btakeshi
２次関数の軸と頂点、最大最小そしてグラフを書く際に必ず必要になる「平方完成」をマスターしましょう。このページでは２次の係数が１ではない問題を取り上げます。最初に因数分解をするステップが入るだけで、あとはレベル①と同じです。きっちりと手順を理解して使えるようにしましょう。
次の２次関数を y=a(x-p)^2+q の形に変形し，グラフの軸と頂点を求めよ。
↓この問題へのリンクはこちら（右クリックで保存）
【解答】
\begin{align*}
& \scriptsize\qquad\color{red}\Darr\bf\ ２次の係数が１以外！\\
y &= \colorbox{mistyrose}{$\displaystyle2$} x^2 -12 x +22\\
& \scriptsize\qquad\qquad\quad\color{red}\Darr\ -12 \div \left( 2 \right) = \colorbox{mistyrose}{$\displaystyle-6$}\\
&= \colorbox{white}{$2$} \left( x^2 \colorbox{mistyrose}{$\displaystyle-6$} x \right) +22\\
& \scriptsize\qquad\qquad\quad\color{orange}\Darr\ -6 \times \dfrac12 = \colorbox{bisque}{$\displaystyle-3$}\\
&= \colorbox{white}{$2$}\left\{ \left( x \colorbox{bisque}{$\displaystyle-3$} \right)^2 \colorbox{palegreen}{$\displaystyle - 9$} \right\} +22\\
& \scriptsize\qquad\qquad\qquad\qquad\quad\color{green}\Uarr\ 2乗した\ 9\ を引く\\
&= \colorbox{white}{$2$} \left( x -3 \right)^2 \colorbox{violet}{$\displaystyle-18$} +22\\
& \scriptsize\qquad\qquad\qquad\qquad\quad\color{magenta}\Uarr\ 2 \times \left( -9 \right)\\
&= \colorbox{white}{$2$} \left( x -3 \right)^2 +4\\
よって\\
& \scriptsize\color{deepskyblue}x - 3=0\ を解いて\ x = \colorbox{lightcyan}{$\displaystyle3$}\\
& 軸\ x = \colorbox{lightcyan}{$\displaystyle3$}\\
& 頂点\ \left( \colorbox{lightcyan}{$\displaystyle3$},\ 4 \right)
\end{align*}
↓この問題へのリンクはこちら（右クリックで保存）
【解答】
\begin{align*}
& \scriptsize\qquad\color{red}\Darr\bf\ ２次の係数が１以外！\\
y &= \colorbox{mistyrose}{$\displaystyle-$} x^2 +4 x -3\\
& \scriptsize\qquad\qquad\quad\color{red}\Darr\ +4 \div \left( -1 \right) = \colorbox{mistyrose}{$\displaystyle-4$}\\
&= \colorbox{white}{$-$} \left( x^2 \colorbox{mistyrose}{$\displaystyle-4$} x \right) -3\\
& \scriptsize\qquad\qquad\quad\color{orange}\Darr\ -4 \times \dfrac12 = \colorbox{bisque}{$\displaystyle-2$}\\
&= \colorbox{white}{$-$}\left\{ \left( x \colorbox{bisque}{$\displaystyle-2$} \right)^2 \colorbox{palegreen}{$\displaystyle - 4$} \right\} -3\\
& \scriptsize\qquad\qquad\qquad\qquad\quad\color{green}\Uarr\ 2乗した\ 4\ を引く\\
&= \colorbox{white}{$-$} \left( x -2 \right)^2 \colorbox{violet}{$\displaystyle+4$} -3\\
& \scriptsize\qquad\qquad\qquad\qquad\quad\color{magenta}\Uarr\ -1 \times \left( -4 \right)\\
&= \colorbox{white}{$-$} \left( x -2 \right)^2 +1\\
よって\\
& \scriptsize\color{deepskyblue}x - 2=0\ を解いて\ x = \colorbox{lightcyan}{$\displaystyle2$}\\
& 軸\ x = \colorbox{lightcyan}{$\displaystyle2$}\\
& 頂点\ \left( \colorbox{lightcyan}{$\displaystyle2$},\ 1 \right)
\end{align*}
↓この問題へのリンクはこちら（右クリックで保存）
