Parabola t=ax kuadrat +bx+c melalui titik (0,1),(1,0),dan (3,0). Jika titik minimum parabola tersebut adalah (p,q) maka q...

t = ax²+bx+c

titik (0,1)
1 = a(0)²+b(0)+c
c = 1

titik (1,0)
0 = a+b+c
a+b = -c
a+b = -1

titik (3,0)
0 = 9a+3b+c
3(3a+b) = -c
3(2a+a+b) = -c
6a-3 = -1
a = 2/6
a = ⅓

⅓+b = -1
b = -4/3

parabola :
t = ⅓x²-(4/3)x -1

titik minimum → (x,y)

x = -b/2a = -(-4/3) / 2(⅓)
x = 4/3 ÷ 2/3
x = 2

y = -D/4a
y = -(b²-4ac)/4a
y = -([-4/3]²-4.⅓.-1)/4(⅓)
y = -(16/9 + 12/9) / (4/3)
y = -28/9 ÷ 4/3
y = -7/3

titik minimum → (2,-7/3)

q = -7/3