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AQA MathsAlgebra

Quadratic graphs

Plot and interpret curved quadratic graphs.

AQAGCSE MathsAlgebraFoundation and Higher
Visual model

Quadratic graphs are U-shaped curves

y=x2y = x^{2}turning point
Gold-standard guide
20 mins

What you will learn

Plot and interpret curved quadratic graphs.
Use a clear step-by-step method for quadratic graphs.
Check your answer and avoid the most common exam mistake.
Useful before you start
Core number skillsEarlier algebra skillsShowing clear working
Core knowledge

Know the rule, then use it

These are the short notes. Read each one, then check you can use it in the worked example below.

Method

Method

For y = x2 + bx + c: y-intercept is c, x-intercepts come from factorising, turning point x-coordinate is −b/2a

Step 1

Find the y-intercept by setting x = 0

y = 0 − 0 + 6 = 6, so the graph crosses the y-axis at (0, 6)

Step 2

Find x-intercepts by factorising the quadratic

x2 − 5x + 6 = (x − 2)(x − 3) = 0, giving x = 2 and x = 3

Step 3

Find the turning point x-coordinate using symmetry

x = (2 + 3)/2 = 2.5, or use x = −b/2a = 52\frac{5}{2} = 2.5

Watch out

Watch out

Students find the x-intercepts but forget to calculate the y-coordinate of the turning point

f
Roots

The roots are where the graph crosses the x-axis.

f
Turning point

Completing the square helps find the minimum or maximum.

Worked example

Sketch y = x2 − 5x + 6, showing the y-intercept, x-intercepts and turning point.

1

Find the y-intercept by setting x = 0: y = 0 − 0 + 6 = 6, so the graph crosses the y-axis at (0, 6).

2

Find x-intercepts by factorising the quadratic: x2 − 5x + 6 = (x − 2)(x − 3) = 0, giving x = 2 and x = 3.Points (2, 0) and (3, 0).

3

Find the turning point x-coordinate using symmetry: x = (2 + 3)/2 = 2.5, or use x = −b/2a = 52\frac{5}{2} = 2.5.

4

25. 25).

Final answer

y-intercept (0, 6), x-intercepts (2, 0) and (3, 0), turning point (2.5, −0.25)

Question ladder

Build up to the hardest questions

Do them in order. If you miss a step, read the solution, then redo the question without looking.

Workedreasoning

Sketch y = x2 − 5x + 6, showing the y-intercept, x-intercepts and turning point.

4 marks4 minsquadratic-graphs-worked
Show solution
Worked solution
  1. 1.Find the y-intercept by setting x = 0: y = 0 − 0 + 6 = 6, so the graph crosses the y-axis at (0, 6).
  2. 2.Find x-intercepts by factorising the quadratic: x2 − 5x + 6 = (x − 2)(x − 3) = 0, giving x = 2 and x = 3.Points (2, 0) and (3, 0).
  3. 3.Find the turning point x-coordinate using symmetry: x = (2 + 3)/2 = 2.5, or use x = −b/2a = 52\frac{5}{2} = 2.5.
  4. 4.25. 25).
Final answer

y-intercept (0, 6), x-intercepts (2, 0) and (3, 0), turning point (2.5, −0.25)

Mark points
  • M1: find the y-intercept by setting x = 0
  • M1: find x-intercepts by factorising the quadratic
  • M1: find the turning point x-coordinate using symmetry
  • M1: calculate the turning point y-coordinate
  • A1: y-intercept (0, 6), x-intercepts (2, 0) and (3, 0), turning point (2.5, −0.25)
Watch out

Students find the x-intercepts but forget to calculate the y-coordinate of the turning point.Substituting x = −b/2a back into the equation is essential — never just mark the turning point on the axis.

Diagnosticrecall

Find the y-intercept and x-intercepts of y = x2 − 7x + 10

1 mark2 minsquadratic-graphs-q1
Show solution
Worked solution
  1. 1.Spot the skill: For y = x2 + bx + c: y-intercept is c, x-intercepts come from factorising, turning point x-coordinate is −b/2a.
  2. 2.Use the find the y-intercept by setting x = 0 stage first, then find x-intercepts by factorising the quadratic.
  3. 3.Keep the final answer visible: y-intercept (0, 10), x-intercepts (2, 0) and (5, 0).
Final answer

y-intercept (0, 10), x-intercepts (2, 0) and (5, 0)

Mark points
  • M1: use the correct for y = x2 + bx + c: y-intercept is c, x-intercepts come from factorising, turning point x-coordinate is −b/2a.the parabola always has a line of symmetry exactly halfway between the two roots.
  • A1: y-intercept (0, 10), x-intercepts (2, 0) and (5, 0)
Watch out

Students find the x-intercepts but forget to calculate the y-coordinate of the turning point.Substituting x = −b/2a back into the equation is essential — never just mark the turning point on the axis.

Easyprocedure

Find the coordinates of the turning point of y = x2 − 6x + 5

2 marks3 minsquadratic-graphs-q2
Show solution
Worked solution
  1. 1.Spot the skill: For y = x2 + bx + c: y-intercept is c, x-intercepts come from factorising, turning point x-coordinate is −b/2a.
  2. 2.Use the find x-intercepts by factorising the quadratic stage first, then find the turning point x-coordinate using symmetry.
  3. 3.Keep the final answer visible: (3, −4).
Final answer

(3, −4)

Mark points
  • M1: use the correct for y = x2 + bx + c: y-intercept is c, x-intercepts come from factorising, turning point x-coordinate is −b/2a.the parabola always has a line of symmetry exactly halfway between the two roots.
  • A1: (3, −4)
Watch out

Students find the x-intercepts but forget to calculate the y-coordinate of the turning point.Substituting x = −b/2a back into the equation is essential — never just mark the turning point on the axis.

