3.3bHigher tier4MA1 · Section 3: Sequences, functions and graphs

Quadratic graphs (sketch, solve, turning point)

Key learnings

A quadratic graph is a parabola
$y = ax^2 + bx + c$ gives a U-shape ( $a > 0$ ) or ∩-shape ( $a < 0$ ). It crosses the x-axis at the solutions of $ax^2+bx+c=0$ and the y-axis at $(0, c)$ .
Completed square gives the turning point
Writing $y = (x+p)^2 + q$ shows the turning point is at $(-p, q)$ — the minimum (if $a > 0$ ) or maximum (if $a < 0$ ) of the curve.
Solve by reading roots from the graph
Where the curve crosses the x-axis ( $y = 0$ ) are the solutions; symmetry puts the turning point exactly halfway between the two roots.

Common mistakes

Drawn from real examiner reports.

No sketch for a quadratic inequality

Students who drew a sketch were far more likely to write the correct final inequality (a single middle region vs two outer regions); those who did not often gave the wrong direction.

Highlighted on November 2024 Paper 2H (Q20) and the June 2024 Paper 2H summary.

Reading factors off a calculator

Factors read from a calculator gave forms like $(x + 2.1)(x - 1)$ that do not expand back to the given quadratic, and scored nothing where algebraic working was required.

Seen on November 2024 Paper 2H (Q20).

Misreading the graph scale

Marks were lost reading values off the axes when each square was not one unit — always check the scale before reading a root or a turning point.

Noted in the June 2024 Paper 1H summary (reading graph scales accurately).

Sign slip completing the square

For $y = x^2 - 6x + 5$ you get $(x - 3)^2 - 9 + 5 = (x - 3)^2 - 4$ . Forgetting to subtract the $9$ , or dropping the $+5$ , gives the wrong constant and so the wrong turning point.

Turning point x has the opposite sign to p

From $y = (x + p)^2 + q$ the turning point is $(-p, q)$ : the x-coordinate is $-p$ , the opposite sign to $p$ . So $(x - 3)^2 - 4$ turns at $(3, -4)$ , not $(-3, -4)$ .

Exam tips

Plot a smooth curve and find symmetry
Join plotted points with a single smooth curve (never ruled segments) and use the line of symmetry $x = -\tfrac{b}{2a}$ to locate the turning point.
Sketch first for an inequality
For a quadratic inequality, sketch the parabola and mark the roots. Then $> 0$ is where the curve is above the axis (the two outer regions) and $< 0$ is the single region between the roots.
Check roots by expanding back
After factorising, multiply the brackets out in your head to confirm they give the original quadratic. Calculator-style factors such as $(x + 2.1)(x - 1)$ will fail this check.

Revision notes

$y = ax^2 + bx + c$ is a parabola: a U-shape if $a > 0$ , an ∩-shape if $a < 0$ .

y-intercept: put $x = 0 \Rightarrow (0, c)$ .
Roots (x-intercepts): solve $ax^2 + bx + c = 0$ (factorise, formula, or read off the graph where $y = 0$ ).
Turning point: the lowest (or highest) point.

Flashcards

Foundation

Where does y = x² + bx + c cross the y-axis?

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Quadratic graphs (sketch, solve, turning point)

Key learnings

Common mistakes

Exam tips

Revision notes

Shape and key features

Finding the turning point by completing the square

Solving from a graph

Quadratic inequalities — always sketch

Flashcards

Practice questions