Koordinat titik balik maksimum dari fungsi f(x) = -6 – ( x – 4 )^2

Jawab

f(x) = – 6 – (x – 4)²
f(x) = – 6 – (x² -8x + 16)
f(x) = – 6 – x² +8x – 16
f(x) = – x² + 8x – 22

maksimum untuk y' =0
-2x +8 =0
-2x  = -8
x = 4
f(x) = – x² + 8x – 22
f(4) = – (4)² +8(4) – 22
f(4) = – 16 + 32 -22
f(4) = – 6

titik balik  { x,  f(x) } = {4, -6 }