X2 2 4x 1 0 July 8, 2019 16 32 64 128 − 2 4x 5 )( x2 1 )= 0 X2 k 4x k 1 )+ 2 0 X2 m 4x m 1 )+ 2 0 Sumbu simetris dari x2 4x 1 2 0 X x2 4x 5 2 x 1 0 X2 y2 4x 2 0 m =- 1
