fie functia de gradul 2 f(x)=ax²+bx+c.
V(1,1) => [tex] \frac{-b}{2a}=1 \\ \frac{-delta}{4a} =1
[/tex]
din prima relatie =>-b=2a=> b=-2a
A(2,2)∈ Gf⇒ f(2)=2
a*2²+b*2+c=2
4a+2b+c=2.
4a+2*(-2a)+c=2
4a-4a+c=2 ⇒c=2
4a+2b+2=2
4a+2b=0 |:2
2a+b=0
-Δ/4a=1 =>
[tex] \frac{-(b^{2}-4ac )}{4a}=1 =\ \textgreater \ -b^{2}+4ac=4a \\
-(-2a)^{2} +4a*2=4a \\
-4a^{2}+8a-4a=0 \\ -4a^{2} +4a=0 \\ 4a(-a+1)=0 \\ 4a=0 =\ \textgreater \ a=0 \\ -a+1=0=\ \textgreater \ a=-1
[/tex]
caz 1: a=0 => b=-2a=> b=-2*0=0 => f(x)=2 nu e buna solutia fiindca nu e functie de gr 2 cum imi cere
caz 2: a=-1 => b=-2*(-1)=2 => f(x)=-x²+2x+2. asta e solutia finala. a=-1 si b=2. c-ul a fost aflat deja din relatii,este 2
