Jawab:

Penjelasan dengan langkah-langkah:

Titik potong sumbu X = (2, 0) dan (-3, 0)

Titik potong sumbu Y = (0, -6)

f(x) = ax² + bx + c

(2, 0) ⇒ a(2)² + b(2) + c = 0

4a + 2b + c = 0

(-3, 0) ⇒ a(-3)² + b(-3) + c = 0

9a - 3b + c = 0

(0, -6) ⇒ a(0)² + b(0) + c = -6

c = -6

4a + 2b + c = 0

4a + 2b + (-6) = 0

4a + 2b = 6

9a - 3b + c = 0

9a - 3b + (-6) = 0

9a - 3b = 6

4a + 2b = 6 (x (-1/2))

9a - 3b = 6 (x (1/3))

-2a - b = -3

3a - b = 2     _

-5a = -5

a = -5/(-5)

a = 1

3(1) - b = 2

b = 3 - 2

b = 1

Fungsi kuadratnya adalah :

f(x) = x² + x - 6

Titik balik fungsi kuadrat : (-b/2a, -(b² - 4ac)/4a)

(-1/2(1), -(1² - 4(1)(-6))/4(1)) = (-1/2, -(1 + 24)/4)

= (-1/2, -25/4)

= (-1/2, -6  1/4)