【解答】
\begin{align*}
& \scriptsize\qquad\color{red}\Darr\bf\ ２次の係数が１以外！\\
y &= \colorbox{mistyrose}{$\displaystyle-2$} x^2 +4 x +1\\
& \scriptsize\qquad\qquad\quad\color{red}\Darr\ +4 \div \left( -2 \right) = \colorbox{mistyrose}{$\displaystyle-2$}\\
&= \colorbox{white}{$-2$} \left( x^2 \colorbox{mistyrose}{$\displaystyle-2$} x \right) +1\\
& \scriptsize\qquad\qquad\quad\color{orange}\Darr\ -2 \times \dfrac12 = \colorbox{bisque}{$\displaystyle-1$}\\
&= \colorbox{white}{$-2$}\left\{ \left( x \colorbox{bisque}{$\displaystyle-1$} \right)^2 \colorbox{palegreen}{$\displaystyle - 1$} \right\} +1\\
& \scriptsize\qquad\qquad\qquad\qquad\quad\color{green}\Uarr\ 2乗した\ 1\ を引く\\
&= \colorbox{white}{$-2$} \left( x -1 \right)^2 \colorbox{violet}{$\displaystyle+2$} +1\\
& \scriptsize\qquad\qquad\qquad\qquad\quad\color{magenta}\Uarr\ -2 \times \left( -1 \right)\\
&= \colorbox{white}{$-2$} \left( x -1 \right)^2 +3\\
よって\\
& \scriptsize\color{deepskyblue}x - 1=0\ を解いて\ x = \colorbox{lightcyan}{$\displaystyle1$}\\
& 軸\ x = \colorbox{lightcyan}{$\displaystyle1$}\\
& 頂点\ \left( \colorbox{lightcyan}{$\displaystyle1$},\ 3 \right)
\end{align*}
↓この問題へのリンクはこちら（右クリックで保存）
【解答】
\begin{align*}
& \scriptsize\qquad\color{red}\Darr\bf\ ２次の係数が１以外！\\
y &= \colorbox{mistyrose}{$\displaystyle2$} x^2 -8 x -1\\
& \scriptsize\qquad\qquad\quad\color{red}\Darr\ -8 \div \left( 2 \right) = \colorbox{mistyrose}{$\displaystyle-4$}\\
&= \colorbox{white}{$2$} \left( x^2 \colorbox{mistyrose}{$\displaystyle-4$} x \right) -1\\
& \scriptsize\qquad\qquad\quad\color{orange}\Darr\ -4 \times \dfrac12 = \colorbox{bisque}{$\displaystyle-2$}\\
&= \colorbox{white}{$2$}\left\{ \left( x \colorbox{bisque}{$\displaystyle-2$} \right)^2 \colorbox{palegreen}{$\displaystyle - 4$} \right\} -1\\
& \scriptsize\qquad\qquad\qquad\qquad\quad\color{green}\Uarr\ 2乗した\ 4\ を引く\\
&= \colorbox{white}{$2$} \left( x -2 \right)^2 \colorbox{violet}{$\displaystyle-8$} -1\\
& \scriptsize\qquad\qquad\qquad\qquad\quad\color{magenta}\Uarr\ 2 \times \left( -4 \right)\\
&= \colorbox{white}{$2$} \left( x -2 \right)^2 -9\\
よって\\
& \scriptsize\color{deepskyblue}x - 2=0\ を解いて\ x = \colorbox{lightcyan}{$\displaystyle2$}\\
& 軸\ x = \colorbox{lightcyan}{$\displaystyle2$}\\
& 頂点\ \left( \colorbox{lightcyan}{$\displaystyle2$},\ -9 \right)
\end{align*}
↓この問題へのリンクはこちら（右クリックで保存）
【解答】
\begin{align*}
& \scriptsize\qquad\color{red}\Darr\bf\ ２次の係数が１以外！\\
y &= \colorbox{mistyrose}{$\displaystyle-$} x^2 +2 x +3\\
& \scriptsize\qquad\qquad\quad\color{red}\Darr\ +2 \div \left( -1 \right) = \colorbox{mistyrose}{$\displaystyle-2$}\\
