已知二次函数y=f(x)的最大值等于13,且f(3)=f(1)=5,求f(x)解析式
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f(3)=f(1)=5,最大值等于13说明x^2的因数为-
且中轴为x = 2
则f(x) = m(x-2)^2 +13 ,m<0
则f(1) = m+13 = 5 m=-8
f(x) =-8(x-2)^2 +13
且中轴为x = 2
则f(x) = m(x-2)^2 +13 ,m<0
则f(1) = m+13 = 5 m=-8
f(x) =-8(x-2)^2 +13
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let
f(x)= ax^2+ bx + c
f(3)= 9a+3b+c = 5 --(1)
f(1)= a+ b +c = 5 ---(2)
(1)-(2)
8a+2b=0
b= -4a ---(3)
f'(x)= 2ax+b
f'(x)=0 => x=-b/2a
f(-b/2a)= f(2)
= 4a+2b+c = 13 ---_(4)
(1)- (4)
5a+b= -8
5a-4a=-8
a = -8
b = 32
c = -19
f(x)= -8x^2+32x-19 #
f(x)= ax^2+ bx + c
f(3)= 9a+3b+c = 5 --(1)
f(1)= a+ b +c = 5 ---(2)
(1)-(2)
8a+2b=0
b= -4a ---(3)
f'(x)= 2ax+b
f'(x)=0 => x=-b/2a
f(-b/2a)= f(2)
= 4a+2b+c = 13 ---_(4)
(1)- (4)
5a+b= -8
5a-4a=-8
a = -8
b = 32
c = -19
f(x)= -8x^2+32x-19 #
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