A11: Quadratic Graphs
Identify roots, intercepts and turning points of quadratic functions; completing the square (Higher)
📋 Key Concepts
Quadratic graph: A U-shaped curve called a parabola. The equation is y = ax² + bx + c.
Key features:
- Roots (x-intercepts): Where the curve crosses the x-axis (y = 0)
- y-intercept: Where the curve crosses the y-axis (x = 0)
- Turning point: The vertex (minimum or maximum point)
- Line of symmetry: Vertical line through the turning point
📝 Plotting Quadratic Graphs
Example 1
Draw the graph of y = x² - 3x - 4 for -2 ≤ x ≤ 5
| x | -2 | -1 | 0 | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|---|---|---|
| y | 6 | 0 | -4 | -6 | -6 | -4 | 0 | 6 |
Plot points and draw a smooth curve through them.
Shape: If a > 0, the parabola opens upward (U-shape). If a < 0, it opens downward (∩-shape).
📝 Finding the y-intercept
Rule: Substitute x = 0 into the equation. The y-intercept is always (0, c).
Example 2
Find the y-intercept of y = x² - 5x + 6
Solution:
Substitute x = 0: y = 0² - 5(0) + 6 = 6
y-intercept is (0, 6)
📝 Finding the Roots (x-intercepts)
Method: Set y = 0 and solve the quadratic equation.
Example 3
Find the roots of y = x² - 3x - 4
Solution:
Set y = 0: x² - 3x - 4 = 0
Factorise: (x - 4)(x + 1) = 0
x = 4 or x = -1
Roots are (4, 0) and (-1, 0)
📝 Finding the Turning Point
Method 1 (using symmetry): The x-coordinate of the turning point is halfway between the roots.
x = root₁ + root₂⁄2
Example 4
y = x² - 3x - 4 has roots at x = -1 and x = 4. Find the turning point.
Solution:
x-coordinate = -1 + 4⁄2 = 3⁄2 = 1.5
y-coordinate = (1.5)² - 3(1.5) - 4 = 2.25 - 4.5 - 4 = -6.25
Turning point: (1.5, -6.25)
Method 2 (completing the square - Higher):
y = a(x + p)² + q has turning point (-p, q)
Example 5
Find the turning point of y = x² + 6x + 5 by completing the square.
Solution:
y = (x² + 6x) + 5
y = (x + 3)² - 9 + 5
y = (x + 3)² - 4
Turning point: (-3, -4)
📝 Line of Symmetry
Rule: The line of symmetry passes through the turning point.
For y = ax² + bx + c: Line of symmetry is x = -b⁄2a
Example 6
Find the line of symmetry of y = x² - 6x + 5
Solution:
a = 1, b = -6
x = -(-6)⁄2(1) = 6⁄2 = 3
Line of symmetry: x = 3
🧮 Worked Examples — Exam Style
Quadratic graphs: parabola (U-shape). Key features: y-intercept (constant term), roots (factorise or read the graph), turning point (midway between the roots for symmetric parabolas).
Find the roots and turning point of y = x² − 6x + 8.
Roots: (x − 2)(x − 4) = 0 → x = 2, 4
Turning point x = midway = 3; y = 9 − 18 + 8 = −1 → (3, −1)
M1 for factorising, A1 for roots, A1 for (3, −1). Symmetry: the turning point sits halfway between the roots.
For y = x² − 4x + 1, complete a table for x = 0, 1, 2, 3, 4 and describe the shape.
y: 1, −2, −3, −2, 1 — symmetric values confirm the axis x = 2, minimum (2, −3)
M1 for the table, A1 for symmetry/minimum. Table symmetry is a quick check on your arithmetic.
Exam wording to expect: "Complete the table and draw the graph", "estimate the solutions of …", "write down the coordinates of the turning point".
❓ Practice Questions
Q1: Find the y-intercept of y = x² - 4x + 3.
Q2: Find the roots of y = x² - 9.
Q3: Find the roots of y = x² - 5x + 6.
Q4: A quadratic has roots at x = 1 and x = 5. What is the x-coordinate of the turning point?
Q5: Find the turning point of y = (x - 2)² + 3.
Q6: Find the line of symmetry of y = x² + 4x + 1.
Q7: Sketch y = x² − 2x − 8, marking the y-intercept and roots
Q8: For which values of x is y = x² − 2x − 8 negative?
✅ Answers
- (0, 3)
- x = 3 and x = -3, so roots are (3, 0) and (-3, 0)
- x = 2 and x = 3, so roots are (2, 0) and (3, 0)
- x = 3
- (2, 3)
- x = -2
📋 Model Answers — Full Working
Every mark comes from the working, not just the answer. These model answers show the lines a mark scheme expects to see.
