f(x) = -x²-4x+5 tentukan nilai maximum nya

f(x)=-x²-4x+5
y=x²-4x+5
misal y=0
-x²-4x+5=0
(-x-5)(x-1)=0
x1=-5 atau x2=1
(-5,0) dan (1,0)

misal x=0
y=-0²-4(0)+5
y=5
(0,5)

nilai maks
x puncak/maks
xp=-b/2a
xp=-(-4)/2(-1)
xp=4/-2
xp=-2

y puncak/maks
yp=f(xp)
f(xp)=-xp²-4xp+5
f(-2)=-(-2)²-4(-2)+5
f(-2)=-4+8+5
f(-2)=9
yp=9

jadi nilai maksnya (-2,9)

==Jawaban==
Titik balik diperoleh ketika f'(x) = 0
f(x) = -x² - 4x + 5 (kalikan -1)
f(x) = x² + 4x - 5
f(x) = (x + 5)(x - 1)
f(x) = x1 = -5 atau x2 = 1

Nilai balik:
f(-5) = -x² - 4x + 5
f(-5) = -(5)² - 4(-5) + 5
f(-5) = -25 + 20 + 5
f(-5) = 0

f(1) = -x² - 4x + 5
f(1) = -(1)² - 4(1) + 5
f(1) = -1 - 4 + 5
f(1) = 0

Jadi, nilai maksimum fungsi -x² - 4x + 5 = 0