yr 11 quadratics question?

a) y = ax(x - 50)....for some constant a

Now, y = 4.5 when x = 25 so,

a(25)(-25) = 4.5

so, a = -9/1250

Hence, y = (-9/1250)x(x - 50)

b) you can surely do this

c) We require when (-9/1250)x(x - 50) = 3

so, x(x - 50) = -1250/3

i.e. 3x³ - 150x + 1250 = 0

Using the quadratic formula we get:

x = (150 ± 50√3)/6 => (75 ± 25√3)/3

so, at x = 10.57 metres and at x = 39.43

Note: 39.43 m is 10.57 m before C

d) When x = 12 we have:

y = (-9/1250)(12)(-38) => 3.28 metres

e) The width of the platform is 10 m from the centre in either direction. So, including the space of 0.3 m we have:

10.3 metres from the centre

i.e. 14.7 metres or 35.3 metres from A

so, with x = 14.7 we get:

y = (-9/1250)(14.7)(-35.3) => 3.74 metres

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a.

Assign the water level as the x-axis, and the vertical line passing through A as the y-axis.
Take 1 unit = 1 m
Then, A(0, 0), B(25, 4.5) and C(50, 0)

Let y = ax² + bx + c be the formula of the curve.

A(0, 0) lies on the curve:
0 = a(0)² + b(0) + c
c = 0
Then, equation of the curve: y = ax² + bx

B(25, 4.5) lies on the curve:
4.5 = a(25)² + b(25)
625a + 25b = 4.5 …… [1]

C(50, 0) lies on the curve:
0 = a(50)² + b(50)
50b = -(50)²a 
b = -50a …… [2]

Substitute [2] into [1]:
625a + 25(-50a) = 4.5
625a - 1250a = 4.5
-625a = 4.5
a = -9/1250

Substitute a = -9/1250 into [2]:
b = -50 × (-9/1250)
b = 9/25

Hence, the formula of the curve:
y = -(9/1250)x² + (9/25)x

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c.
When y = 3:
3 = -(9/1250)x² + (9/25)x
1250 = -3x² + 150x
3x² - 150x + 1250 = 0
x = [150 ± √(150² - 4*3*1250)]/(2*3)
x = (75 ± 25√3)/3
x = 10.57, 39.43

The horizontal distance from A is 10.57 m or 39.43 m.