The parabola is the map: its x-axis crossings are the roots, and the stretch above or below is the inequality's solution.
A quadratic function y = ax² + bx + c, the quadratic equation ax² + bx + c = 0, and the quadratic inequality ax² + bx + c > 0 are not three separate topics — they are one picture read three ways. The graph is the map. The points where the curve touches the ground are the equation's roots. The stretches of road where the curve floats above the ground or sinks below it are the inequality's solution set. This lesson ties the three together, with the discriminant as the probe that tells you, before you draw, how many crossings to expect.
Run your finger along the x-axis. Every point on it has the same height: y = 0. So when a parabola y = ax² + bx + c crosses the x-axis, its output is zero there. The x-coordinates of those crossings are therefore exactly the numbers that make ax² + bx + c = 0 — the roots of the quadratic equation.
Take y = x² − 2x − 3. To find where it meets the ground, set y = 0:
x² − 2x − 3 = 0 ⇒ (x + 1)(x − 3) = 0 ⇒ x = −1 or x = 3.
The curve crosses at (−1, 0) and (3, 0) — and those very same numbers, −1 and 3, are the roots of x² − 2x − 3 = 0. The graph and the equation agree, because they are asking the same question.
An x-intercept of y = ax² + bx + c is a root of ax² + bx + c = 0, and the other way round. To find one, find the other: set y = 0. "Solving the equation" and "finding where the graph crosses" are the same act.
Before you even draw, you can predict how the bowl will meet the ground. The number that probes this is the discriminant, Δ = b² − 4ac — the part of the quadratic formula that lives under the square root. It comes in three flavours:
| Discriminant | Crossings | Roots of ax² + bx + c = 0 |
|---|---|---|
| Δ = b² − 4ac > 0 | two | two distinct real roots |
| Δ = 0 | one (tangent) | one repeated root |
| Δ < 0 | none | no real roots |
This is exactly the discriminant you met with the quadratic formula in Stage 11 — but now you can see what it is counting.
To solve ax² + bx + c = 0 you have three honest paths, and they all land in the same place. You can factor, you can use the quadratic formula, or you can simply read the x-coordinates where the parabola crosses the x-axis. The graph turns the answer into something you can point at.
Solve x² − x − 6 = 0 by graphing. Draw y = x² − x − 6 and find where it meets the axis. It crosses at x = −2 and x = 3. Check with factoring: x² − x − 6 = (x + 2)(x − 3), which is zero at −2 and 3 — the picture and the algebra agree.
A tangent parabola (Δ = 0) only kisses the axis at its vertex. That single touch point is one repeated root — for example y = x² − 6x + 9 = (x − 3)² touches at x = 3 only, where Δ = 36 − 36 = 0. Don't report it as "two roots that happen to be equal" and don't miss it as "no root."
Now ask not where the curve hits the ground but where it is above or below it. ax² + bx + c > 0 asks for the x's where the curve floats above the x-axis; ax² + bx + c < 0 asks where it dips below. The roots split the x-axis into stretches, and on each stretch the curve is entirely above or entirely below — so you just label the stretches.
Take an upward bowl (a > 0) with roots x₁ < x₂. Between the roots the curve dips below the axis; outside them it rises above. Reading the picture:
The brackets matter. A strict sign (< or >) excludes the roots, so the endpoints are open ( ). An inclusive sign (≤ or ≥) includes them — at a root the curve equals zero, which satisfies "≥ 0" — so the endpoints are closed [ ]. (This is the interval and union language from intervals & neighborhoods, and the same inequalities you met algebraically in quadratic inequalities.)
With y = x² − 2x − 3 (roots −1 and 3, opening up):
x² − 2x − 3 > 0 ⇒ x < −1 or x > 3 = (−∞, −1) ∪ (3, ∞)
x² − 2x − 3 ≤ 0 ⇒ −1 ≤ x ≤ 3 = [−1, 3] (closed — the roots are included)
"Above means outside, below means between" is the rule only for an upward bowl (a > 0). When a < 0 the bowl opens down, so it is above the axis between the roots and below outside them — flip the inside/outside wording. And when Δ ≤ 0 there are no two roots to split the axis: see 23.5.5 for what happens then.
