jika puncak parabola y=x^2-(a+1)x+(2a+4) terletak pada garis y=3x, maka a =

y = x^2 - (a + 1)x + (2a + 4)
Absis puncak (sumbu simetri) : x = -b/2a = -(-(a + 1))/2(1) = (a + 1)/2
y = 3x
y = y
x^2 - (a + 1)x + (2a + 4) = 3x
x^2 - ax - x + (2a + 4) - 3x = 0
x^2 - ax - 4x + (2a + 4) = 0
x^2 - (a + 4)x + (2a + 4) = 0 ====> masukkan x = (a + 1)/2
((a + 1)/2)^2 - (a + 4)(a + 1)/2 + (2a + 4) = 0
(a^2 + 2a + 1)/4 - (a^2 + 5a + 4)/2 + (2a + 4) = 0 ===> kali 4
a^2 + 2a + 1 - 2a^2 - 10a - 8 + 8a + 16 = 0
-a^2 + 9 = 0
a^2 - 9 = 0
(a + 3)(a - 3) = 0
a = -3 atau a = 3