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Turning Points and Symmetry of Quadratics Revision Notes | GCSE AQA Higher Maths

Write the quadratic as a(x + p)² + q, and the turning point is (-p, q). Free GCSE AQA Maths notes with worked examples and 11 questions answered.

Updated 24 Mar 2026 · Revise Right Team

Turning Points and Symmetry of Quadratics is part of Graphs in GCSE AQA Higher Maths.

4 min

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39

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GCSE AQA Higher

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What this lesson covers

Write the quadratic as a(x + p)² + q, and the turning point is (-p, q). Free GCSE AQA Maths notes with worked examples and 11 questions answered.

  • Deduce the turning point of a quadratic by completing the square
  • State the coordinates of the turning point directly from completed square form
  • Explain the symmetrical property of a quadratic and identify the axis of symmetry
  • Sketch a quadratic graph labelling the turning point, axis of symmetry, and intercepts

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Key ideas to keep in view

Use this lesson to stay inside Graphs while connecting the detail back to the wider Algebra topic.

  • Turning point: The point on a quadratic curve where the graph changes direction. A minimum turning point is the lowest point of a U-shaped parabola; a maximum turning point is the highest point of an inverted parabola.
  • Completed square form: A rewritten form of a quadratic expression $a(x+p)^2 + q$, used to reveal the turning point coordinates $(-p,\, q)$ and the axis of symmetry $x = -p$.
  • Axis of symmetry: A vertical line that passes through the turning point of a parabola, dividing it into two mirror-image halves. For $a(x+p)^2 + q$, the axis of symmetry is $x = -p$.
  • Parabola: The U-shaped (or inverted U-shaped) curve produced by any quadratic function. All parabolas have exactly one turning point and one axis of symmetry.

Revision notes

Read the core explanations from Turning Points and Symmetry of Quadratics before testing yourself from memory.

1. Why Turning Points Matter

Every quadratic graph has one special point where the curve stops falling and starts rising — or vice versa. That point is the turning point, and knowing exactly where it sits is one of the most powerful things you can do with a quadratic.

On Higher-tier papers, you will be asked to deduce the turning point — meaning you must derive it algebraically, not read it off a graph. The method the specification requires is completing the square. Once you see how it works, you will also unlock the symmetry of every parabola for free.

2. The Symmetrical Property of a Quadratic

Before diving into algebra, notice something visual. A parabola is perfectly symmetrical: fold it along a vertical line through its turning point and both halves match exactly. That vertical line is called the axis of symmetry.

The axis of symmetry always passes through the turning point. This means: once you know the turning point, you know the axis of symmetry instantly — they share the same $x$-coordinate.

Here is the key insight: the symmetry of a parabola means that if you know where the curve crosses the $x$-axis at two points, the axis of symmetry — and therefore the turning point — lies exactly halfway between those two roots. That connection becomes very useful when sketching.

3. Completing the Square to Find the Turning Point

The standard form $y = ax^2 + bx + c$ hides the turning point. Completing the square rewrites the quadratic into the form $a(x+p)^2 + q$, which reveals the turning point coordinates directly as $(-p,\, q)$.

The process of completing the square works by creating a perfect square bracket and then adjusting with a constant to keep the expression equivalent to the original. Work through the steps below carefully — each one earns method marks in an exam.

Worked example: Completing the Square: $y = x^2 + 6x + 7$

Write $y = x^2 + 6x + 7$ in completed square form and hence state the turning point and axis of symmetry.

  1. Coefficient of $x$ is $6$. Half of $6$ is $3$. — This value goes inside the bracket: $(x + 3)^2$. [M1 — starting the completing-the-square process]
  2. $(x+3)^2 = x^2 + 6x + 9$ — $(x+3)^2$ gives an extra $+9$ that was not in the original expression, so we must subtract $9$ to compensate.
  3. $y = (x+3)^2 - 9 + 7$ — Replace $x^2 + 6x$ with $(x+3)^2 - 9$, then bring in the original constant $+7$. [M1 — correct bracket and adjustment]
  4. $y = (x+3)^2 - 2$ — $-9 + 7 = -2$. The expression is now in completed square form. [A1 — correct simplified form]
  5. Turning point: $(-3,\, -2)$ — From $a(x+p)^2 + q$: here $p = 3$ so $x = -3$; $q = -2$ so $y = -2$. [A1 — correct coordinates]
  6. Axis of symmetry: $x = -3$ — The axis of symmetry is the vertical line through the turning point. Since $a = 1 > 0$, this is a minimum. [A1]

Answer: $y = (x+3)^2 - 2$; turning point $(-3, -2)$; axis of symmetry $x = -3$; minimum point.

