Answer
$T$ is a linear transformation.
Work Step by Step
A transformation is said to be linear if
$T(av+w)=aT(v)+T(w)$, for all $v,w$ in the given domain and for all scalars $a$.
Let $a_0+a_1x+a_2x^2,b_0+b_1x+b_2x^2$ be arbitrary elements of the domain $P_2$ and $k$ be any arbitrary scalar. Then
$\begin{array}{l}
T(k({a_0} + {a_1}x + {a_2}{x^2}) + ({b_0} + {b_1}x + {b_2}{x^2}))\\
= T((k{a_0} + k{a_1}x + k{a_2}{x^2}) + ({b_0} + {b_1}x + {b_2}{x^2}))\\
= T((k{a_0} + {b_0}) + (k{a_1} + {b_1})x + (k{a_2} + {b_2}){x^2})
\end{array}$
$ = [(k{a_0} + {b_0}) + (k{a_1} + {b_1}) + (k{a_2} + {b_2})] + [(k{a_1} + {b_1}) \quad + (k{a_2} + {b_2})]x + (k{a_2} + {b_2}){x^2}$
$ = (k{a_0} + k{a_1} + k{a_2} + {b_0} + {b_1} + {b_2}) + (k{a_1} + k{a_2} + {b_1} + {b_2})x +\quad (k{a_2} + {b_2}){x^2}$
$ = (k{a_0} + k{a_1} + k{a_2}) + ({b_0} + {b_1} + {b_2}) + (k{a_1} + k{a_2})x + ({b_1} + {b_2})x +\quad k{a_2}{x^2} + {b_2}{x^2}$
$ = k({a_0} + {a_1} + {a_2}) + ({b_0} + {b_1} + {b_2}) + k({a_1} + {a_2})x + ({b_1} + {b_2})x + \quad k{a_2}{x^2} + {b_2}{x^2}$
$ = [k({a_0} + {a_1} + {a_2}) + k({a_1} + {a_2})x + k{a_2}{x^2}] +\quad [({b_0} + {b_1} + {b_2}) + ({b_1} + {b_2})x + {b_2}{x^2}]$
$ = k[({a_0} + {a_1} + {a_2}) + ({a_1} + {a_2})x + {a_2}{x^2}] +\quad [({b_0} + {b_1} + {b_2}) + ({b_1} + {b_2})x + {b_2}{x^2}]$
$ = kT({a_0} + {a_1}x + {a_2}{x^2}) + T({b_0} + {b_1}x + {b_2}{x^2})$
Hence, $T$ is a linear transformation.