&= \colorbox{white}{$-$} \left( x^2 \colorbox{mistyrose}{$\displaystyle-2$} x \right) +3\\
& \scriptsize\qquad\qquad\quad\color{orange}\Darr\ -2 \times \dfrac12 = \colorbox{bisque}{$\displaystyle-1$}\\
&= \colorbox{white}{$-$}\left\{ \left( x \colorbox{bisque}{$\displaystyle-1$} \right)^2 \colorbox{palegreen}{$\displaystyle - 1$} \right\} +3\\
& \scriptsize\qquad\qquad\qquad\qquad\quad\color{green}\Uarr\ 2乗した\ 1\ を引く\\
&= \colorbox{white}{$-$} \left( x -1 \right)^2 \colorbox{violet}{$\displaystyle+1$} +3\\
& \scriptsize\qquad\qquad\qquad\qquad\quad\color{magenta}\Uarr\ -1 \times \left( -1 \right)\\
&= \colorbox{white}{$-$} \left( x -1 \right)^2 +4\\
よって\\
& \scriptsize\color{deepskyblue}x - 1=0\ を解いて\ x = \colorbox{lightcyan}{$\displaystyle1$}\\
& 軸\ x = \colorbox{lightcyan}{$\displaystyle1$}\\
& 頂点\ \left( \colorbox{lightcyan}{$\displaystyle1$},\ 4 \right)
\end{align*}
↓この問題へのリンクはこちら（右クリックで保存）
【解答】
\begin{align*}
& \scriptsize\qquad\color{red}\Darr\bf\ ２次の係数が１以外！\\
y &= \colorbox{mistyrose}{$\displaystyle2$} x^2 -4 x -3\\
& \scriptsize\qquad\qquad\quad\color{red}\Darr\ -4 \div \left( 2 \right) = \colorbox{mistyrose}{$\displaystyle-2$}\\
&= \colorbox{white}{$2$} \left( x^2 \colorbox{mistyrose}{$\displaystyle-2$} x \right) -3\\
& \scriptsize\qquad\qquad\quad\color{orange}\Darr\ -2 \times \dfrac12 = \colorbox{bisque}{$\displaystyle-1$}\\
&= \colorbox{white}{$2$}\left\{ \left( x \colorbox{bisque}{$\displaystyle-1$} \right)^2 \colorbox{palegreen}{$\displaystyle - 1$} \right\} -3\\
& \scriptsize\qquad\qquad\qquad\qquad\quad\color{green}\Uarr\ 2乗した\ 1\ を引く\\
&= \colorbox{white}{$2$} \left( x -1 \right)^2 \colorbox{violet}{$\displaystyle-2$} -3\\
& \scriptsize\qquad\qquad\qquad\qquad\quad\color{magenta}\Uarr\ 2 \times \left( -1 \right)\\
&= \colorbox{white}{$2$} \left( x -1 \right)^2 -5\\
よって\\
& \scriptsize\color{deepskyblue}x - 1=0\ を解いて\ x = \colorbox{lightcyan}{$\displaystyle1$}\\
& 軸\ x = \colorbox{lightcyan}{$\displaystyle1$}\\
& 頂点\ \left( \colorbox{lightcyan}{$\displaystyle1$},\ -5 \right)
\end{align*}
↓この問題へのリンクはこちら（右クリックで保存）
【解答】
\begin{align*}
& \scriptsize\qquad\color{red}\Darr\bf\ ２次の係数が１以外！\\
y &= \colorbox{mistyrose}{$\displaystyle2$} x^2 +8 x +5\\
& \scriptsize\qquad\qquad\quad\color{red}\Darr\ +8 \div \left( 2 \right) = \colorbox{mistyrose}{$\displaystyle+4$}\\
&= \colorbox{white}{$2$} \left( x^2 \colorbox{mistyrose}{$\displaystyle+4$} x \right) +5\\
& \scriptsize\qquad\qquad\quad\color{orange}\Darr\ +4 \times \dfrac12 = \colorbox{bisque}{$\displaystyle+2$}\\
