F.4 Maths - Variation

(a)Let y = k(x+p)(x+3), where k is a non-zero constant

From the given, we get

70 = k(2+p)(2+3)
k(2+p) = 14----------(1)

96 = k(3+p)(3+3)
k(3+p) = 16-----------(2)

(1)/(2),

(2+p)/(3+p) = 14/16

16(2+p) = 14(3+p)

p = 5

(b)Substitute p = 5 into (1), we get k = 2

Therefore, the equation connectiong x and y is

y = 2 (x+5) (x+3)

(c) When y = 160,

2 (x+5) (x+3) = 160

x^2+8x+15 = 80

x^2+8x-65 = 0

(x-5) (x+13) = 0

x = 5 or -13

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Let y = k(x+p)(x+3) = k[x^2 + (p+3)x +3p]

when x = 2, y = 70
70 = k ( 4 + 2p + 6 + 3p )
70 = k ( 5p + 10 )
14 = k ( p + 2 ) ................... [1]

when x = 3, y = 96
96 = k ( 9 + 3p + 9 + 3p )
96 = k ( 6p + 18 )
16 = k ( p + 3 ) ...................... [2]

[2] - [1],
k = 2
p = 16/2 - 3 = 5

Thus, y = 2 [x^2 + (5+3)x +3(5)] = 2 ( x^2 + 8x + 15)

when y = 160,
x^2 + 8x + 15 = 80
x^2 + 8x - 65 = 0
(x + 13) ( x - 5) = 0
x = -13 or 5