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Sketching Quadratics

HigherHigher tier onlyAQAEdexcelOCR

Get to grips with sketching quadratics using these Higher GCSE Maths practice questions. The worksheet focuses on sketching quadratic graphs and turning points, and the accompanying mark scheme breaks down each solution clearly. Suitable for AQA, Edexcel and OCR. Download the questions and answers for free. Mark the turning point and the roots as the key features.

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These worksheets and mark schemes are original, written for Virtus Academy and checked against the current AQA, Edexcel and OCR specifications. Every worksheet comes with a full mark scheme.

This is a Higher tier only topic, so there's no Foundation paper — only the Higher worksheet and mark scheme below.

Topic overview

Sketching a quadratic means drawing its general shape and key features without plotting a table of points. It tests whether you can read the important information straight from the equation.

Three features are needed: the direction of the curve, the roots, and the \(y\)-intercept. The direction comes from the sign of the \(x^2\) term, the roots from factorising, and the intercept from setting \(x = 0\).

The turning point can be found from the completed-square form, or by using the symmetry: it lies exactly halfway between the two roots. Labelling these features is what earns the marks — a sketch without labels shows nothing.

Revision notes

Finding the roots

Factorise and set each bracket to zero.

For \(y = x^2 - 5x + 6\), factorising gives \((x-2)(x-3)\), so the roots are \(x = 2\) and \(x = 3\). These are where the curve meets the horizontal axis.

Finding the y-intercept

Substitute \(x = 0\) into the equation. The constant term is the answer.

For \(y = x^2 - 5x + 6\), setting \(x = 0\) gives \(y = 6\), so the curve crosses the vertical axis at \((0, 6)\).

Finding the turning point

The turning point sits midway between the roots, so average them to find its \(x\) value, then substitute to find \(y\).

Here the midpoint of \(2\) and \(3\) is \(2.5\), and substituting gives \(y = -0.25\), so the minimum is at \((2.5, -0.25)\).

Key points

  • A sketch shows shape and key features, not plotted points.
  • The sign of \(x^2\) gives the direction.
  • Factorise to find the roots.
  • Set \(x = 0\) to find the y-intercept.
  • The turning point lies midway between the roots.
  • Label every feature you find.

Worked examples

Example 1

Find the roots of \(y = x^2 - 7x + 10\).

Working

\[(x - 2)(x - 5)\]factorise the quadratic
\[x = 2 \text{ and } x = 5\]set each bracket equal to zero

Example 2

Find the y-intercept of \(y = x^2 + 3x - 4\).

Working

\[y = 0^2 + 3(0) - 4\]substitute x = 0
\[(0, -4)\]the intercept is the constant term

Example 3

Find the turning point of \(y = x^2 - 6x + 8\).

Working

\[(x - 2)(x - 4) \Rightarrow x = 2, 4\]factorise to find the roots
\[\frac{2 + 4}{2} = 3\]the turning point is midway between them
\[y = 9 - 18 + 8 = -1 \text{, so } (3, -1)\]substitute to find the y value

Common mistakes

  • Plotting a full table when a sketch is asked for.

    A sketch needs the shape and labelled features, not accurate plotting.

  • Forgetting to label the roots and intercept.

    The marks are for the labelled features, not the curve itself.

  • Getting the direction wrong.

    A negative x² coefficient opens downwards with a maximum.

  • Assuming the turning point is on the y-axis.

    It sits midway between the roots, which is rarely at x = 0.

Exam tips

  • Factorise first — the roots do most of the work.
  • Label the roots, the y-intercept and the turning point.
  • Check the direction against the sign of the x² term.
  • Use the symmetry to locate the turning point quickly.

Key terms

Sketch
A diagram showing shape and key features without exact plotting.
Root
Where the curve crosses the x-axis.
y-intercept
Where the curve crosses the y-axis.
Symmetry
The mirror property of a parabola about its turning point.

Written and reviewed against the current AQA, Edexcel and OCR specifications. Spotted an error? Let us know.