Yg nomer 4 ya kak bang ​

nilai maksimumnya = 18

Pembahasan :

y = –x² – 2px + p²

bentuk ax² + bx + c

a = – 1

b = –2p

sumbu simetri x = 3

syarat maksimum, x = –b/(2a)

x = –(–2p)/(2(–1))

3 = 2p/(–2)

3 = –p

p = –3

y = -x² – 2px + p²

y = –x² –2(–3)x + (–3)²

y = –x² + 6x + 9

substitusi x = 3

y = –3² + 6(3) + 9

y = –9 + 18 + 9

y = 18

nilai maksimumnya = y = 18