f(x) = -1/2 x² - 2x

a = -1/2, b = -2, c = 0

a. D = b² - 4ac

D = (-2)² - 4(-1/2)(0)

= 4

b. a = -1/2

c. karna a = -1/2 < 0, maka grafik menghadap ke arah bawah

d. tipot sumbu x ---> y = 0

0 = -1/2 x² - 2x

0 = x(-1/2 x - 2)

x = 0 atau x = -4

titik potong sumbu X --> (0,0) dan (-4,0)

e. tipot sumbu Y ---> x = 0

y = -1/2(0)² - 2(0)

y = 0

titik potong sumbu Y ---> (0,0)

f. sumbu simetri

x = -b/2a

x = -(-2) / 2(-1/2)

x = -2

g. koordinat titik puncak

xp = -2

yp = f(xp)

= -1/2(-2)² - 2(-2)

= -2 + 4

yp = 2

titik puncak (-2,2)

h. terlampir

i. daerah asal = {x| -5 < x < 2, x € R}

= {-4, -3, -2, -1, 0, 1}

f(-4) = -1/2(-4)² - 2(-4) = -8 + 8 = 0

f(-3) = -1/2(-3)² - 2(-3)= -9/2 + 6 = 3/2

f(-2) = -1/2(-2)² - 2(-2) = -2 + 4 = 2

f(-1) = -1/2(-1)² - 2(-1) = -1/2 + 2 = 3/2

f(-0) = -1/2(0)² - 2(0) = 0

f(1) = -1/2(1)² - 2(1) = -1/2 - 2 = -5/2

daerah hasil = {-5/2, 0, 3/2, 2}

j. daerah hasil = {y| -6 < y < 0}

= {-5, -4, -3, -2, -1}

f(x) = -1/2 x² - 2x

-6 = -1/2 x² - 2x

------------------------ × 2

-12 = -x² - 4x

x² + 4x - 12 = 0

(x - 2)(x + 6) = 0

x = 2 atau x = -6