1The idea in one paragraph

A quadratic function is one whose highest power of x is 2: f(x) = ax² + bx + c, with a ≠ 0. Its graph is a symmetric U-shaped curve called a parabola, opening upwards when a > 0 and downwards when a < 0. Every parabola has an axis of symmetry, a vertical line, and a vertex, the turning point that sits on it. The same quadratic can be written in three forms. The standard form ax² + bx + c shows the y-intercept. The factorised form a(x − p)(x − q) shows the x-intercepts. The vertex form a(x − h)² + k shows the vertex. All three describe the same parabola, so the skill of this subtopic is moving between them and knowing which one answers the question in front of you.

2The parabola and the coefficient a

Start with the simplest quadratic, y = x². Squaring never gives a negative answer, and it gives the same answer for 3 and −3, so the graph sits on or above the x-axis and is a mirror image of itself in the y-axis. Its lowest point, the vertex, is the origin.

Multiply by a number a and the shape changes in two ways, as Figure 1 shows.

The sign of a sets the direction. If a > 0 the parabola opens upwards, like a cup, and the vertex is a minimum. If a < 0 every output is turned negative, the parabola opens downwards, like a cap, and the vertex is a maximum. A quick way to remember: a positive a smiles.

The size of a sets the width. y = 3x² climbs three times as fast as y = x², so it looks narrower; y = 0.4x² climbs more slowly and looks wider. Formally, a is a vertical stretch of y = x², which you meet in full in 2.11.

The value of a is the same in all three forms. That is useful: whichever form you are given, you know which way up the graph is before you do any work.

3The standard form, ax² + bx + c

The standard form f(x) = ax² + bx + c gives you two things directly.

The y-intercept is (0, c). Put x = 0 and the first two terms vanish, leaving c.

The axis of symmetry is x = −b/2a. This is in the formula booklet. Once you have the axis, substitute its value into f to get the y-coordinate of the vertex.

Take f(x) = 2x² − 8x + 6, drawn in Figure 2. Here a = 2, b = −8 and c = 6.

Since a = 2 > 0, the vertex is a minimum and the least value of f is −2.

Symmetry gives you extra points for nothing. The y-intercept (0, 6) is 2 units left of the axis x = 2, so its mirror image, 2 units to the right, is (4, 6). On a sketch this matters: plotting the y-intercept and its mirror point makes the parabola symmetric without any more working.

Where does x = −b/2a come from? A parabola's x-intercepts sit equally either side of its axis, so the axis is their average. In 2.7 you will see that the quadratic formula gives the roots as −b/2a plus and minus the same amount, so their average is exactly −b/2a. That also explains why the formula works even when there are no x-intercepts at all.

4The factorised form, a(x − p)(x − q)

The factorised form f(x) = a(x − p)(x − q), also called the intercept form, shows where the graph crosses the x-axis. A product is zero only when one of its factors is zero, so f(x) = 0 exactly when x = p or x = q. The x-intercepts are (p, 0) and (q, 0).

Watch the signs. The bracket (x − 5) gives the intercept x = 5, and the bracket (x + 1) = (x − (−1)) gives x = −1. The number in the bracket appears with its sign flipped.

The axis of symmetry is halfway between the intercepts:

In a(x − p)(x − q), the x-intercepts are (p, 0) and (q, 0), and the axis of symmetry is x = (p + q)/2.

Take f(x) = −(x + 1)(x − 5), drawn in Figure 3. Here a = −1, so it opens downwards.

If the parabola touches the x-axis at one point instead of crossing it, then p = q and the factorised form is a(x − p)², with a single intercept (p, 0) that is also the vertex. And some parabolas never meet the x-axis at all: y = x² + 1 is an example. Those have no factorised form with real numbers, which is exactly what the discriminant in 2.7 detects.

5The vertex form, a(x − h)² + k

The vertex form f(x) = a(x − h)² + k shows the vertex: it is (h, k), and the axis of symmetry is x = h.

Here is why. Take f(x) = (x − 3)² − 4. The squared bracket (x − 3)² can never be negative, and it is zero only when x = 3. So f(x) is at least 0 − 4 = −4, and it equals −4 exactly when x = 3. The lowest point is (3, −4). If a were negative, the same reasoning would give a highest point instead.

