Răspuns:
Explicație pas cu pas:
f(x)=ax²+bx+c; a,b,c=?
A(0,-1)∈Gf, ⇒f(0)=-1, ⇒c=-1.
B(3,-1)∈Gf, ⇒f(3)=-1, ⇒a·3²+b·3-1=-1, ⇒9a+3b=0 |:3, ⇒3a+b=0, deci b=-3a.
Daca dreapta y=2 este tangenta la Gf, ⇒ varful parabolei e situat pe dreapta y=2 si a<0.
Abscisa varfului parabolei [tex]x=-\dfrac{b}{2a}.~Din~relatia~b=-3a,~ -\dfrac{b}{a}=3,~:2,~-\dfrac{b}{2a}=\dfrac{3}{2}=x_{V}.~Deci~f(x_{V})=2,~a* (\dfrac{3}{2})^{2}+b*\dfrac{3}{2}-1=2,~a*\dfrac{9}{4} +b*\dfrac{3}{2}-1=2~|*4,~9a+6b-4=8,~9a+6b=12~|:3,~3a+2b=4,~inlocuim,~3a+2*(-3a)=4,~3a-6a=4,~-3a=4,~a=-\frac{4}{3}.~Deci~b=-3*(-\frac{4}{3})=4.\\Deci~f(x)=-\frac{4}{3}x^{2}+4x-1.[/tex]
ca confirmare, anexez graficul.... Succese!
