Worksheet Ch 1: Quadratic Functions & Equations

Section A • Bahagian A [12 Marks]

Question 1 [4 Marks]

Solve the quadratic equation by factorization:

$$\frac{2x^2 + 5x}{3} = 4 - x$$

Step 1: Multiply both sides by 3

$2x^2 + 5x = 12 - 3x$

Step 2: Rearrange to standard general form $ax^2 + bx + c = 0$

$2x^2 + 8x - 12 = 0 \implies x^2 + 4x - 6 = 0$ (or keeping coefficients)

Step 3: Quadratic factorization / formula

$(2x - 3)(x + 4) = 0 \text{ for equivalent standard factors}$

Step 4: Final exact roots

$x = 1.16 \text{ or } x = -5.16$ (or exact surd form)

Question 2 [4 Marks]

The diagram shows the sketch of the quadratic curve $f(x) = -2x^2 + 4x + 6$.

(a) State the coordinates of the $y$-intercept.
(b) Determine the axis of symmetry and the maximum coordinates of the turning point.

(a) When $x = 0$, $y = 6$

$(0, 6)$

(b) Axis of symmetry: $x = -\frac{b}{2a} = -\frac{4}{2(-2)} = 1$

$x = 1$

Maximum $y$-value: $f(1) = -2(1)^2 + 4(1) + 6 = 8$

Maximum turning point $= (1, 8)$

Question 3 [4 Marks]

Form a quadratic equation in general form $ax^2 + bx + c = 0$ having roots $-\frac{2}{3}$ and $4$.

Step 1: Set up factors $(x - r_1)(x - r_2) = 0$

$(3x + 2)(x - 4) = 0$

Step 2: Expand to general form

$3x^2 - 10x - 8 = 0$

Section B • Bahagian B (HOTS / KBAT) [18 Marks]

Question 4 (Real-World Area Application) [9 Marks]

Puan Aida wants to lay a uniform decorative pebble border of width $x\text{ m}$ around a rectangular swimming pool measuring $12\text{ m}$ by $8\text{ m}$. The total area of the pool and the border combined is $140\text{ m}^2$.

(a) Form a quadratic equation in terms of $x$ to represent the situation. [3 marks]
(b) Calculate the width of the border, $x$, in meters. [4 marks]
(c) If the cost of laying pebbles is RM45 per square meter, calculate the total cost for the border. [2 marks]

(a) Total Length $= 12 + 2x$, Total Width $= 8 + 2x$

$(12 + 2x)(8 + 2x) = 140$

$96 + 40x + 4x^2 = 140 \implies 4x^2 + 40x - 44 = 0$

$x^2 + 10x - 11 = 0$

K1 N2

(b) Factorize $(x + 11)(x - 1) = 0$

$x = 1 \text{ or } x = -11$

Since width $x > 0$, reject $x = -11$.

Width of border, $x = 1\text{ m}$

K2 N2

$\text{Total Cost} = 44 \times \text{RM } 45 = \text{RM } 1,980$

K1 N1

Question 5 (Projectile Rocket Motion) [9 Marks]

A model water rocket is launched from a platform $4\text{ m}$ above the ground. Its height $h(t)$ in meters after $t$ seconds is given by:

$$h(t) = -5t^2 + 20t + 4$$

(a) Find the time taken for the rocket to reach its maximum height. [3 marks]
(b) What is the maximum height attained by the rocket? [2 marks]
(c) Calculate the time $t$, in seconds, when the rocket strikes the ground. [4 marks]

(a) Axis of symmetry $t = -\frac{b}{2a} = -\frac{20}{2(-5)} = 2\text{ seconds}$

$t = 2\text{ s}$

K2 N1

(b) Substitute $t = 2$: $h(2) = -5(2)^2 + 20(2) + 4 = -20 + 40 + 4 = 24\text{ m}$

$\text{Max Height} = 24\text{ m}$

K1 N1

$t = \frac{-20 \pm \sqrt{20^2 - 4(-5)(4)}}{2(-5)} = \frac{-20 \pm \sqrt{400 + 80}}{-10} = \frac{-20 \pm \sqrt{480}}{-10}$

$t = 4.19\text{ s} \quad (\text{reject } t = -0.19\text{ s})$

K2 N2