4. When the Coefficient of $x^2$ is Not 1

The process above assumed the coefficient of $x^2$ is $1$. When it is not, you must factor out $a$ from the $x^2$ and $x$ terms first, then complete the square inside the bracket.

Worked example: Completing the Square: $y = 2x^2 - 8x + 3$

Write $y = 2x^2 - 8x + 3$ in the form $a(x+p)^2 + q$ and hence find the turning point and axis of symmetry.

  1. $y = 2(x^2 - 4x) + 3$ — Factor $2$ from $2x^2 - 8x$. The constant $+3$ stays outside. [M1 — correct factoring]
  2. Coefficient of $x$ inside is $-4$. Half of $-4$ is $-2$. — The bracket will be $(x - 2)^2$.
  3. $(x-2)^2 = x^2 - 4x + 4$ — $(x-2)^2$ introduces an extra $+4$ inside the bracket, so we subtract $4$ inside to compensate.
  4. $y = 2\left[(x-2)^2 - 4\right] + 3$ — Replace $x^2 - 4x$ with $(x-2)^2 - 4$ inside the factor of $2$. [M1 — correct bracket and inner adjustment]
  5. $y = 2(x-2)^2 - 8 + 3 = 2(x-2)^2 - 5$ — $2 \times (-4) = -8$, then $-8 + 3 = -5$. [A1 — correct simplified form]
  6. Turning point: $(2,\, -5)$ Axis of symmetry: $x = 2$ — From $2(x-2)^2 - 5$: $p = -2$ so $x = -(-2) = 2$; $q = -5$. Since $a = 2 > 0$, this is a minimum. [A1]

Answer: $y = 2(x-2)^2 - 5$; turning point $(2, -5)$; axis of symmetry $x = 2$; minimum point.

5. Minimum and Maximum Turning Points

The sign of $a$ — the coefficient of $x^2$ — tells you immediately whether the turning point is a minimum or a maximum. This is a one-second check that examiners expect you to state.

  • $a > 0$: parabola opens upward (U-shape); turning point is a minimum. The $y$-value at the turning point is the smallest value the function takes.
  • $a < 0$: parabola opens downward (inverted U-shape); turning point is a maximum. The $y$-value at the turning point is the largest value the function takes.

Worked example: Maximum Turning Point: $y = -x^2 + 4x + 1$

Write $y = -x^2 + 4x + 1$ in completed square form. State the turning point, axis of symmetry, and whether it is a minimum or maximum.

  1. $y = -(x^2 - 4x) + 1$ — Factor $-1$ from $-x^2 + 4x$. The constant $+1$ stays outside. [M1]
  2. Coefficient of $x$ inside is $-4$. Half of $-4$ is $-2$. — The bracket will be $(x - 2)^2$.
  3. $(x-2)^2 = x^2 - 4x + 4$ — An extra $+4$ is introduced inside the bracket, so subtract $4$ inside to compensate.
  4. $y = -\left[(x-2)^2 - 4\right] + 1$ — Replace $x^2 - 4x$ with $(x-2)^2 - 4$ inside the factor of $-1$. [M1]
  5. $y = -(x-2)^2 + 4 + 1 = -(x-2)^2 + 5$ — $-1 \times (-4) = +4$, then $4 + 1 = 5$. [A1]
  6. Turning point: $(2,\, 5)$ Axis of symmetry: $x = 2$ Type: maximum (since $a = -1 < 0$) — From $-(x-2)^2 + 5$: $p = -2$ so $x = 2$; $q = 5$. The negative $a$ means the parabola opens downward — maximum. [A1]

Answer: $y = -(x-2)^2 + 5$; turning point $(2, 5)$; axis of symmetry $x = 2$; maximum point.

6. Sketching a Quadratic Using the Turning Point

A sketch question asks you to draw the approximate shape of the curve, labelling key features. You do not need to plot every point — but you must show the turning point, the axis of symmetry, and any intercepts with the axes.

The symmetrical property of the parabola is particularly useful here. If you find where the curve crosses the $y$-axis (substitute $x = 0$), you can reflect that point across the axis of symmetry to find a second point on the curve at the same height — giving your sketch more accuracy with minimal extra work.

Worked example: Sketching $y = x^2 + 6x + 7$ Using Completed Square Form

Sketch the graph of $y = x^2 + 6x + 7$, labelling the turning point, axis of symmetry, and $y$-intercept. (We already found $y = (x+3)^2 - 2$.)