Step back and the whole picture snaps into focus. The function is the entire curve — every input paired with its output. The equation asks only for the landmarks: where is the curve at ground level, y = 0? Those are the roots. The inequality asks for a stretch of road: where along the x-axis is the curve above the ground (y > 0) or below it (y < 0)? One map, three kinds of question.
| You have… | You ask… | You read off the graph… |
|---|---|---|
| function y = ax²+bx+c | what is the output at each x? | the whole curve |
| equation ax²+bx+c = 0 | where is y = 0? | the x-intercepts (roots) |
| inequality ax²+bx+c > 0 | where is y above/below 0? | the intervals above (or below) |
The discriminant ties the edge cases together too. Suppose a > 0 and Δ < 0: the whole bowl floats above the axis, so it never reaches zero. Then ax² + bx + c > 0 holds for all x (solution set ℝ), the equation has no real root, and ax² + bx + c < 0 is never true (no solution). A "negative discriminant" does not mean the inequality has no answer — it may mean the answer is everything.
Function, equation, inequality are one object seen three ways. Draw the parabola once: its crossings solve the equation, the stretches above/below solve the inequalities, and Δ = b² − 4ac tells you, in advance, how many crossings there will be — and so which story you are in.
One parabola y = ax² + bx + c carries the answers to a function question, an equation question, and an inequality question all at once. The link is the line y = 0 — the x-axis.
Next, in 23.6 Extremes & Modeling, we put the vertex to work: because it is the single highest or lowest point, it answers every "as much as possible / as little as possible" question — fences, prices, and fountains.
Where does the graph of y = x² − 4 cross the x-axis?
Set y = 0: x² − 4 = 0 ⇒ x² = 4 ⇒ x = −2 and x = 2. The crossings (−2, 0) and (2, 0) are exactly the roots of x² − 4 = 0.
How many real roots does x² + x + 1 = 0 have? Explain using the discriminant.
Δ = b² − 4ac = 1² − 4(1)(1) = 1 − 4 = −3 < 0, so there are no real roots — the bowl floats clear above the x-axis and never touches it.
For which x is x² − 4 < 0?
The bowl y = x² − 4 opens up with roots −2 and 2, and "< 0" means below the axis, which is between the roots. Strict sign ⇒ open ends: −2 < x < 2 = (−2, 2).
For which x is x² − 4 ≥ 0?
"≥ 0" means above or on the axis — that is outside the roots, and inclusive includes them. So x ≤ −2 or x ≥ 2 = (−∞, −2] ∪ [2, ∞), with closed brackets at −2 and 2.
The parabola y = x² − 6x + 9 just touches the x-axis. At what x, and what is Δ?
Δ = (−6)² − 4(1)(9) = 36 − 36 = 0, so it is tangent — one repeated root. x² − 6x + 9 = (x − 3)² = 0 ⇒ x = 3. It touches at (3, 0), the vertex.
Solve x² − x − 6 ≥ 0.
Roots first: x² − x − 6 = (x + 2)(x − 3) = 0 ⇒ x = −2, 3. The bowl opens up, so "≥ 0" is the outside, inclusive of the roots: x ≤ −2 or x ≥ 3 = (−∞, −2] ∪ [3, ∞).
Six questions to lock it in. Tap the answer you think is right.
This lesson is the keystone of the quadratic unit: it shows that a quadratic function, the matching equation, and the matching inequality are one graph read three different ways. The unifying line is y = 0. Where the parabola meets it are the roots of the equation; the stretches where the curve is above or below it are the solution sets of the inequalities. Encouraging students to draw the bowl first — then read landmarks for the equation and intervals for the inequality — replaces a pile of separate rules with a single mental picture they can reconstruct anytime.
Four misconceptions deserve attention. First, confusing "above the axis" with "right of the axis": the inequality is about the curve's height, not which side of the y-axis it is on — drag a finger up-and-down, not left-and-right. Second, forgetting to flip the inside/outside rule when a < 0; a downward bowl is positive between its roots, the reverse of an upward one. Third, mixing up open and closed endpoints: a strict </> excludes the roots (open dots), while ≤/≥ includes them (filled dots), because at a root the value is exactly 0. Fourth — and most subtle — reading Δ < 0 as "no solution to the inequality"; in fact a positive-leading parabola that never touches the axis makes ax²+bx+c > 0 true for all x and < 0 true for none.
Alignment with the US Common Core: A-REI.B.4b (solve quadratic equations; recognize when there are no real solutions), A-REI.D.11 (a solution is where two graphs meet — here, the curve and the line y = 0), the quadratic-inequality reasoning of A-CED.A.1 and A-REI.B, and F-IF.C.7a (graph quadratics; show intercepts). The two interactive figures let students slide the discriminant through its three regimes and toggle a relation to watch the solution interval rebuild itself — exactly the self-checking these standards reward.