The diagram shows the cross section of a tunnel for vehicles to pass. The shape of the arched tunnel can be modelled by the equation $y = -1.76x^2 + 4.4x$, where $y$ m represents the height above the ground and $x$ m represents the horizontal distance from $O$.

1. Factor out the coefficient of $x^2$ from the quadratic expression.

$y = -1.76(x^2 - 2.5x)$

2. Complete the square for the expression inside the bracket.

$y = -1.76 \left[ \left(x - \frac{2.5}{2}\right)^2 - \left(\frac{2.5}{2}\right)^2 \right]$
$y = -1.76 \left[ (x - 1.25)^2 - 1.5625 \right]$

3. Expand to express in the required form.

$y = -1.76(x - 1.25)^2 - 1.76(-1.5625)$
$y = -1.76(x - 1.25)^2 + 2.75$

1. Express the quadratic function in completed square form (result from part i)

$y = -1.76(x^2 - 2.5x)$
$y = -1.76[(x - 1.25)^2 - 1.25^2]$
$y = -1.76(x - 1.25)^2 + 2.75$

2. State the coordinates of the highest point

The maximum value of $y$ is $2.75$, which occurs when $x - 1.25 = 0 \implies x = 1.25$.
The coordinates of the highest point are $(1.25, 2.75)$.

1. Determine the horizontal position of the lorry's edges when it passes through the center of the tunnel

The center of the tunnel is at $x = 1.25$ m (from part ii).
For a lorry of width $1.8$ m to pass through the center, its edges will be at $x = 1.25 - \frac{1.8}{2}$ and $x = 1.25 + \frac{1.8}{2}$.
$x = 1.25 - 0.9 = 0.35$ m and $x = 1.25 + 0.9 = 2.15$ m.

2. Calculate the height of the tunnel at the edge of the lorry

Substitute $x = 0.35$ into the equation of the tunnel:
$y = -1.76(0.35)^2 + 4.4(0.35)$
$y = -0.2156 + 1.54 = 1.3244$ m

3. Compare the tunnel height with the lorry height to conclude

Since the height of the tunnel at the edge of the lorry ($1.3244$ m) is less than the height of the lorry ($2$ m), the lorry will hit the tunnel and cannot pass.