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artinya x' = x - 3 maka x = x' + 3
dan
y' = y + 4 maka y = y' - 4
f(x) = -2(x-1)^2 + 2
y-4 = -2(x+2)^2 + 2
y-4 = -2(x^2 + 4x + 4) + 2
y = -2x^2 + 8x - 8 + 2 + 4
y = -2x^2 + 8x - 2
titik balik = (-b/2a , -d/4a)
-b/2a = -8/-4
= 2
subsitusikan x = 2
y = -2(2)^2 + 8(2) - 2
= -8 + 16 - 2
= 6
titik balik = ( 2,6)
soal no 6 ) kekiri 3 satuan kebawah 4 satuan
berarti nilai x' = x - 3 maka x = x' + 3
dan
nilai y' = y - 4 maka y = y' + 4
y+4 = -2(x+2)^2 + 2
y = -2x^2 - 8x - 8 + 2 - 4
y = -2x^2 - 8x - 10
titik baliknya
-b/2a = 8/-4
= -2
y = -2(-2)^2 - 8(-2) - 10
= -8 + 16 - 10
= -2
titik balik = (-2,-2)
soal no 7) kekanan 3 keatas 4
berarti x-3 dan y-4
y-4 = -2(x-4)^2 + 2
y = -2x^2 +16x - 32 + 2 + 4
y = -2x^2 + 16x - 26
titik balik
x = -b/2a
= -16/-4
= 4
y = -2(4)^2 +16(4) -26
= -32 + 64 - 26
= 6
titik balik = (4,6)
no 8) kekanan 3 kebawa 4
berarti x - 3 dan y + 4
y+4 = -2(x-4)^2 + 2
y = -2x^2 + 16x -32 + 2 - 4
= -2x^2 + 16x -34
titik balik
x = -b/2a
= -16/-4
= 4
y = -2(4)^2 +16(4) -34
= -32 + 64 - 34
= -2
titik balik = (4,-2)