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.
تحتاج إلى العدد فقط؟ حلّال المعادلات
القراءات
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.
- x = (−b ± √(b² − 4ac)) / 2a
- y = a(x − p)² + q, p = −b/2a
- Δ = b² − 4ac
في المنهج
- 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
التحدي
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.
تعلّم هذا جيدًا
دروس من GCSE Maths حول هذا الموضوع: خطوات محلولة، ثم أسئلة بأسلوب الامتحان بأرقام جديدة في كل مرة. الدروس باللغة الإنجليزية.
- Solving by factorisingQuadraticsمجاني
- Quadratic graphsQuadraticsPlus
- Completing the squareQuadraticsPlus
عرض الدورة كاملةالدورة كاملة واختبارات الموضوعات والامتحانات التجريبية متاحة مع Plus. اطّلع على Plus
الأسئلة الشائعة
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.