  1. Turning point: $(-3,\, -2)$ — Read directly from $(x+3)^2 - 2$: $p = 3$ so $x = -3$; $q = -2$. Plot and label this point. [B1]
  2. Axis of symmetry: $x = -3$ — Draw a dashed vertical line at $x = -3$ through the turning point. [B1]
  3. Substitute $x = 0$: $y = 0^2 + 6(0) + 7 = 7$ $y$-intercept: $(0,\, 7)$ — Plot $(0, 7)$ on the sketch. [B1]
  4. $(0, 7)$ is $3$ units to the right of $x = -3$. Reflected point is $3$ units to the left: $(-6,\, 7)$. — The symmetrical property means $(-6, 7)$ is also on the curve. This gives the sketch a balanced shape.
  5. Draw a smooth U-shape through $(-6, 7)$, $(-3, -2)$, and $(0, 7)$, opening upward. — Since $a = 1 > 0$, the parabola opens upward. Label the turning point $(-3, -2)$ and the $y$-intercept $(0, 7)$. [B1]

Answer: U-shaped parabola with turning point $(-3, -2)$, axis of symmetry $x = -3$, and $y$-intercept $(0, 7)$.

7. Exam Technique and Common Mistakes

Completing the square is a reliable method — but there are a handful of places where marks are regularly dropped. The guidance below covers the errors that appear most often in mark schemes.

A strong way to describe a quadratic is to link each feature to its algebra. The turning point tells you the maximum or minimum value, the axis of symmetry tells you where the graph balances, and the roots show where the expression equals zero. When a quadratic is written in completed-square form, the turning point and symmetry line can be read directly; when it is written in factorised form, the roots are immediate. In the exam, move flexibly between these forms so you can explain not just what the graph looks like, but why those features appear where they do.

Frequently asked questions

What is a turning point on a quadratic graph?

A turning point is the point where a quadratic curve stops falling and starts rising, or vice versa. For a U-shaped parabola it is a minimum, the smallest value the function takes, and the axis of symmetry always passes through it.

How do you find the turning point by completing the square?

Write the quadratic in the form a(x + p)² + q; the turning point is then (-p, q). For y = x² + 6x + 7, halve 6 to get (x + 3)², which adds 9, so y = (x + 3)² - 9 + 7 = (x + 3)² - 2 and the turning point is (-3, -2).

How do you know if a turning point is a maximum or minimum?

The sign of a, the coefficient of x², shows the type: if a is more than 0 the parabola is U-shaped and the turning point is a minimum; if a is less than 0 it is a maximum. For y = -x² + 4x + 1 = -(x - 2)² + 5, the maximum is (2, 5).

How do you complete the square for x² + 6x + 7?

x² + 6x + 7 in completed square form is (x + 3)² - 2. Half of 6 is 3, and (x + 3)² = x² + 6x + 9, so subtract 9 and add 7 to get (x + 3)² - 2, giving a turning point at (-3, -2).

What is the turning point of y = 2x² - 8x + 3?

The turning point of y = 2x² - 8x + 3 is (2, -5), a minimum. Factoring out 2 gives 2(x² - 4x) + 3 = 2[(x - 2)² - 4] + 3 = 2(x - 2)² - 5, and the axis of symmetry is x = 2.

How do you read the turning point from completed square form?

For a quadratic in the form a(x + p)² + q, the turning point is (-p, q), and the axis of symmetry is x = -p. For example, (x + 3)² - 2 has p = 3 and q = -2, so the turning point is (-3, -2).

How do you use symmetry to find another point on a parabola?

A point is reflected across the axis of symmetry to find another point at the same height. For y = x² + 6x + 7, the y-intercept (0, 7) is 3 units right of x = -3, so (-6, 7) is also on the curve.

Why is the x-coordinate of the turning point -p?

The x-coordinate of the turning point is -p because the bracket (x + p)² equals zero when x = -p, giving the smallest or largest value of the expression. A common slip is to forget the sign change, so (x - 2)² - 5 has its turning point at x = 2, not -2.

What is the minimum value of a quadratic?

The minimum value of a quadratic with a > 0 is the y-value at its turning point, q in the form a(x + p)² + q. For y = (x + 3)² - 2, the minimum value is -2, when x = -3.

What does it mean to deduce a turning point?

Deducing a turning point means deriving it algebraically rather than reading it off a graph. The method required is completing the square, which rewrites the quadratic as a(x + p)² + q so the turning point (-p, q) can be stated directly.

What happens when you complete the square for -x² + 4x + 1?

-x² + 4x + 1 in completed square form is -(x - 2)² + 5. Factoring out -1 gives -(x² - 4x) + 1 = -[(x - 2)² - 4] + 1, and since a = -1 is negative, the turning point (2, 5) is a maximum.

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