SOAL MATEMATIKA SMP

1. Fungsi kuadrat yang grafiknya berpuncak dititik (2, 3) dan melalui titik (-2, 1) adalah …
   
   

A. y = -1/8(x – 2)² + 3

B. y = -1/8(x – 2)² – 3

C. y = 1/8(x + 2)² – 3

D. y = 1/8(x + 2)² + 3

E. y = 1/8(x – 2)² + 3

PEMBAHASAN :

f(x) = ax² + bx + c

f'(x) = 2ax + b

0 = 2a.2 + b

0 = 4a + b

-b = 4a … (i)

nilai fungsi pada titik puncak

f(2) = a(2)² + b.2 + c

3 = 4a + 2b + c

3 = -b + 2b + c

3 = b + c … (ii)

f(-2) = a(-2)² + b(-2) + c

1 = 4a – 2b + c

1 = -b – 2b + c

1 = -3b + c … (iii)

eliminasi persamaan (ii) dan (iii)

b + c = 3

-3b + c = 1 –

4b = 2

b = 1/2

substitusi b = 1/2 ke persamaan (ii)

1/2 + c = 3

c = 5/2

substitusi b = 1/2 ke persamaan (i)

-1/2 = 4a

a = -1/8

f(x) = (-1/8)x² + 1/2 x + 5/2

= (-1/8)x² + 4/8 x + 5/2

= -1/8(x² – 4x) + 5/2

= -1/8(x – 2)² + 4/8 + 5/2

= -1/8(x – 2)² + 4/8 + 20/8

= -1/8(x – 2)² + 3

JAWABAN : A