Răspuns :
forma canonica
f(x)=a(x+b/2a)²+(-Δ/4a)
a=-7
b=0
c=8
f(x)=-7x²+8
d)a=4
b=-5
c=0
f(x)=4(x-5/8)-25/16
f(x)=a(x+b/2a)²+(-Δ/4a)
a=-7
b=0
c=8
f(x)=-7x²+8
d)a=4
b=-5
c=0
f(x)=4(x-5/8)-25/16
a(x+b/2a)²-Δ/4a
c)a=-7; b=0 c=8 Δ=b²-4ac=0+4*7*8; -Δ=-4*7*8
-7(x+0)² -4*7*8/(4*(-7))=-7x²+8 aceeasi forma
d) a=4 b=-5 c=0
b/2a=-5/8
Δ=25-0 -Δ=-25
4(x-5/8)²-25/16
c)a=-7; b=0 c=8 Δ=b²-4ac=0+4*7*8; -Δ=-4*7*8
-7(x+0)² -4*7*8/(4*(-7))=-7x²+8 aceeasi forma
d) a=4 b=-5 c=0
b/2a=-5/8
Δ=25-0 -Δ=-25
4(x-5/8)²-25/16