f(x)=x^2-2x+a
Stim ca Varful parabolei are coordonatele:
[tex]V \Big(-\dfrac{b}{2a'} , -\dfrac{\Delta}{4a'}\Big) [/tex]
Cadranul IV este in partea din dreapta jos a sistemului de coordonate.
(in dreapta axei Oy si sub axa Ox)
Conditiile noastre sunt:
[tex]V(x_0, y_0) \\ \\
\begin{document}
\[ \left\{
\begin{array}{Il}
\ x_0\ \textgreater \ 0 & &y_0\ \textless \ 0 \ \end{array} \Rightarrow \begin{document}
\[ \left\{
\begin{array}{Il}
\ -\dfrac{b}{2a'} \ \textgreater \ 0 & & & &-\dfrac{\Delta}{4a'}\ \textless \ 0 \ \end{array} \Rightarrow
\right. \]
\right. \]
\end{document}
\Rightarrow [/tex]
[tex]\boxed{1} \quad -\dfrac{b}{2a'}\ \textgreater \ 0 \Rightarrow \dfrac{-(-2)}{2\cdot 1} \ \textgreater \ 0 \Rightarrow 1 \ \textgreater \ 0,\quad \forall $ $ x\in \mathbb_{R} $ \\ \\ $ \boxed{2} -\dfrac{\Delta}{4a'} \ \textless \ 0 \Rightarrow -\dfrac{b^2-4a'c}{4a'}\ \textless \ 0\Rightarrow - \dfrac{4-4a}{4} \ \textless \ 0\Big|\cdot(-4) \Rightarrow \\ \\ \Rightarrow 4-4a\ \textgreater \ 0 \RIghtarrow -4a \ \textgreater \ -4 \Rightarrow 4a \ \textless \ 4 \Rightarrow a\ \textless \ 1 \\ \\$ Din \boxed{1} \cap $ $ \boxed{2} \Rightarrow a \ \textless \ 1 \Rightarrow \boxed{a\in (-\infty, 1)}[/tex]
