Quadratic graphs: roots and vertex
Change a, b and c and watch the parabola move: its roots, its turning point and its line of symmetry, with the completed square and factorised forms worked out exactly.
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What's happening
Every quadratic y = ax² + bx + c draws a parabola: a smile when a > 0, a frown when a < 0. It is symmetric about the vertical line x = −b/2a, and the turning point sits on that line. Completing the square rewrites it as y = a(x − p)² + q, which hands you the turning point (p, q) directly, because the squared bracket can never be negative.
The roots are where the curve meets the x-axis, and the quadratic formula finds them. The discriminant b² − 4ac decides how many there are: positive gives two real roots, zero gives one repeated root where the curve just touches the axis, and negative gives none, because the formula would need the square root of a negative number (the roots are then complex). When the discriminant is a perfect square the roots are rational and the quadratic factorises.
- Quadratic formulax = (−b ± √(b² − 4ac)) / 2a
- Completed square formy = a(x − p)² + q, p = −b/2a
- DiscriminantΔ = b² − 4ac
On the syllabus
- GCSE Maths Higher AQA, Edexcel, OCRquadratic graphs, roots and turning points, factorising, completing the square, the quadratic formula and quadratic inequalities
- A-Level Mathsthe discriminant
Challenge
Predict first: for y = 2x² − 8x + 3, where is the turning point, and how many times does the curve cross the x-axis? Type the x-coordinate of the turning point, press Check, then set a, b and c to see.
The line of symmetry is x = −(−8)/(2 × 2) = 2, and y = 2(2)² − 8(2) + 3 = −5, so the turning point is (2, −5), a minimum. The discriminant is 64 − 24 = 40 > 0, so there are two real roots, (4 ± √10)/2, about 0.419 and 3.581.
Learn this properly
Lessons from GCSE Maths on this topic: worked steps, then exam-style questions with new numbers every time.
- Solving by factorisingQuadraticsFree
- Quadratic graphsQuadraticsPlus
- Completing the squareQuadraticsPlus
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FAQ
How do I find the turning point of a quadratic?
Complete the square: y = a(x − p)² + q has its turning point at (p, q). Or use p = −b/2a and put it back into the equation to get q. For y = x² − 2x − 3, p = 1 and q = −4.
What does the discriminant tell you?
b² − 4ac > 0 means two real roots, b² − 4ac = 0 means one repeated root (the curve touches the x-axis), and b² − 4ac < 0 means no real roots (the curve never meets the x-axis).
When does a quadratic factorise?
When its roots are rational, which happens exactly when the discriminant is a perfect square (of a rational number). x² − 2x − 3 has discriminant 16 = 4², so it factorises as (x − 3)(x + 1).
How do I solve a quadratic inequality like x² − 2x − 3 > 0?
Find the roots (−1 and 3), sketch the curve, and read off where it is above the axis. For a smile-shaped curve that is outside the roots: x < −1 or x > 3. For < 0 it is between them: −1 < x < 3.
How do I find a quadratic from its roots?
Write y = a(x − r₁)(x − r₂) and use one more fact to fix a. With roots 1 and 3 through (0, 6): 6 = a(0 − 1)(0 − 3) = 3a, so a = 2 and y = 2x² − 8x + 6. In Solve for, pick Roots and a point to see the same working.