F.4 Functions and Graphs

Part (C) ;
x^2 + 6x + k = 0
so α+ β = - 6 and αβ = k.
For the function y = -2x^2 + bx + c
Since the curve also passes through α and β, so α and β are also the roots of y = 0,
so α + β = - b/(-2) = - 6 , that is b = - 12
and αβ = c/(-2) = k, that is c = - 2k.
So the function is y = -2x^2 - 12x - 2k
From part (B), axis of symmetry is x = - 3.
When x = - 3, y = -2(-3)^2 - 12(-3) - 2k = 18 - 2k
So height of triangle ABC = 18 - 2k.
(α - β)^2 = (α + β)^2 - 4αβ = (-6)^2 - 4(k) = 36 - 4k
So base of triangle ABC = (α - β) = sqrt(36 - 4k) [take the positive answer because α > β]
So area of triangle ABC = [(18 - 2k)sqrt(36 - 4k)]/2
For area to be > 128
[(18 - 2k)sqrt(36 - 4k)]/2 > 128
(18 - 2k)sqrt(9 - k) > 128
2(9 - k)sqrt(9 - k) > 128
(9 - k)sqrt(9 - k) > 64
Square both sides
(9 - k)^3> 64^2
(9 - k)^3 > 16^3
9 - k > 16
k < -7
So Jodie's claim is correct.