Math 20-1 · Inequalities and graphs
Regions, not points
An equation asks where something is exactly true; an inequality asks where it is true at all. The answer is a region of the plane or a stretch of the number line — and the fastest way to find it is to look at the graph.
- 1. The words, first
- 2. Inequalities in two variables
- 3. Quadratic inequalities in one variable
- 4. Absolute value and reciprocal graphs
- 5. What costs marks
The words, first
The idea: Two new graph shapes, and a vocabulary for talking about regions instead of points.
| Word | What it means |
|---|---|
| Boundary | The curve or line where the two sides are equal. Drawn solid when it is included (≤, ≥) and dashed when it is not (<, >). |
| Test point | Any point not on the boundary, substituted to find out which side is the solution region. The origin is the usual choice. |
| Solution region | The shaded part of the plane. Every point in it satisfies the inequality. |
| Absolute value function | y = |f(x)|. The graph of f with everything below the x-axis reflected upward, producing a corner at each x-intercept. |
| Reciprocal function | y = 1/f(x). Large values become small, small values become large, and a zero of f becomes a vertical asymptote. |
| Asymptote | A line the graph approaches without reaching. A vertical one at each zero of f; a horizontal one at y = 0. |
| Invariant point | A point unchanged by the transformation. For reciprocals these sit where y = 1 or y = −1, the only numbers equal to their own reciprocal. |
Inequalities in two variables
The idea: Draw the boundary, decide solid or dashed, then test one point. That is the whole procedure.
Worked example: y ≥ x² − 4.
- Boundary: the parabola y = x² − 4, drawn solid because of the ≥.
- Test (0, 0): is 0 ≥ −4? Yes.
- So shade the region containing the origin — the inside of the parabola.
Only choose a different test point if the origin lies on the boundary. Reasoning about which side “looks bigger” is slower and less reliable than substituting one point.
Checking a point. Is (1, 5) a solution of y > x² + 3? Substitute: x² + 3 = 4, and 5 > 4 ✓. Yes.
Quadratic inequalities in one variable
The idea: Find the roots, sketch the parabola, and read off where it is above or below the axis. Never divide by a bracket whose sign you do not know.
Worked example: x² − x − 6 ≤ 0. Factor to (x − 3)(x + 2) ≤ 0, so the roots are −2 and 3. The parabola opens upward, so it is at or below zero between the roots: −2 ≤ x ≤ 3.
The other direction: x² > 16. Rewrite as (x − 4)(x + 4) > 0, roots ±4. An upward parabola is above zero outside its roots, so x < −4 or x > 4 — two separate pieces. Taking a square root and writing x > 4 quietly loses every negative solution.
Opens up, ≤ 0 → between the roots · opens up, ≥ 0 → outside them. Flip both if it opens down.
No solution is a real answer. x² + 1 < 0 has none: a square is never negative, and the discriminant, −4, agrees.
Absolute value inequalities. |x − 4| < 3 means “less than 3 away from 4”, so 1 < x < 7. Greater-than gives the two outside pieces instead.
Absolute value and reciprocal graphs
The idea: Both are built from a graph you already have. Draw the original first, lightly, then transform it.
y = |f(x)|. Keep everything on or above the x-axis; reflect everything below it upward. For y = |2x − 6|, the line is negative left of x = 3, so that part flips, leaving a V with its corner at (3, 0).
y = 1/f(x). Work through the features in order:
- Wherever f(x) = 0, the reciprocal has a vertical asymptote. For f(x) = x² − 4, they are at x = 2 and x = −2.
- Wherever f(x) = 1 or −1, the point does not move — the invariant points.
- Where f is large, the reciprocal is near zero, so y = 0 is a horizontal asymptote.
- A minimum of f at a negative value becomes a maximum of the reciprocal. For y = 1/(x² − 4), f has a minimum of −4 at x = 0, so the reciprocal has a local maximum at (0, −¼).
What costs marks
The idea: Mostly boundaries and missing pieces.
- Solid versus dashed. A strict inequality excludes its boundary, and a solid line claims points that are not solutions.
- Giving one interval when there are two. Greater-than quadratic and absolute value inequalities almost always split.
- Dividing an inequality by a bracket. If its sign is unknown you cannot know whether to flip the inequality. Factor and look at the graph instead.
- Forgetting to flip when multiplying or dividing both sides by a negative number.