Kuadrat Fungsi

f(x) = x² + 2x - 3

a = 1

b = 2

c = -3

Df = {x| -4 < x < 2 , x ∈ R}

a.

titik potong dg sb x → y = 0

x² + 2x - 3 = 0

(x + 3)(x - 1) = 0

x = -3 atau x = 1

titik potong → (-3,0) dan (1,0)

b.

titik potong dg sb y → x = 0

f(0) = y = c = -3

titik potong → (0,-3)

c.

sumbu simetri x = -b / 2a

x = -2/2.1

x = -1

d.

titik balik

y = f(-1) = (-1)² + 2(-1) - 3

y = -4

titik balik → (-1 , -4)

e..

-4 < x < 2

f(-4) = (-4)² + 2(-4) - 3 = 5

f(2) = 2² + 2.2 - 3 = 5

Range fungsi, Rf :

Rf = {y| -4 ≤ x < 5, y ∈ R}