&= \colorbox{white}{$2$}\left\{ \left( x \colorbox{bisque}{$\displaystyle+2$} \right)^2 \colorbox{palegreen}{$\displaystyle - 4$} \right\} +5\\
& \scriptsize\qquad\qquad\qquad\qquad\quad\color{green}\Uarr\ 2乗した\ 4\ を引く\\
&= \colorbox{white}{$2$} \left( x +2 \right)^2 \colorbox{violet}{$\displaystyle-8$} +5\\
& \scriptsize\qquad\qquad\qquad\qquad\quad\color{magenta}\Uarr\ 2 \times \left( -4 \right)\\
&= \colorbox{white}{$2$} \left( x +2 \right)^2 -3\\
よって\\
& \scriptsize\color{deepskyblue}x + 2=0\ を解いて\ x = \colorbox{lightcyan}{$\displaystyle-2$}\\
& 軸\ x = \colorbox{lightcyan}{$\displaystyle-2$}\\
& 頂点\ \left( \colorbox{lightcyan}{$\displaystyle-2$},\ -3 \right)
\end{align*}
↓この問題へのリンクはこちら（右クリックで保存）
【解答】
\begin{align*}
& \scriptsize\qquad\color{red}\Darr\bf\ ２次の係数が１以外！\\
y &= \colorbox{mistyrose}{$\displaystyle-3$} x^2 +6 x +2\\
& \scriptsize\qquad\qquad\quad\color{red}\Darr\ +6 \div \left( -3 \right) = \colorbox{mistyrose}{$\displaystyle-2$}\\
&= \colorbox{white}{$-3$} \left( x^2 \colorbox{mistyrose}{$\displaystyle-2$} x \right) +2\\
& \scriptsize\qquad\qquad\quad\color{orange}\Darr\ -2 \times \dfrac12 = \colorbox{bisque}{$\displaystyle-1$}\\
&= \colorbox{white}{$-3$}\left\{ \left( x \colorbox{bisque}{$\displaystyle-1$} \right)^2 \colorbox{palegreen}{$\displaystyle - 1$} \right\} +2\\
& \scriptsize\qquad\qquad\qquad\qquad\quad\color{green}\Uarr\ 2乗した\ 1\ を引く\\
&= \colorbox{white}{$-3$} \left( x -1 \right)^2 \colorbox{violet}{$\displaystyle+3$} +2\\
& \scriptsize\qquad\qquad\qquad\qquad\quad\color{magenta}\Uarr\ -3 \times \left( -1 \right)\\
&= \colorbox{white}{$-3$} \left( x -1 \right)^2 +5\\
よって\\
& \scriptsize\color{deepskyblue}x - 1=0\ を解いて\ x = \colorbox{lightcyan}{$\displaystyle1$}\\
& 軸\ x = \colorbox{lightcyan}{$\displaystyle1$}\\
& 頂点\ \left( \colorbox{lightcyan}{$\displaystyle1$},\ 5 \right)
\end{align*}
↓この問題へのリンクはこちら（右クリックで保存）
【解答】
\begin{align*}
& \scriptsize\qquad\color{red}\Darr\bf\ ２次の係数が１以外！\\
y &= \colorbox{mistyrose}{$\displaystyle3$} x^2 +18 x +29\\
& \scriptsize\qquad\qquad\quad\color{red}\Darr\ +18 \div \left( 3 \right) = \colorbox{mistyrose}{$\displaystyle+6$}\\
&= \colorbox{white}{$3$} \left( x^2 \colorbox{mistyrose}{$\displaystyle+6$} x \right) +29\\
& \scriptsize\qquad\qquad\quad\color{orange}\Darr\ +6 \times \dfrac12 = \colorbox{bisque}{$\displaystyle+3$}\\
&= \colorbox{white}{$3$}\left\{ \left( x \colorbox{bisque}{$\displaystyle+3$} \right)^2 \colorbox{palegreen}{$\displaystyle - 9$} \right\} +29\\
