fungsi kuadrat f(x)=x^2+2px+p mempunyai nilai minimum -p dengan p tidak sama dengan nol. jika sumbu simetri kurva f adalah x=a,maka nilai a+fA=??

f(x)= x² + 2px + p
nilai min y = -p
x² + 2px + p = - p
x² + 2px + 2p = 0

y = -p --> D/-4a = -p
(2p)² - 4(1)(p)= - p
4p² - 4p + p = 0
4p² - 3p = 0
p(4p -3) =0
p = 0 (tm) atau p = 3/4

f(x)= x² +2(3/4) x + 3/4
f(x)= x² + 3/2 x + 3/4
sumbu simetri  2x + 3/2 = 0
2x = - 3/2
x = - 3/4 --> x = a --> a = - 3/4

nilai a + f(a) = a + a² + 3/2 a + 3/4 = a² + 5/2 a + 3/4
a= - 3/4 --> a + f(a) = (-3/4)² + 5/2(-3/4) + 3/4
a+ f(a)= 9/16 - 15/8 + 3/4
a + f(a)= 51/16