How this instrument works
Behind every shaded region on a parabola graph sits one plain subtraction. Take a candidate point (px, py), evaluate the parabola's own height at that same x, namely a·px² + b·px + c, and subtract it from py. The sign of that difference — test = py − (a·px² + b·px + c) — carries the whole answer: positive means the point sits above the curve, negative means below, and zero means the point lies exactly on the boundary.
This works because a parabola written as y = ax² + bx + c is really a rule that assigns one height to every x. A point (px, py) satisfies y > ax² + bx + c precisely when its own py exceeds that assigned height, and satisfies y < ax² + bx + c precisely when py falls short. Subtracting turns a picture — is this dot above or below that curve — into a single signed number, which is why the test doubles as the closed-form check behind shading the correct half of a graphed quadratic inequality without drawing anything.
A useful edge case: when the point sits precisely on the parabola, py equals a·px² + b·px + c exactly and the test returns zero, neither above nor below. That boundary matters for strict versus non-strict inequalities — y ≥ ax² + bx + c includes the curve itself, while y > ax² + bx + c excludes it, and a zero result signals a point sitting right on the line between the two, needing the stricter reading of the inequality symbol to decide whether it counts.
- Set the parabola's shape in Parabola: a (from y=ax²+bx+c), b, and c.
- Enter the location to check in Test point: x and Test point: y.
- Read Test value (positive = above the curve, negative = below) for the verdict.
- Zero in that field means the point sits exactly on the parabola, the boundary itself.
Worked example — testing the point (2, 5)
Take the parabola y = x² (so a=1, b=0, c=0) and the candidate point (2, 5). Plug x=2 into the parabola itself: 1×2² + 0×2 + 0 = 4, the curve's own height at that x. Subtract that from the point's y-value of 5: test = 5 − 4 = 1, a positive number. The point (2, 5) therefore sits above the parabola, exactly the region an inequality like y > x² describes — and no graph paper was needed to reach that conclusion.
Nudge the same point down to (2, 4) and the test collapses to zero, since 4 is exactly the parabola's height at x=2 — the point then sits precisely on the curve, the line between the shaded and unshaded halves. Nudge it further, to a y-value below 4, and the test turns negative, placing the point in the region satisfying y < x² instead. The single subtraction slides smoothly from one region to the boundary to the other, which is the entire trick behind shading a quadratic inequality without ever drawing the curve.
Questions
What does a positive test value mean?
A positive result means the point sits above the parabola: its y-coordinate exceeds the curve's own height at that x. Formally, test = py − (a·px² + b·px + c), and when that subtraction comes out positive the point satisfies an inequality of the form y > ax² + bx + c. For the parabola y = x² and the point (2, 5), the curve sits at height 4 when x = 2, the point's y of 5 is one unit higher, and the test returns +1 — comfortably above.
What does a negative test value mean?
A negative result places the point below the parabola — its y-coordinate falls short of the curve's height at that x, satisfying an inequality of the form y < ax² + bx + c. For the parabola y = (x−1)², entered here as a=1, b=−2, c=1, the point (0, 0) tests at −1: the curve reaches height 1 at x=0, and the point's y of 0 sits one unit under it, squarely in the lower region.
Why does a test value of zero matter?
Zero means the point lies exactly on the parabola rather than in either shaded region — py equals a·px² + b·px + c precisely, with no gap either way. That distinction decides how a strict inequality like y > ax² + bx + c is read at the boundary: the curve itself belongs to y ≥ ax² + bx + c but not to the strict version, so a zero result flags a point that needs the ≥ or ≤ symbol to be included.
Do I need to graph the parabola to use this?
No — that is the entire point of the subtraction. Plugging px into a·px² + b·px + c gives the curve's height at that single x without sketching anything, and comparing it to py answers the above-or-below question directly. Graphing still helps for intuition and for seeing a whole shaded region at once, but deciding any one candidate point only takes this one calculation.
Does the sign of a change which side counts as above?
No — 'above' and 'below' always refer to the y-axis direction, regardless of whether the parabola opens upward (a positive) or downward (a negative one). The test formula, py minus the parabola's height, stays the same either way; what changes is where the curve itself sits, since a negative a flips it into a dome shape rather than a bowl. A positive result still means the point's y is the larger of the two values compared.
Can this check work for a straight line instead of a parabola?
Yes — setting a to zero turns ax² + bx + c into the straight line bx + c, and the same subtraction, py minus that line's height at px, tells you whether a point sits above or below a line rather than a curve. Nothing in the formula requires a to be nonzero; it simply stops being a parabola in the strict sense and becomes a linear inequality test instead.