1. Diketahui fungsi kuadrat berikut f(x): X² – 4x – 12

Jawab:

1. a. T.P terhadap sb-x, syarat jika y = 0

       x² – 4x – 12 = 0

      (x + 2)(x – 6) = 0

     x + 2 = 0      atau x – 6 = 0

      x = -2          atau x = 6

T.P(-2,0), (6,0)

b. TP. terhadap sb-y, syarat x = 0

f(x) = y

 y = 0² – 4(0) – 12 = -12

TP.(0, -12

c. nilai ekstrim diperoleh jika x= -b/2a dimana diketahui a = 1, b = -4 , c = -12

  x = -(-4)/2.1 = 4/2 = 2

  f(2) = 2² – 4(2) – 12

        = 4 – 8 – 12

        = -16

 jadi, nilai ekstrimnya adalah -16

d. Titik ekstrimnya (2, -16)

2. P(1, -2)                 x = -b/2a  ⇒ 1 = -b/2a ⇔ -b = 2a ⇔ b = -2a

   f(x) = ax² + bx + c

   -2 = a(1)² + b(1) + c

    -2 = a + b +c

  A(3,0)

0 = a(3)² + b(3) + c

0 = 9a + 3b + c

 9a + 3b + c = 0

  a + b + c     = -2  -

   8a + 2b   = 2

⇔ 8a + 2(-2a) = 2

⇔ 8a – 4a = 2

⇔ 4a = 2

⇔ a = 1/2

b = -2(1/2) = -1

  a + b + c = -2

⇒ 1/2 + (-1) + c = -2

⇒ 1/2 – 1 + c = -2

⇒ 1/2 + c = -2 + 1

⇒ 1/2 + c = -1

⇒ c = -1/2 -1 = -1/2 – 2/2 = -3/2

jadi fungsi kuadratnya:

f(x) = 1/2x² – x – 3/2