Maka:
y = ax + b
-1 = a(3) + b
-1 = 3a + b
b = -1 – 3a
x2 – 4x + 2 = ax + b
x2 – 4x – ax + 2 – b = 0
x2 – (4 + a) x + 2 – b = 0
Jadi
a = 1
b = -(4 + a)
c = 2 – b
0 = b2 – 4ac
0 = (-(4 + a))2 – 4(1) (2 – b)
0 = 16 + 8a + a2 – 8 + 4b
0 = a2 + 8a + 4b + b
Diperoleh substitusi b = -1 – 3a
a2 + 8a + 4b + 8 = 0
a2 + 8a + 4(-1 – 3a) + 8 = 0
a2 + 8a – 4 – 12a + 8 + 0
a2 – 4a + 4 = 0
(a – 2)2 = 0
a – 2 = 0
a = 2
b = -1 – 3a
b = -1 – 3(2)
b = -7
Jadi, nilai :
a = 2
b = -7
10. Diketahui y = 2x2 – 12x + 16
Maka dari y = ax2 + bx + c diperoleh
a = 2, b = -12, c = 16
y = 0