Quadratic functions, y = ax² + bx + c with a ≠ 0, are the first topic in Sec 3 A-Math. Later chapters lean on them: the discriminant work in A2 (Quadratic Equations & Inequalities), curve sketching in coordinate geometry, and modelling questions about a ball thrown upwards, a profit that peaks or a fence enclosing the largest area.
The key skill is completing the square to rewrite the general form as y = a(x − h)² + k, then reading off the turning point (h, k). Sketching, finding the greatest or least value and showing that y is always positive all follow from this one rearrangement.
What these notes cover
- •Vertex form and the turning point
- •Completing the square when a ≠ 1
- •Reading off the maximum or minimum value
- •Writing the final answer
- •What a sketch means in A-Math
- •The four-step sketch checklist
- •Symmetry shortcut for a fourth point
- •Quadratic versus reciprocal graphs
- •Always positive, always negative
- •The discriminant method (a preview of A2)
- •Modelling: projectiles and enclosed areas
- •Common mistakes and exam technique
Key definitions
- Parabola
- The graph of any quadratic: a single smooth U-shape (a > 0) or n-shape (a < 0) with one turning point, no asymptotes, and symmetry about the vertical line through its vertex.
- Vertex (completed-square) form
- The form y = a(x − h)² + k, obtained by completing the square, from which the turning point (h, k) and the line of symmetry x = h can be read off.
- Turning point (vertex)
- The point (h, k). It is a minimum point when a > 0 and a maximum point when a < 0; k is the minimum or maximum value of y, occurring at x = h.
- Always positive / always negative
- A quadratic is always positive if y > 0 for all real values of x, so the whole curve lies strictly above the x-axis without even touching it; it is always negative if y < 0 for all real values of x.
- Sketch
- A freehand curve of the correct shape, with axes drawn with a ruler, on which every important feature is labelled with its coordinates or equation. It is not a plotted graph.
Key formulas
Vertex form
y = a(x − h)² + k
(h, k) is the turning point and x = h is the line of symmetry. If a > 0, k is the minimum value; if a < 0, k is the maximum value; both occur at x = h.
Completed square of the general quadratic
Obtained by factoring a out of the x² and x terms only, then adding and subtracting (b ÷ (2a))² inside the bracket.
Line of symmetry / x-coordinate of vertex
A fast check on your answer. Completing the square is the expected working in the exam.
Maximum or minimum value
The value of y at the turning point. Use it to check a completed-square answer.
x-intercepts from vertex form
Found by setting y = 0. Real intercepts exist only when −k ÷ a ≥ 0, i.e. k and a have opposite signs (or k = 0).
Always positive (discriminant)
Always negative needs a < 0 and b² − 4ac < 0. In vertex form: always positive ⇔ a > 0 and k > 0; always negative ⇔ a < 0 and k < 0. Use strict inequalities unless the question says “never negative” or “y ≥ 0”.
Worked example
Express y = −2x² + 12x − 7 in the form a(x − h)² + k. Hence state the maximum value of y and the value of x at which it occurs.
- 1Factor −2 out of the x² and x terms only; since 12x ÷ (−2) = −6x, the sign flips: y = −2(x² − 6x) − 7.
- 2Half the coefficient of x inside the bracket is −3, and its square is 9, so add and subtract 9: y = −2(x² − 6x + 9 − 9) − 7.
- 3The first three terms form a perfect square: y = −2[(x − 3)² − 9] − 7.
- 4Multiply the −9 by the −2 outside to get +18, then combine with −7: y = −2(x − 3)² + 18 − 7 = −2(x − 3)² + 11.
- 5Since −2(x − 3)² ≤ 0 for all real x, y ≤ 11, so the maximum value of y is 11, occurring when x = 3 (maximum point (3, 11), line of symmetry x = 3).
- 6Check with the formulas: h = −12 ÷ (2 × (−2)) = 3 and k = −7 − 144 ÷ (4 × (−2)) = −7 + 18 = 11.
Watch out: Multiplying the −9 by the −2 outside the bracket gives +18, not −18 — a very common sign slip.
Common mistakes
Not factoring out a before completing the square
Wrong: 2x² − 12x + 5 = (x − 6)² − 31
Right: Take out the coefficient of x² first: 2x² − 12x + 5 = 2(x² − 6x) + 5 = 2(x − 3)² − 18 + 5 = 2(x − 3)² − 13.
Wrong sign of h
Wrong: From y = 2(x + 3)² − 7, stating the turning point as (3, −7).
Right: Set the bracket to zero: x + 3 = 0 gives x = −3, so the turning point is (−3, −7).
Proving “always positive” with the discriminant alone
Wrong: “b² − 4ac = −7 < 0, so y > 0 for all x.”
Right: b² − 4ac < 0 only shows the curve never crosses the x-axis; it could lie entirely below it. Also state a > 0, then conclude “y > 0 for all real values of x”.
Questions students ask
How do you complete the square when a is not 1?
Factor a out of the x² and x terms only, leaving the constant c outside the bracket. Inside the bracket, add and subtract (b ÷ (2a))² and write the first three terms as a perfect square. Then multiply the leftover term by a and combine it with c.
How do you tell if the turning point of a quadratic is a maximum or minimum?
Look at the sign of a in y = a(x − h)² + k. If a > 0 the curve opens upwards and k is the minimum value; if a < 0 it opens downwards and k is the maximum value. Either way it occurs at x = h, and the answer should state both the value and where it occurs.
What must a quadratic graph sketch show in A-Math?
The correct orientation, the turning point labelled with coordinates, the line of symmetry drawn dashed and labelled as an equation such as x = 3, and the y-intercept. Include the x-intercepts if they exist and are needed. A correct curve with no labels scores almost nothing.
How do you show a quadratic is always positive for all real x?
Either complete the square and show that a > 0 and k > 0, or show that a > 0 and b² − 4ac < 0. Both conditions are needed — the discriminant alone earns partial marks only. Finish with “y > 0 for all real values of x”.
In the full PDF
- •Labelled sketches of every graph, including the always-positive and always-negative curves
- •Further worked examples: fractional completed squares, a downward-opening sketch with surd intercepts and a range-of-k question
- •Modelling examples on projectile height and a three-sided fenced pen, with valid domains
- •The remaining common mistakes, from swapping maximum and minimum to accepting negative time
- •An eight-point exam technique checklist
- •Section summaries and a full chapter summary