grafik fungsi kuadrat yg mempunyai titik puncak (2,-3) serta melalui titik (1,-2) mempunyai persamaan... Tolong di bantu ya kaka :)

GRAFIK FUNGSI KUADRAT

(xp , yp) = (2 , -3)
(x , y) = (1 , -2)

PK umum
ax² + bx + c = y

xp = -b/2a
2 = -b/2a
-b = 4a
b = -4a

Pknya
y = ax² - 4ax + c

Titik puncak (2 , -3)
-3 = a(2)² - 4a(2) + c
4a - 8a + c = -3

-4a + c = -3


Titik (1 , -2)
-2 = a(1)² - 4a(1) + c
a - 4a + c = -2
-3a + c = -2


-4a + c = -3
-3a + c = -2
------------------- -
-a = -1
a = 1

-3a + c = -2
-3 + c = -2
c = 1

b = -4a
b = -4

PK
y = x² - 4x + 1

[tex]y = a(x - xp) {}^{2} + yp \\ y= a(x - 2) {}^{2} - 3melalui \: (1. - 2) \\ - 2 = a(1 - 2 ) {}^{2} - 3 \\ - 2 = a - 3 \\ a = 1 \\ y =1(x - 2) {}^{2} - 3 \\ y = {x}^{2} - 4x + 4 - 3 \\ y = {x}^{2} - 4x + 1[/tex]
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