Roots and turning points of quadratics - Edexcel GCSE Maths (Calculator) Exam Questions

Practise 28 exam-style questions on Roots and turning points of quadratics for Edexcel GCSE Maths (Calculator). Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with an AI tutor on hand if you get stuck.

28 questions
1–5 marks each
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20 questions, marked as you go

Sample questions with model answers

Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.

Question 1

2 marks

Here is the graph of \(y = x^2 - 2x - 4\)

Use the graph to find estimates for the solutions of \(x^2 - 2x - 4 = 0\)

A coordinate grid with x from −2 to 4 and y from −6 to 5; each grid square is 1 unit across (with 5 small squares along each side of a large square), and the axes cross at the origin O. The curve labelled "y = x² − 2x − 4" is a U-shaped parabola: it meets the x-axis twice; these points can be read from the grid; it crosses the y-axis at (0, −4); it has a minimum point at (1, −5).

Model answer

The solutions are where the curve crosses the x-axis (\(y = 0\)).
\(x \approx -1.2\) and \(x \approx 3.2\)

Explanation

Read the x-values where the curve crosses the x-axis, using the small squares to estimate the decimals. The curve is symmetrical about \(x = 1\), so the two solutions should be the same distance either side of 1 - a useful check.

Mark scheme

1 mark
B1 for a value from −1.35 to −1.15
exact value \(1 - \sqrt{5} = -1.236...\); correct answer with no working scores full marks
1 mark
B1 for a value from 3.15 to 3.35
exact value \(1 + \sqrt{5} = 3.236...\)

Question 2

3 marks

Find the x-coordinates of the points where the curve \(y = 3x^2 - 5x - 4\) crosses the x-axis.
Give your answers correct to 2 decimal places.

Model answer

The curve crosses the x-axis where \(3x^2 - 5x - 4 = 0\)
\(x = \dfrac{5 \pm \sqrt{25 + 48}}{6} = \dfrac{5 \pm \sqrt{73}}{6}\)
\(x = \dfrac{5 + 8.5440...}{6} = 2.2573...\) or \(x = \dfrac{5 - 8.5440...}{6} = -0.5906...\)
\(x = 2.26\) or \(x = -0.59\) (2 d.p.)

Explanation

On the x-axis \(y = 0\), so solve \(3x^2 - 5x - 4 = 0\). It does not factorise, so use the quadratic formula with \(a = 3\), \(b = -5\), \(c = -4\). Careful: \(-4ac = -4 \times 3 \times (-4) = +48\), and divide the whole numerator by \(2a = 6\).

Mark scheme

1 mark
M1 for a correct substitution into the quadratic formula, e.g. \(\dfrac{5 \pm \sqrt{(-5)^2 - 4 \times 3 \times (-4)}}{2 \times 3}\)
allow one sign error in the substitution; correct answer with no working scores full marks
1 mark
M1 for \(\dfrac{5 \pm \sqrt{73}}{6}\)
oe, e.g. \(\dfrac{5 \pm 8.544...}{6}\)
1 mark
A1 for 2.26 and −0.59
awrt 2.26 and −0.59 (2.257... and −0.590...)

Question 3

3 marks

Here is the graph of \(y = x^2 - 2.5x - 3.5\)

Use the graph to write down the roots of \(x^2 - 2.5x - 3.5 = 0\)
Use the symmetry of the curve to find the exact coordinates of its turning point.

A coordinate grid with x from −2 to 5 and y from −6 to 5; each grid square is 1 unit across (with 2 small squares along each side of a large square), and the axes cross at the origin O. The curve labelled "y = x² − 2.5x − 3.5" is a U-shaped parabola: it meets the x-axis twice; these points can be read from the grid; it crosses the y-axis at (0, −3.5); it has a minimum point, which can be read from the grid.

Model answer

Roots: \(x = -1\) and \(x = 3.5\)
The line of symmetry is halfway between the roots:
\(x = \dfrac{-1 + 3.5}{2} = 1.25\)
\(y = 1.25^2 - 2.5 \times 1.25 - 3.5 = 1.5625 - 3.125 - 3.5 = -5.0625\)
Turning point: (1.25, −5.0625)

Explanation

A parabola is symmetrical about a vertical line through its turning point, which is halfway between the roots. The y-coordinate cannot be read exactly from the graph, so substitute \(x = 1.25\) into the equation.

Mark scheme

1 mark
B1 for roots \(x = -1\) and \(x = 3.5\)
both needed (exact values); correct answer with no working scores full marks
1 mark
M1 for the line of symmetry \(x = \dfrac{-1 + 3.5}{2} = 1.25\)
ft their roots
1 mark
A1 for (1.25, −5.0625)
cao; an estimate read from the graph such as (1.25, −5.1) scores M1 A0

Question 4

4 marks

The curve with equation \(y = x^2 + bx + c\) has a minimum point at \((3.2, -1.44)\).

Find the coordinates of the points where the curve crosses the x-axis and the y-axis.

Model answer

\(y = (x - 3.2)^2 - 1.44\)
x-axis: \((x - 3.2)^2 = 1.44\), \(x - 3.2 = \pm 1.2\)
\(x = 2\) or \(x = 4.4\): points (2, 0) and (4.4, 0)
y-axis: \(y = (-3.2)^2 - 1.44 = 10.24 - 1.44 = 8.8\): point (0, 8.8)

Explanation

Because the coefficient of \(x^2\) is 1, the minimum point gives the completed-square form \(y = (x - 3.2)^2 - 1.44\) directly. The roots are equally spaced either side of \(x = 3.2\). Check by expanding: \(x^2 - 6.4x + 8.8 = (x - 2)(x - 4.4)\).

Mark scheme

1 mark
M1 for \((x - 3.2)^2 - 1.44 = 0\)
oe, e.g. \(x^2 - 6.4x + 8.8 = 0\); correct answer with no working scores full marks
1 mark
A1 for (2, 0) and (4.4, 0)
cao
1 mark
M1 for substituting \(x = 0\), e.g. \((0 - 3.2)^2 - 1.44\)
or expanding to find \(c\)
1 mark
A1 for (0, 8.8)
cao

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