State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.
Which description correctly defines boundary?
The line or curve separating points that satisfy an inequality from those that do not.
Work through a checked case
- Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.
For y≥x²−2 AND y<x+4, draw the parabola solid and the line dashed. The boundaries meet when x²−2=x+4: (x−3)(x+2)=0, giving (−2,2) and (3,7). Between −2<x<3, the line is above the parabola. The allowed points satisfy x²−2≤y<x+4. At x=0, the vertical interval is −2≤y<4. The point (0,0) is included, (0,4) is excluded and (0,−3) is excluded. The intersection points are excluded because the line inequality is strict.
Regions between linear and quadratic boundaries
Draw each equality boundary first
Match solution-set and boundary decisions to their conditions.
Find the left intersection x-coordinate for y=x²−2 and y=x+4.
Solve x²−x−6=0=(x+2)(x−3); the left root is −2.
Test a tempting shortcut
- A dashed boundary is not part of the region, even where it meets a solid boundary. Solving for intersection x-values gives the horizontal extent, not all allowed coordinates. A region for OR is the union, which can be much larger than the overlap.
- When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.
Every intersection point is included if either boundary is solid. This claim is false. Explain which definition or assumption it violates.
Find the right intersection x-coordinate.
The other root of x²−x−6=0 is 3.
Every intersection point is included if either boundary is solid.
A dashed boundary is not part of the region, even where it meets a solid boundary. Solving for intersection x-values gives the horizontal extent, not all allowed coordinates. A region for OR is the union, which can be much larger than the overlap.
Interpret a new situation
- For a different inequality such as y>x²−2, the permitted side is above the parabola and the curve itself is excluded. If a context also requires x≥0, clip the already combined region to that half-plane. Verify an interior point and a point outside each boundary separately.
- A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.
At x=0, find the included lower y-bound of the combined region.
At x=0 the parabola has y=−2; ≥ includes this lower boundary.
Match each part of a complete solution to its purpose.
An assumption justifies the model; a check tests the result; interpretation connects it to the question.
Use this in your course
- 7357 · A-level · B. Match the target tier and specification before assigning extensions.
- Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.
The line or curve separating points that satisfy an inequality from those that do not. Choose the relationship, show the method, check its assumptions and interpret the result.
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