(a)
Roots at \( x = -4 \), \( x = 6 \), so \( f(x) = a(x + 4)(x – 6) \)
Given form: \( f(x) = -10(x + 4)(x – 6) \)
Verify: \( f(0) = -10(4) \times (-6) = 240 \), matches \( y \)-intercept
Result: \( f(x) = -10(x + 4)(x – 6) \) [2]

(b)
Vertex \( x \)-coordinate: \( h = \frac{-4 + 6}{2} = 1 \)
Vertex \( y \)-coordinate: \( f(1) = -10(1 + 4)(1 – 6) = -10 \times 5 \times (-5) = 250 \)
Vertex form: \( f(x) = -10(x – 1)^2 + 250 \)
Verify: \( f(0) = -10(0 – 1)^2 + 250 = 240 \), \( f(-4) = -10(-4 – 1)^2 + 250 = 0 \), \( f(6) = 0 \)
Result: \( f(x) = -10(x – 1)^2 + 250 \) [2]

(c)
Expand: \( f(x) = -10(x + 4)(x – 6) = -10(x^2 – 2x – 24) = -10x^2 + 20x + 240 = 240 + 20x – 10x^2 \)
Alternatively: \( -10(x – 1)^2 + 250 = -10(x^2 – 2x + 1) + 250 = 240 + 20x – 10x^2 \)
Result: \( f(x) = 240 + 20x – 10x^2 \) [2]

(d)(i)
Velocity: \( v = 240 + 20t – 10t^2 \)
Maximum speed at vertex: \( t = -\frac{20}{2 \times (-10)} = 1 \)
Alternatively: \( v'(t) = 20 – 20t = 0 \implies t = 1 \)
Result: \( t = 1 \) [2]

(d)(ii)
Speed zero: \( v = -10(t – 6)(t + 4) = 0 \implies t = 6 \text{ or } t = -4 \)
For \( 0 \leq t \leq 6 \): \( t = 6 \)
Acceleration: \( a(t) = v'(t) = 20 – 20t \)
At \( t = 6 \): \( a(6) = 20 – 20 \times 6 = -100 \)
Result: Acceleration = \(-100 \, \text{m/s}^2\) [2]