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

Practise 28 exam-style questions on Roots and turning points of quadratics for Edexcel GCSE Maths (Non-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 \(y = x^2 + 2x - 3\).

Write down the equation of the line of symmetry of the curve and the coordinates of its turning point.

A coordinate grid with x from −4 to 2 and y from −5 to 6; each grid square is 1 unit across, and the axes cross at the origin O. The curve labelled "y = x² + 2x − 3" is a U-shaped parabola: it crosses the x-axis at (−3, 0) and (1, 0); it crosses the y-axis at (0, −3); it has a minimum point, which can be read from the grid.

Model answer

Line of symmetry: \(x = -1\)
Turning point: \((-1, -4)\)

Explanation

The line of symmetry is the vertical line through the turning point, halfway between the roots \(-3\) and 1. A vertical line has an equation of the form \(x = k\), so write \(x = -1\), not just \(-1\) or \(y = -1\).

Mark scheme

1 mark
B1 for \(x = -1\)
must be an equation; correct answer with no working scores full marks
1 mark
B1 for \((-1, -4)\)
cao

Question 2

2 marks

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

The equation \(x^2 - 4x = k\), where \(k\) is a constant, has two different solutions.
Use the graph to find the range of possible values of \(k\).

A coordinate grid with x from −1 to 5 and y from −5 to 6; each grid square is 1 unit across, and the axes cross at the origin O. The curve labelled "y = x² − 4x" is a U-shaped parabola: it crosses the x-axis at (0, 0) and (4, 0); it has a minimum point at (2, −4).

Model answer

The solutions are where the horizontal line \(y = k\) meets the curve.
The minimum point of the curve is \((2, -4)\).
The line meets the curve twice when it is above the minimum point.
So \(k > -4\).

Explanation

Think of the horizontal line \(y = k\) moving up and down. Below the turning point it misses the curve, at \(y = -4\) it just touches (one solution), and above it cuts the curve twice. So two different solutions need \(k > -4\), with a strict inequality.

Mark scheme

1 mark
M1 for identifying the minimum value \(-4\) or the idea that the line \(y = k\) must cross the curve twice
correct answer with no working scores full marks
1 mark
A1 for \(k > -4\)
do not accept \(k \ge -4\)

Question 3

3 marks

The diagram shows the curve with equation \(y = x^2 + bx + c\), where \(b\) and \(c\) are integers.

Find the value of \(b\) and the value of \(c\).

A coordinate grid with x from −1 to 8 and y from −5 to 13; each grid square is 1 unit across, and the axes cross at the origin O. The curve 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 a point that can be read from the grid; it has a minimum point, which can be read from the grid.

Model answer

The curve crosses the x-axis at \(x = 2\) and \(x = 6\).
\(y = (x - 2)(x - 6) = x^2 - 8x + 12\)
\(b = -8\), \(c = 12\)

Explanation

If the roots are 2 and 6, the quadratic factorises as \((x - 2)(x - 6)\), because the coefficient of \(x^2\) is 1. Expanding gives \(b\) and \(c\). Check with the turning point \((4, -4)\): \(16 - 32 + 12 = -4\).

Mark scheme

1 mark
B1 for reading the roots 2 and 6 from the graph
or the y-intercept 12, or the turning point (4, −4); correct answer with no working scores full marks
1 mark
M1 for \((x - 2)(x - 6)\) or \((x - 4)^2 - 4\) expanded
ft their roots
1 mark
A1 for \(b = -8\) and \(c = 12\)
both needed

Question 4

4 marks

A ball is thrown upwards from the ground.
Its height, \(h\) metres, after \(t\) seconds is given by
\[h = 20t - 5t^2\]
By completing the square, find the maximum height of the ball and the time at which it reaches this height.
Find also the time at which the ball hits the ground.

Model answer

\(h = 20t - 5t^2 = -5(t^2 - 4t)\)
\(= -5[(t - 2)^2 - 4]\)
\(= 20 - 5(t - 2)^2\)
\(5(t - 2)^2 \ge 0\), so the greatest value of \(h\) is 20, when \(t = 2\).
Maximum height 20 m after 2 seconds.
Ground: \(h = 0\), \(5t(4 - t) = 0\), so \(t = 0\) or \(t = 4\). The ball hits the ground after 4 seconds.

Explanation

With a negative \(t^2\) term, take out \(-5\) first, then complete the square inside the bracket. Because a square is never negative, subtracting \(5(t - 2)^2\) can only make \(h\) smaller, so the maximum is 20 at \(t = 2\). The ball lands at the other root, \(t = 4\), which is symmetrical about \(t = 2\).

Mark scheme

1 mark
M1 for \(-5(t^2 - 4t)\)
correct answer with no working scores full marks
1 mark
M1 for \(-5[(t - 2)^2 - 4]\) or \(20 - 5(t - 2)^2\)
1 mark
A1 for maximum height 20 m at \(t = 2\) seconds
both needed
1 mark
B1 for 4 seconds
from \(5t(4 - t) = 0\) or symmetry

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