Bantu jawab ya kakak.. fungsi f dgn rumus F (X) = ax²+bx dgn a dan b bilangn bulat. a. jika f(1) = 1 dan f(3) = 0, hitunglah nilai a dan b ! b. tulislah rumus fungsinya !
Galladeaviero
A) f(1) = 1 --> a+ b = 1....(1) f(3) = 0 --> 9a + 3b = 0 --> b = -3a substitusi ke (1) a - 3a = 1 --> - 2a = 1 --> a = - 1/2 b = - 3a = 3/2 b) rumus fungsi f(x) = -1/2 x² + 3/2 x atau f(x) = -1/2 ( x² - 3)
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gaok
F(X) = ax²+bx dgn a dan b bilangn bulat. f(1) = a(1²) + b(1) = 1 => a + b = 1 => b = 1-a f(3) = a(3²) + b(3) = 0 => 9a + 3b = 0 => 9a + 3(1-a) = 0 => 9a + 3 - 3a = 0 9a -3a = -3 a = -1/2 b = 3/2 fungsinya dalah F (X) = -1/2x²+3/2x
f(3) = 0 --> 9a + 3b = 0 --> b = -3a substitusi ke (1)
a - 3a = 1 --> - 2a = 1 --> a = - 1/2
b = - 3a = 3/2
b) rumus fungsi f(x) = -1/2 x² + 3/2 x
atau f(x) = -1/2 ( x² - 3)
f(1) = a(1²) + b(1) = 1 => a + b = 1 => b = 1-a
f(3) = a(3²) + b(3) = 0 => 9a + 3b = 0 => 9a + 3(1-a) = 0 => 9a + 3 - 3a = 0
9a -3a = -3
a = -1/2
b = 3/2
fungsinya dalah F (X) = -1/2x²+3/2x