Again watch the sign inside the bracket: (x − 3) means h = 3, and (x + 3) means h = −3. The k outside the bracket keeps its sign.

Figure 4 shows another way to see it. The graph of (x − 3)² − 4 is the graph of x² moved 3 units right and 4 units down, so the vertex moves from (0, 0) to (3, −4). The subtopic on transformations (2.11) makes this general: a(x − h)² + k is y = x² stretched vertically by a factor of a and translated by the vector (h, k).

The vertex form also gives the range of a quadratic at once. For f(x) = (x − 3)² − 4 with x ∈ ℝ, the range is f(x) ≥ −4. For f(x) = −(x − 3)² + 4, the range is f(x) ≤ 4.

6Changing from one form to another

The guide expects you to move between all three forms. Figure 5 shows why you would want to: one parabola, y = 2x² − 4x − 6, and the feature each form hands over.

Factorised or vertex form to standard form: expand. This direction is always possible and is the easiest.

Getting the same standard form from both is also a check on your work, and a good one to use whenever a question gives you time.

Standard form to factorised form: factorise. Take out the common factor a first, then factorise what is left.

The x-intercepts are (3, 0) and (−2, 0).

Standard form to vertex form: complete the square. This is the conversion most often examined and the one to practise until it is automatic. When a = 1, halve the coefficient of x, square it, add and subtract it.

When a ≠ 1, take a out of the x² and x terms only, complete the square inside, then multiply back.

The line people get wrong is the fourth: the −9 is inside the bracket that the 2 multiplies, so it becomes −18, not −9. A negative a works the same way, with extra care over signs:

You can check a completed square in two seconds: the vertex x-value must match −b/2a. For 2x² − 12x + 13, −b/2a = 12/4 = 3. It does.

Vertex form to factorised form: set it to zero and solve. From 2(x − 1)² − 8 = 0, you get (x − 1)² = 4, so x − 1 = ±2, giving x = 3 or x = −1. So 2(x − 1)² − 8 = 2(x − 3)(x + 1). This only works when the vertex and the direction of opening allow the parabola to reach the x-axis; if k and a have the same sign, as in 2(x − 1)² + 8, there are no real intercepts and no factorised form.

7Finding the equation of a parabola

Exam questions often give you a sketch or some facts and ask for the equation. The trick is to pick the form that uses what you are given, so that only a is unknown.

Given the vertex and one other point: use vertex form. A parabola has vertex (−2, 7) and passes through (0, 3).

Given the x-intercepts and one other point: use factorised form. A parabola crosses the x-axis at 1 and 6 and the y-axis at −12.

The vertex follows from the axis x = (1 + 6)/2 = 3.5: y = −2(2.5)(−2.5) = 12.5, so the vertex is (3.5, 12.5).

Given the axis and some points: use the standard form with −b/2a. If f(x) = ax² + bx + 5 has axis of symmetry x = 2, then −b/2a = 2, so b = −4a. That one equation, plus one point on the graph, fixes a and b. Question 4 in Try it is this type.

On Paper 2 you can also check any answer by graphing it and seeing that it passes through the given points.

8Quadratics in context

A parabola is the path of anything thrown, and the graph of any quantity whose best value lies between two extremes. In a context, the vertex is usually the answer: the greatest height, the largest area, the least cost.

A ball is thrown upwards from a height of 1.5 m. Its height in metres after t seconds is modelled by h(t) = −5t² + 20t + 1.5. Complete the square to find the greatest height:

The vertex is (2, 21.5) and a = −5 < 0, so it is a maximum: the ball reaches 21.5 m after 2 seconds. On Paper 2 the GDC maximum tool gives the same point directly. Figure 6 shows the model, and the domain it makes sense on: from the throw at t = 0 to the landing at t ≈ 4.07 s, the positive zero of h (found in 2.7 or with the GDC).

Two habits for context questions. Answer in the context's words and units: "21.5 m", "after 2 seconds". And respect the domain: a negative time or a negative length is a solution of the algebra but not of the problem.