Buatlah grafik titik maksimum dr persamaan

jawb

x² - 4x + y -12 = 0
y = - x² + 4x + 12

titik potong sumbu y, x= 0
y = 12 → (x,y)= (0,12)

titik potong sumbu x, y = 0
0 = -x² + 4x + 12
-(x² - 4x - 12)= 0
-(x - 6)(x +2) = 0
x = 6 ,x  = - 2
(x,y)= (6,0) ,
(x,y) = (-2,0)

titik maksimum
y = - x² + 4x  +12
x = -b/2a = -4/-2   =2
y = - (2)² - 4(2) - 12  = -6

titik minimum (x,y)= ( 2 , -6)