kk tlg jawab ini dong pleasee

A(1 , -1) ; B(2 , -2) ; C(5 , -27)
Melalui titik A dan B

[tex]y-y_1= \frac{y_2-y_1}{x_2-x_1} (x-x_1) [/tex]
[tex] y+1 = \frac{-1}{1} (x-1) [/tex]
x + y = 0
y = -x

f(x) = a(x - 1)(x - 2) - x
Melalui titik C

-27 = a(5 - 1)(5 - 2) - 5
-22 = a(4)(3)
a = -22/12
a = ⁻¹¹/₆

f(x) = ⁻¹¹/₆ (x - 1)(x - 2) - x
f(x) = ⁻¹¹/₆ (x² - 3x + 2) - x
f(x) = -¹¹/₆ x² + ⁹/₂ x - ¹¹/₃

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Titik balik
y = ( -b/2a , -D/4a )

Untuk -b/2a
=> - ⁹/₂ / 2 • - ¹¹/₆
=>²⁷/₂₂ atau 1,227


Untuk -D/4a
=> -(b² - 4ac) / 4a
=> -( [⁹/₂]² - 4[ - ¹¹/₆][ - ¹¹/₃] ) / 4 [- ¹¹/₆]
=> - ²³⁹/₂₆₄ atau -0,905

Maka y = (1,227 , -0,905)

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Memotong sumbu y , x = 0
=> - ¹¹/₆ (0)² + ⁹/₂ (0) - ¹¹/₃
=> - ¹¹/₃ atau - 3,6667