(i) We place the parabola so that the left side of the arch passes through `(0, 0)` and the right side will pass through `(2, 0)`, since the arch is 2 m wide at the bottom. The vertex of the arch is at `(1, 3)`.
General form of a parabola: `y = ax^2+ bx + c`
At `x = 0`, `y = 0` and on substituting, we get `0 = 0 + 0 + c`. So `c = 0`.
At `x = 1`, `y = 3` and on substituting, we get `3 = a + b`.
At `x = 2`, `y = 0` and on substituting, we get `0 = 4a + 2b`.
This gives the simultaneous equations
`a + b = 3`
`2a + b = 0`
Subtracting the 1st line from the 2nd gives `a = -3`.
And so `b = 6`.
So the required parabola is `y = -3x^2+ 6x`, with `x` in metres.
The process of finding the equation is called modeling. It is a very important skill in science and engineering.
(ii) Now for the area:
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