Respuesta :
[tex]f(x)=3(x-1)^{2}+2 \\ \\ f(x)=a(x-p)^{2}+q \\ \\ W=(p,q) \\ \\ W=(1,2) \\ \\ D: x \in R \\ \\ a \ \textgreater \ 0 \\ \\ Range: y \in [2,+\infty)[/tex]
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"Associate with people who are likely to improve you."
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"Associate with people who are likely to improve you."
[tex]f(x)=3(x-1)^2+2\ It's\ a\ quadratic\ function\\\\therefore\ Domain:\boxed{D:x\in\mathbb{R}\ all\ real\ numbers}\\\\vertex:\\the\ vertex\ form:\ y=a(x-h)^2+k\\where\ (h;\ k)h\ and\ k\ are\ the\ x\ and\ y\ coordinates\ of\ the\ vertex\\\boxed{vertex:(1;\ 2)}[/tex]
[tex]range:\\a=3 \ \textgreater \ 0\ therefore\ the\ parabola\ ope ns\ upward\\so\\range\ is\ [vertex\ (y);\ infinity)\\range:\boxed{y\geq2\y\in[2;\ \infty)}\\\\Your\ answer:\\\boxed{The\ vertex\ (1;\ 2);\ domain\ is\ all\ real\ number\ (x\in\mathbb{R});\ range\ y\geq2}[/tex]
[tex]range:\\a=3 \ \textgreater \ 0\ therefore\ the\ parabola\ ope ns\ upward\\so\\range\ is\ [vertex\ (y);\ infinity)\\range:\boxed{y\geq2\y\in[2;\ \infty)}\\\\Your\ answer:\\\boxed{The\ vertex\ (1;\ 2);\ domain\ is\ all\ real\ number\ (x\in\mathbb{R});\ range\ y\geq2}[/tex]