Question

Wzor funkcji kwadratowej
zamien to na postac kanoniczną
a) f(x) =
b) f(x)=2x^2 - 4x + 12

Answer 1

f(x) = 4/3x(x+6)
f(x)= 4/3x*x+4/3x*6
f(x)=4/3x^2+8x

p=-b/2a=-8/2*4/3=-8*3/8=-3
Δ=64
q=-Δ/4a=-64:16/3=-64*3/16=-12
f(x)=a(x-p)²+q
f(x)= 4/3(x+3)²-12
b)f(X)= 2x²-4x+12
p=-b/2a=4/4=1
Δ=(-4)²-4*2*12=16-96=-80
q=80/4a=80/8=10
f(x)=2(x-1)²+10

Answer 2

A)  f(x)= 4/3x(x+6)
4/3x =0  ∨ x+6=0                          y=a(x-p)² +q
x₁ =0  ∨  x₂= -6 
p=(x₁ +x₂):2= (0 -6):2=-3
q=f(p)= 4/3(-3)(-3 +6)= -4 ·3= -12
f(x)= 4/3(x+3)² -12
b)  f(x) =2x² -4x +12
Δ= 16-4·2·12= 16 -96= -80
p = -b/2a= 4/4=1
g= -Δ/4a= 80/8=10
f(x(= 2(x -1)² +10