Mediumreasoning

Describe the graph of y = −x2 + 4. State whether it has a maximum or minimum.

3 marks4 minsquadratic-graphs-q3
Show solution
Worked solution
  1. 1.Spot the skill: For y = x2 + bx + c: y-intercept is c, x-intercepts come from factorising, turning point x-coordinate is −b/2a.
  2. 2.Use the find the turning point x-coordinate using symmetry stage first, then calculate the turning point y-coordinate.
  3. 3.Keep the final answer visible: n-shaped parabola, y-intercept (0, 4), maximum point (0, 4).
Final answer

n-shaped parabola, y-intercept (0, 4), maximum point (0, 4)

Mark points
  • M1: use the correct for y = x2 + bx + c: y-intercept is c, x-intercepts come from factorising, turning point x-coordinate is −b/2a.the parabola always has a line of symmetry exactly halfway between the two roots.
  • A1: n-shaped parabola, y-intercept (0, 4), maximum point (0, 4)
Watch out

Students find the x-intercepts but forget to calculate the y-coordinate of the turning point.Substituting x = −b/2a back into the equation is essential — never just mark the turning point on the axis.

Hardproblem solving

Find where y = x2 + 2x − 8 crosses the axes.

3 marks5 minsquadratic-graphs-q4
Show solution
Worked solution
  1. 1.Spot the skill: For y = x2 + bx + c: y-intercept is c, x-intercepts come from factorising, turning point x-coordinate is −b/2a.
  2. 2.Use the calculate the turning point y-coordinate stage first, then find the y-intercept by setting x = 0.
  3. 3.Keep the final answer visible: y-intercept (0, −8), x-intercepts (2, 0) and (−4, 0).
Final answer

y-intercept (0, −8), x-intercepts (2, 0) and (−4, 0)

Mark points
  • M1: use the correct for y = x2 + bx + c: y-intercept is c, x-intercepts come from factorising, turning point x-coordinate is −b/2a.the parabola always has a line of symmetry exactly halfway between the two roots.
  • A1: y-intercept (0, −8), x-intercepts (2, 0) and (−4, 0)
Watch out

Students find the x-intercepts but forget to calculate the y-coordinate of the turning point.Substituting x = −b/2a back into the equation is essential — never just mark the turning point on the axis.

Exam-stylemulti-step

The turning point of y = x2 + bx + 9 is at x = 3. Find b.

4 marks6 minsquadratic-graphs-q5
Show solution
Worked solution
  1. 1.Spot the skill: For y = x2 + bx + c: y-intercept is c, x-intercepts come from factorising, turning point x-coordinate is −b/2a.
  2. 2.Use the find the y-intercept by setting x = 0 stage first, then find x-intercepts by factorising the quadratic.
  3. 3.Keep the final answer visible: b = −6.
Final answer

b = −6

Mark points
  • M1: use the correct for y = x2 + bx + c: y-intercept is c, x-intercepts come from factorising, turning point x-coordinate is −b/2a.the parabola always has a line of symmetry exactly halfway between the two roots.
  • A1: b = −6
Watch out

Students find the x-intercepts but forget to calculate the y-coordinate of the turning point.Substituting x = −b/2a back into the equation is essential — never just mark the turning point on the axis.

Grade 9 stretchproblem solving

Find the minimum point of y = x2 - 6x + 5.

4 marks7 minsquadratic-graph-g9
Show solution
Worked solution
  1. 1.Complete the square.
  2. 2.Read the coordinates of the turning point.
Final answer

(3, -4)

Mark points
  • M1: obtain (x - 3)2 - 4
  • A1: (3, -4)
Watch out

Do not rush straight into arithmetic. Select the relevant method and show a complete chain of working.

Timed checkpoint
12 mins - 9 marks

Switch between skills

Set a timer and attempt all four questions before opening any answers. This is closer to the way skills appear in a real paper.

1Quadratic graphs - 2 marksFind the y-intercept and x-intercepts of y = x2 − 7x + 10Mark answer
Answer

y-intercept (0, 10), x-intercepts (2, 0) and (5, 0)

2Algebraic notation and substitution - 2 marksFind t² + 4t when t = −2Mark answer
Answer

−4

3Simplifying expressions - 2 marksSimplify 2xy + 3x²y − xy + x²yMark answer
Answer

3x²y + xy

4Expanding and factorising - 3 marksFactorise fully 6x² + 9xMark answer
Answer

3x(2x + 3)

Mastery check
  • I can explain the method for quadratic graphs.
  • I can show clear working without skipping key steps.
  • I can avoid this mistake: Students find the x-intercepts but forget to calculate the y-coordinate of the turning point.Substituting x = −b/2a back into the equation is essential — never just mark the turning point on the axis.
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Related topics
Algebraic notation and substitutionSimplifying expressionsExpanding and factorising
Official exam-board sources

This guide follows the AQA GCSE Mathematics 8300 specification. Practice questions are original Learnova questions shaped around official content and exam skills.

Official specification Official past papers
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