Question

1-Koordinat titik minimum dari fungsi f (x) = x² – 4x + 3 adalah.......

2-Titik grafik y = x² – 8x + 12 dengan garis y = x – 2 adalah.....


•Pake langkah cara dan penjelasan

Answer 1

1)
f (x) = x² – 4x + 3

f'(x) = 2x - 4

2x = 4

x = 4/2 = 2

f(2) = 2² - 4(2) + 3 = 7 - 8 = 1

koordinat titik minimum = { 2, 1 }

2)
y = x² – 8x + 12
y = x – 2

<=> x² - 8x + 12 = x - 2

x² - 9x + 14 = 0

(x - 2)(x - 7) = 0

x1 = 2
x2 = 7

titik grafiknya :
(2, 0) dan (7, 0)

Answer 2

F (x) = x² – 4x + 3
f'(x) = 2x - 4
2x = 4
x = 4/2
= 2
f(2) = 2² - 4(2) + 3
= 7 - 8
= 1
titik minimum = 2, 1

y = x² – 8x + 12
y = x – 2
x² - 8x + 12 = x - 2
x² - 9x + 14 = 0
(x - 2)(x - 7) = 0
x1 = 2
x2 = 7

titik grafiknya :
(2, 0) dan (7, 0)