ますどら

\[ \begin{aligned} y &= \colorbox{aqua}{$-$}x^2+4x+1 && \text{\scriptsize \color{aqua}：$x^2$の係数がある！} \\ &= \colorbox{aqua}{$-$}(x^2\colorbox{misque}{$-4$}x)+1 && \text{\scriptsize \color{aqua}（$x^2$の係数 \colorbox{aqua}{\color{blue}$-1$} でくくる）} \\ &= -\lbrace(x\colorbox{misque}{$-2$})^2-2^2\rbrace+1 && \text{\scriptsize \color{gray}（$x$の符号なし係数 $4 \times \tfrac12 = 2$を引いて, の半分の$2$を使って平方完成）} \\ &= -\lbrace(x-2)^2-4\rbrace+1 && \text{\scriptsize \color{gray}（$2^2$ を計算）} \\ &= -(x-2)^2+4+1 && \text{\scriptsize \color{gray}（中括弧をはずす）} \\ &= -(x-2)^2+5 && \text{\scriptsize \color{gray}（定数項を整理）} \end{aligned} \]

\[ \begin{aligned} y &= x^2-4x \\ &= (x-2)^2-2^2 \\ &= (x-2)^2-4 \end{aligned} \]

\[ \begin{aligned} y &= 2x^2-8x+6 \\ &= 2(x^2-4x)+6 \\ &= 2\lbrace(x-2)^2-2^2\rbrace+6 \\ &= 2\lbrace(x-2)^2-4\rbrace+6 \\ &= 2(x-2)^2-8+6 \\ &= 2(x-2)^2-2 \end{aligned} \]

\[ \begin{aligned} y &= 3x^2-6x+2 \\ &= 3(x^2-2x)+2 \\ &= 3\lbrace(x-1)^2-1^2\rbrace+2 \\ &= 3\lbrace(x-1)^2-1\rbrace+2 \\ &= 3(x-1)^2-3+2 \\ &= 3(x-1)^2-1 \end{aligned} \]