Dau coronita!!!! Urgent!!!

I)
[tex]\it f:\mathbb{R} \longrightarrow \mathbb{R},\ f(x)=x^2-2x+m\\ \\ a)\ \ P(3,\ 3 )\in Gf \Rightarrow f(3)=3 \Rightarrow 9-6+m=3 \Rightarrow 3+m=3 \Rightarrow m=0\\ \\ b) Coeficientul\ lui\ x^2\ este\ 1>0 \Rightarrow f\ admite\ un\ minim, de\ abscis\breve a\\ \\ \left.\begin{aligned} \it -\dfrac{b}{2a}=-\dfrac{-2}{2}=1\\ \\ \it f(1)=1-2=-1 \end{aligned}\right\} \Rightarrow V(1,\ -1)\ este\ v\hat arful\ parabolei\ (punct\ de\ minim)[/tex]