Reprezentarea grafică a funcției este o parabolă.
Vârful parabolei este :
[tex]\it \ V\left(-\dfrac{b}{2a},\ -\dfrac{\Delta}{4a}\right).[/tex]
Din datele problemei rezultă:[tex]\it -\dfrac{b}{2a} =\dfrac{3}{4} \Rightarrow -\dfrac{b}{a} =\dfrac{3}{2} \Rightarrow \dfrac{b}{a} = -\dfrac{3}{2} \ \textless \ 0 \Rightarrow a,\ b\ au\ semne\ diferite\ \ \ (*)[/tex]
[tex]\it \dfrac{b}{a} = -\dfrac{3}{2} \Rightarrow 2b = -3a \Rightarrow 4b^2=9a^2 \ \ \ \ (1)
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-\dfrac{\Delta}{4a} = -\dfrac{5}{4} \Rightarrow \dfrac{b^2-4a\cdot1}{4a} =\dfrac{5}{4} \Rightarrow 4b^2-16a=20a \Rightarrow 4b^2=36a\ \ (2)
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(1), \ (2) \Rightarrow 9a^2=36a |_{:a} \Rightarrow 9a=36 \Rightarrow a = 4\ \ \ \ (3)[/tex]
[tex]\it (2),\ (3) \Rightarrow 4b^2= 36\cdot4 \Rightarrow b^2=36 \Rightarrow b=\pm6 \stackrel{(*)}{\Longrightarrow} b=-6[/tex]
b)
[tex]\it Gf\cap Oy = A(0,\ y) \Rightarrow y=f(0) =1 \Rightarrow Gf\cap Oy = A(0,\ 1)
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[tex]\it Gf\cap Ox = \{B(x_1,\ 0),\ C(x_2,\ 0)\},\ unde\ x_1,\ x_2 \ sunt\ zerourile\ func\c{\it t}iei[/tex]
[tex]\it \mathcal{A}_{ABC} = \dfrac{AO\cdot BC}{2} = \dfrac{1\cdot(x_2-x_1)}{2}= \dfrac{x_2-x_1}{2}.[/tex]
