📚 Back to Contents
📚 MATHSKILLER TEXTBOOK SERIES
02

Quadratic Functions

and Their Graphs
📖 Detailed Explanations
📝 DSE Past Papers
📈 Graph Analysis
For HKDSE Mathematics (Compulsory Part)
© 2025 MathsKiller | mathskiller.pro

📋 Table of Contents

2.1 Three Forms of Quadratic FunctionsP.3
2.2 Vertex Form and Vertex FormulaP.5
2.3 Graph Features and Coefficient SignsP.7
2.4 Completing the SquareP.9
2.5 Graph TransformationsP.11
2.6 DSE Past Paper QuestionsP.13
2.7 Practice QuestionsP.18
CHAPTER 2

Quadratic Functions and Their Graphs

🎯 Learning Objectives

2.1
Three Forms of Quadratic Functions
Form Formula Key Features
General Form $y = ax^2 + bx + c$ y-intercept = $c$
Vertex Form $y = a(x-h)^2 + k$ Vertex = $(h, k)$ ⭐ Most tested!
Factored Form $y = a(x-p)(x-q)$ x-intercepts = $p$, $q$

⚡ Quick Recognition

• See $(x-h)^2$ → Vertex form, vertex = $(h, k)$

• See $(x-p)(x-q)$ → Factored form, x-intercepts = $p$, $q$

2.2
Vertex Form and Vertex Formula
Vertex Formula (from General Form)
Vertex $= \left(-\dfrac{b}{2a}, f\left(-\dfrac{b}{2a}\right)\right)$

🧠 Practical Method (Recommended!)

x-coordinate: $x = -\dfrac{b}{2a}$
y-coordinate: Substitute x into the function
Example 1

Find the vertex of $y = 2x^2 - 8x + 5$.

📝 Solution

Step 1: Find x-coordinate

$x = -\dfrac{-8}{2(2)} = \dfrac{8}{4} = 2$

Step 2: Substitute to find y-coordinate

$y = 2(2)^2 - 8(2) + 5 = 8 - 16 + 5 = -3$

∴ Vertex = $(2, -3)$

⚡ Common Trap!

$y = (x + 3)^2 - 5$

Vertex = $(3, -5)$ ❌ WRONG!

Vertex = $(-3, -5)$ ✓ CORRECT!

Remember: $(x+3)^2 = (x-(-3))^2$, so $h = -3$

2.3
Graph Features and Coefficient Signs

📖 Effect of Coefficient $a$

Value of $a$Opening DirectionVertex Type
$a > 0$Opens upward ∪Minimum point
$a < 0$Opens downward ∩Maximum point

📖 Determining Coefficient Signs from Graph

ObservationConclusion
Opens upward$a > 0$
Opens downward$a < 0$
Axis of symmetry to the RIGHT of y-axis$a$ and $b$ have OPPOSITE signs
Axis of symmetry to the LEFT of y-axis$a$ and $b$ have SAME sign
y-intercept ABOVE x-axis$c > 0$
y-intercept BELOW x-axis$c < 0$
2.4
Completing the Square
Example 2

Convert $y = 2x^2 - 12x + 13$ to vertex form.

📝 Solution

Step 1: Factor out $a$ from x terms

$y = 2(x^2 - 6x) + 13$

Step 2: Complete the square (add and subtract $(\frac{6}{2})^2 = 9$)

$y = 2(x^2 - 6x + 9 - 9) + 13$

$y = 2[(x-3)^2 - 9] + 13$

Step 3: Simplify

$y = 2(x-3)^2 - 18 + 13 = 2(x-3)^2 - 5$

∴ Vertex form: $y = 2(x-3)^2 - 5$, Vertex = $(3, -5)$

2.6
DSE Past Paper Questions

📋 Paper 2 Multiple Choice

DSE 2020 Q14

If the minimum value of $y = x^2 - 4x + k$ is 5, find $k$.

A. 1   B. 5   C. 9   D. 13

📝 Quick Solution

x-coordinate of vertex: $x = \frac{4}{2} = 2$

Minimum = $f(2) = 4 - 8 + k = k - 4 = 5$

$k = 9$

Answer: C

DSE 2022 Q18

The vertex of the parabola $y = -2x^2 + 8x - 3$ is

A. $(-2, -27)$   B. $(-2, 5)$   C. $(2, -27)$   D. $(2, 5)$

📝 Quick Solution

$x = -\frac{8}{2(-2)} = 2$

$y = -2(4) + 16 - 3 = 5$

Answer: D

DSE 2023 Q21

The figure shows the graph of $y = ax^2 + bx + c$. The graph opens upward, axis of symmetry is to the right of y-axis, y-intercept is negative. Which is correct?

A. $a > 0$, $b > 0$, $c > 0$

B. $a > 0$, $b < 0$, $c < 0$

C. $a < 0$, $b > 0$, $c > 0$

D. $a < 0$, $b < 0$, $c < 0$

📝 Quick Solution

• Opens upward → $a > 0$

• Axis on right → $a$ and $b$ opposite signs → $b < 0$

• y-intercept negative → $c < 0$

Answer: B

2.7
Practice Questions

📝 Self Test

1. Find the vertex of $y = x^2 + 6x + 5$.
2. Convert $y = x^2 - 4x + 7$ to vertex form.
3. If the minimum of $y = 2(x-3)^2 + k$ is $-4$, find $k$.
4. Find the maximum value of $y = -x^2 + 2x + 3$.
5. The graph of $y = ax^2 + bx + c$ opens downward, has axis of symmetry $x = 2$, and positive y-intercept. Determine the signs of $a$, $b$, $c$.

📋 Answers

1. $(-3, -4)$   2. $y = (x-2)^2 + 3$   3. $k = -4$   4. 4   5. $a < 0$, $b > 0$, $c > 0$

📘 MathsKiller Textbook Series

Chapter 2: Quadratic Functions

© 2025 MathsKiller | mathskiller.pro