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untuk soal no 1 digeser kekiri 3 satuan , artinya x' = x-3 maka x = x' + 3
f(x) = -2(x'+3-1)^2 + 2
= -2(x+2)^2 + 2
= -2(x^2 +4x + 4) + 2
= -2x^2 + 8x - 8 + 2
= -2x^2 + 8x - 6
titik puncaknya ( -b/2a , -d/4a )
-b/2a = -8/2(-2)
= -8/-4
= 2
-d/4a bisa juga dg subsitusi nilai x = 2
-2x^2 + 8x - 6
= -2(2)^2 + 8(2) - 6
= -8 + 16 - 6
= 2
jd titik puncaknya di (2,2)
untuk soal no 2 kekanan 3 satuan artinya x' = x + 3 maka x = x' -3
jadi f(x) = -2(x-1)^2 + 2
= -2(x-3-1)^2 + 2
= -2(x-4)^2 + 2
= -2(x^2 - 8x + 16) + 2
= -2x^2 + 16x - 32 + 2
= -2x^2 + 16x - 30
titik puncak (-b/2a .-d/4a)
-b/2a = -16/-4
= 4
-d/4a kita gunakan subsitusi saja
-2x^2 + 16x - 30
= -2(4)^2 + 16(4) - 30
= -32 + 64 - 30
= 2
titik puncak = ( 4, 2)