Q1. x = 0 → (0, 3) — the constant term
Q2. x² = 9 → x = ±3, points (3, 0) and (−3, 0)
Q3. (x − 2)(x − 3) = 0 → x = 2, x = 3
Q4. Midway between 1 and 5 → x = 3
Q5. Vertex form directly: (2, 3) — minimum (positive x²)
Q6. x = −b/(2a) = −4/2 = x = −2
Q7. (x − 4)(x + 2) = 0 → roots 4 and −2; y-intercept (0, −8); turning point (1, −9)
Q8. Below the x-axis BETWEEN the roots: −2 < x < 4
📌 Key Facts to Memorise
- Positive x² → U-shape (minimum); negative x² → ∩-shape (maximum)
- y-intercept = the constant term c
- Roots: solve ax² + bx + c = 0 by factorising, formula, or graph-reading
- Turning point x = midway between roots = −b/(2a)
- Line of symmetry passes through the turning point, vertical
🗺️ In the Exam
Quadratic graphs: 4–6 marks across tables, drawing, reading roots and turning points. "Estimate solutions from your graph" links the drawing to equation solving.
Links to: solving quadratics (A18), transformations of graphs (A13), inequalities (A22).
🧭 Board Focus — AQA
Spec 8300. Three papers, each 80 marks and 1h30: Paper 1 non-calculator, Papers 2 and 3 calculator. Foundation/Higher tiers. Roughly a third of the marks sit under AO2 (reasoning) and AO3 (problem-solving), so show every line of working — method marks survive arithmetic slips.
Always check: specification details change — confirm paper structure and set texts against the current specification for your exam year.
🎯 Exam Tips
- y-intercept is always (0, c) - just read off the constant
- Find roots by factorising or using the quadratic formula
- Turning point x-coordinate is between the two roots
- Use completing the square for turning point (Higher)
- Check if parabola opens up (a > 0) or down (a < 0)
🧠 Problem-Solving Strategies
Problem-Solving
Key features of a parabola: y-intercept (set x=0), roots (set y=0 and factorise), turning point (midpoint of roots or complete the square). If a > 0 the parabola opens upward (minimum); if a < 0 it opens downward (maximum).
Multi-Step Problem
A ball is thrown upward. Its height h metres after t seconds is h = -5t² + 20t + 1. (a) Find the initial height. (b) Find when the ball hits the ground. (c) Find the maximum height.
Solution:
(a) h = -5(0)² + 20(0) + 1 = 1 metre
(b) -5t² + 20t + 1 = 0 → t = -20 ± √(400+20)⁄-10 → t ≈ 4.05 seconds
(c) Turning point at t = -20⁄2(-5) = 2. h = -5(4) + 20(2) + 1 = 21 metres
⚠️ Common Errors
Watch Out!
1. Wrong: The y-intercept of y = x² - 5x + 6 is 6 at point (6, 0) Correct: The y-intercept is (0, 6) — it's on the y-axis where x = 0
2. Wrong: The turning point is always a minimum Correct: It's a minimum when a > 0 (opens up), a maximum when a < 0 (opens down)
3. Wrong: Completing the square: x² + 6x + 5 = (x + 6)² + 5 Correct: x² + 6x + 5 = (x + 3)² - 9 + 5 = (x + 3)² - 4 (halve the coefficient of x)
✍️ 6-Mark Exam Question
Extended Answer
6 marks: y = x² - 6x + 5. (a) Find the y-intercept. (b) Find the roots. (c) Find the coordinates of the turning point. (d) Sketch the curve, labelling all key points.
(a) When x = 0: y = 5. y-intercept = (0, 5)
(b) x² - 6x + 5 = 0 → (x - 1)(x - 5) = 0 → x = 1 or x = 5. Roots: (1, 0) and (5, 0)
(c) By completing the square: (x - 3)² - 9 + 5 = (x - 3)² - 4. Turning point: (3, -4)
(d) U-shaped parabola passing through (0, 5), (1, 0), (3, -4), (5, 0). Minimum at (3, -4). Line of symmetry x = 3.
Mark scheme: (a) 1 mark. (b) 1 mark. (c) 2 marks. (d) 2 marks for sketch with labelled points.
📊 AO3: Reason & Interpret
Reasoning and Interpretation
The profit P (in £1000s) from selling x items is P = -x² + 20x - 64.
(a) Find the profit when 10 items are sold.
(b) How many items give zero profit (break-even)?
(c) What is the maximum profit and how many items achieve it?
Answers: (a) P = -100 + 200 - 64 = 36 (£36,000). (b) -x² + 20x - 64 = 0 → (x-4)(x-16) = 0 → x = 4 or x = 16 items. (c) Turning point at x = 10: P = 36, so max profit is £36,000 at 10 items.
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🎬 Video Resources
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