For f(x) = a*(x + b)^2 - 8 to have a tangent y = 2x at (4,8) two conditions have to be met:
1. f(4) = 8 [the parabola passes through (4,8)]
2. f'(4) = 2 [the tangent of the parabola at (4,8) is 2]
To satisfy condition 1,
a*(4 + b)^2 - 8 = 8
a*(4 + b)^2 = 16 -- eq.1
To satisfy condition 2,
Expanding,
f(x) = a*(x^2 + 2bx + b^2) - 8
= ax^2 + 2abx + ab^2 - 8
f'(x) = 2ax + 2ab
2a(4) + 2ab = 2
a(4 + b) = 1
a = 1/(4 + b) -- eq.2
Substituting this value for a into eq.1 gives:
[(4 + b)^2]/(4 + b) = 16
4 + b = 16
b = 12
Substituting this value for b into eq.2 gives:
a = 1/(4 + 12)
a = 1/16
Therefore, the parabola that is tangential with y = 2x at (4,8) is given by the equation:
f(x) = (1/16)*(x + 12)^2 - 8
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Are you in Yr 11 now? If you are then I don't think you would have learnt differentiation yet so the above solution would mean nonsense to you but I can't see how this question can be solved without calculus.
