Answer
See below
Work Step by Step
Assume $(a_1,b_1,c_1)\\
(a_2,b_2,c_2) \in R^3$
$T((a_1,b_1,c_1)+(a_2,b_2,c_2))\\
=T(a_1+a_2,b_1+b_2,c_1+c_2)\\
=(a_1+a_2)x^2+[2(b_1+b_2)-(c_1+c_2)]x+[(a_1+a_2)-2(b_1+b_2)+(c_1+c_2)]\\
=a_1x^2+a_2x^2+(2b_1-c_1)x+(2b_2-c_2)x+(a_1-2b_1+c_1)+(a_2-2b_2+c_2)\\
=T(a_1,b_1,c_1)+T(a_2,b_2,c_2)$
$T(\alpha (a_1,b_1,c_1))\\
=T(\alpha a_1,\alpha b_1,\alpha c_1)\\
=\alpha a_1x^2+x(2\alpha b_1-\alpha c_1)+(\alpha a_1-2\alpha b_1+\alpha c_1)\\
=\alpha [a_1x^2+(2b_1-c_1)+(a_1-2b_1+c_1)]\\
=\alpha T(a_1,b_1,c_1)$
Hence, $T$ is a linear transformation
Obtain: $Ker(T)=\{(a,b,c) \in R^3:T (a,b,c)=0\}\\
=\{(a,b,c) \in R^3:ax^2+(2b-c)x+(a-2b+c)=0\}\\
\rightarrow a=0,c=2b\\
\rightarrow Ker(T)=\{(0,b,2b)\}=\{b(0,1,2)\}=span \{(0,1,2)\}\\
\rightarrow \dim Ker(T)=1$
then $\dim [Ker(T)]=1 \ne \{0\}$
$T$ is not one-to-one
Apply Rank Nullity Theorem:
$\dim [Ker(T)]+\dim [Rng(T)]=\dim R^3\\
1+\dim [Rng(T)]=3\\
\dim [Rng(T)]=2$
Since $\dim [Rng(T)]=2 \ne 3=\dim P_2(R)$,
hence, $T$ is onto.
Consequently, $Rng(T)=\{T(a,b,c):(a,b,c) \in R^3\}\\
=\{ax^2+(2b-c)x+(a-2b+c):a,b,c \in R\}\\
=\{ax^2+2bx-cx+a-2b+c: a,b,c \in R^3\}\\
=\{a(x^2+1)+2b(x-1)-c(x-1): a,b,c \in R^3\}\\
=span \{x^2-1,x-1,x-1\}\\
=span \{x^2-1,x-1\}$
Hence, basic for $Rng(T)$ is $\{x^2-1,x-1\}$
