MATHS f.4 快,,,THZ

(a) y = 2(x-a)^2 + b
for the y-intercept (0, 22),
22 = 2(0-a)^2 + b
22 = 2a^2 + b -------(1)
for the minimum point
c = 2(3-a)^2 + b
c = 18 - 12a + a^2 + b -----------(2)
sub (1) into (2):
c = 18 - 12a + 22 -----------(3)
differentiate the equation w.r.t. x
dy/dx = 4(x-a)
for the minimum point (3, c), dy/dx = 0
4(3-a) = 0
a = 3
sub a = 3 into eq(2):
c = 18 - 12(3) + 22 = 4
sub a = 3 into eq(1):
22 = 2(3)^2 + b
b = 4
The equation is y = 2(x-3)^2 + 4
(b) when x = 3, y = 4, the curve is at the minimum
the minimum value, y = 4

2008-11-16 16:15:33 補充：
Question
x + y = 10 -----(1)
Set x^2 + y^2 = A = x^2 + (x - 10)^2 = 2x^2 - 20x + 100
differentiate w.r.t. x
dA/dx = 4x - 20
at minimum, dA/dx = 0
4x-20 = 0
x = 5
when x > 5, dA/dx > 0
when x< 5, dA/dx < 0
It is minimum.
sub x = 5 into (1): y = 5

根據y=a(x-h)^2+k
(h,k) 係max or min pt
a=3
b=c
sub (0,22)
22=2(-a)^2+b
22=2a^2+b
22-18=b
b=4

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