Let $V$ be a (real) vector space of all polynomials functions from $\textbf{R}$ to $\textbf{R}$ of degree at most two, that is, the space of all functions of the form \begin{align*} f(x) = c_{0} + c_{1}x + c_{2}x^{2} \end{align*} Let $t$ be a fixed real number and define \begin{align*} g_{1}(x) = 1,\quad g_{2}(x) = x + t,\quad g_{3}(x) = (x+t)^{2} = x^{2} + 2xt + t^{2} \end{align*} Prove that $\mathcal{B} = \{g_{1},g_{2},g_{3}\}$ is a basis for $V$. If $f(x) = c_{0} + c_{1}x + c_{2}x^{2}$, what are the coordinates of $f$ in this ordered basis $\mathcal{B}$?
MY ATTEMPT
At first, we must prove that $\mathcal{B}$ is linear independent. Indeed, this is the case when $t\neq 0$, because \begin{align*} & ag_{1} + bg_{2} + cg_{3} = a + b(x + t) + c(x^{2} + 2xt + t^{2}) = 0 \Longrightarrow\\\\ & (a + bt + ct^{2}) + (b + 2ct)x + cx^{2} = 0 \Longrightarrow c = b = a = 0 \end{align*} The same applies to $t = 0$. Consequently, the coordinates of $f$ in $\mathcal{B}$ are given by \begin{align*} \begin{bmatrix} c'_{0} \\ c'_{1} \\ c'_{2} \end{bmatrix} = \begin{bmatrix} 1 & t & t^{2}\\ 0 & 1 & 2t\\ 0 & 0 & 1 \end{bmatrix}^{-1} \begin{bmatrix} c_{0} \\ c_{1} \\ c_{2} \end{bmatrix} = \ldots \end{align*}
Could someone double-check my solution?