This is one part to the question, the next five parts of the question will be revealed upon answering the previous part correctly Part one: the domain is?

Given:
The parabola equation is,
[tex]f\mleft(x\mright)=-3\mleft(x-3\mright)^2+3[/tex]To find:
domain and range of the graph.
Explanation:
Domain:
The domain of a function is the set of input values for which the function is real and defined.
the function here dose not have any undefined points. So,
the domain is,
[tex]-\infty\: Range;The set of values of the dependent variable for which the function is defined.
for parabola ,
[tex]ax^2+bx+c\: [/tex]with the vertex,
[tex](x_v,\: y_v)[/tex][tex]\begin{gathered} if\: a<0\: \text{ the range is,}f\mleft(x\mright)\le\: y_v \\ \text{if }\: a>0\text{ the range is, }f\mleft(x\mright)\ge\: y_v \end{gathered}[/tex]then,
[tex]\begin{gathered} a=-3 \\ \text{vertices: (}x_v,\: y_v)=(3,\: 3) \end{gathered}[/tex]hence,
[tex]f\mleft(x\mright)\le\: 3[/tex]The maximum point is (3,3).
Final Answer:
Domain of the parabola is,
[tex]-\infty\: Range of the parabola is,[tex]f\mleft(x\mright)\le\: 3[/tex]in interval notation the range is,
[tex]\: \: (-\infty\: ,\: 3\rbrack[/tex]the vertex of the parabola is,
[tex](3,3)[/tex]