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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: soccerboi on November 08, 2012, 10:10:51 am

Title: VCAA 2007 ex2 Q 3d)ii)
Post by: soccerboi on November 08, 2012, 10:10:51 am

I have y=A(x+3)(x+1)(x-1). What point do i use to find A? and why?
Thanks :)

Title: Re: VCAA 2007 ex2 Q 3d)ii)
Post by: Special At Specialist on November 13, 2012, 12:18:31 am

"Same maximum and minimum values of g"

y = A(x+3)(x+1)(x-1)
y = A(x + 3)(x^2 - 1)
TP at dy/dx = 0
x^2 - 1 + 2x(x + 3) = 0
x^2 - 1 + 2x^2 + 3x = 0
3x^2 + 3x - 1 = 0
x = (-3 ± sqrt(9 + 12)) / 6
x = -1/2 ± sqrt(21)/6
We know that the maximum value of g is 32sqrt(3) / 9, so:
At x = -1/2 - sqrt(21)/6, y = 32sqrt(3)/9
32sqrt(3)/9 = A(5/2 - sqrt(21)/6)((1/2 + sqrt(21)/6)^2 - 1)
A = 32sqrt(3) / (4sqrt(21) - 9)

Title: Re: VCAA 2007 ex2 Q 3d)ii)
Post by: BubbleWrapMan on November 13, 2012, 12:28:39 am

Or just apply the transformation you found in d. i. to the function g