& \scriptsize\qquad\qquad\qquad\qquad\quad\color{green}\Uarr\ 2乗した\ 9\ を引く\\
&= \colorbox{white}{$3$} \left( x +3 \right)^2 \colorbox{violet}{$\displaystyle-27$} +29\\
& \scriptsize\qquad\qquad\qquad\qquad\quad\color{magenta}\Uarr\ 3 \times \left( -9 \right)\\
&= \colorbox{white}{$3$} \left( x +3 \right)^2 +2\\
よって\\
& \scriptsize\color{deepskyblue}x + 3=0\ を解いて\ x = \colorbox{lightcyan}{$\displaystyle-3$}\\
& 軸\ x = \colorbox{lightcyan}{$\displaystyle-3$}\\
& 頂点\ \left( \colorbox{lightcyan}{$\displaystyle-3$},\ 2 \right)
\end{align*}
↓この問題へのリンクはこちら（右クリックで保存）
【解答】
\begin{align*}
& \scriptsize\qquad\color{red}\Darr\bf\ ２次の係数が１以外！\\
y &= \colorbox{mistyrose}{$\displaystyle-$} x^2 +4 x \color{lightgray}+0\\
& \scriptsize\qquad\qquad\quad\color{red}\Darr\ +4 \div \left( -1 \right) = \colorbox{mistyrose}{$\displaystyle-4$}\\
&= \colorbox{white}{$-$} \left( x^2 \colorbox{mistyrose}{$\displaystyle-4$} x \right) \color{lightgray}+0\\
& \scriptsize\qquad\qquad\quad\color{orange}\Darr\ -4 \times \dfrac12 = \colorbox{bisque}{$\displaystyle-2$}\\
&= \colorbox{white}{$-$}\left\{ \left( x \colorbox{bisque}{$\displaystyle-2$} \right)^2 \colorbox{palegreen}{$\displaystyle - 4$} \right\} \color{lightgray}+0\\
& \scriptsize\qquad\qquad\qquad\qquad\quad\color{green}\Uarr\ 2乗した\ 4\ を引く\\
&= \colorbox{white}{$-$} \left( x -2 \right)^2 \colorbox{violet}{$\displaystyle+4$} \color{lightgray}+0\\
& \scriptsize\qquad\qquad\qquad\qquad\quad\color{magenta}\Uarr\ -1 \times \left( -4 \right)\\
&= \colorbox{white}{$-$} \left( x -2 \right)^2 +4\\
よって\\
& \scriptsize\color{deepskyblue}x - 2=0\ を解いて\ x = \colorbox{lightcyan}{$\displaystyle2$}\\
& 軸\ x = \colorbox{lightcyan}{$\displaystyle2$}\\
& 頂点\ \left( \colorbox{lightcyan}{$\displaystyle2$},\ 4 \right)
\end{align*}
↓この問題へのリンクはこちら（右クリックで保存）
【解答】
\begin{align*}
& \scriptsize\qquad\color{red}\Darr\bf\ ２次の係数が１以外！\\
y &= \colorbox{mistyrose}{$\displaystyle-$} x^2 -4 x -3\\
& \scriptsize\qquad\qquad\quad\color{red}\Darr\ -4 \div \left( -1 \right) = \colorbox{mistyrose}{$\displaystyle+4$}\\
&= \colorbox{white}{$-$} \left( x^2 \colorbox{mistyrose}{$\displaystyle+4$} x \right) -3\\
& \scriptsize\qquad\qquad\quad\color{orange}\Darr\ +4 \times \dfrac12 = \colorbox{bisque}{$\displaystyle+2$}\\
&= \colorbox{white}{$-$}\left\{ \left( x \colorbox{bisque}{$\displaystyle+2$} \right)^2 \colorbox{palegreen}{$\displaystyle - 4$} \right\} -3\\
& \scriptsize\qquad\qquad\qquad\qquad\quad\color{green}\Uarr\ 2乗した\ 4\ を引く\\
&= \colorbox{white}{$-$} \left( x +2 \right)^2 \colorbox{violet}{$\displaystyle+4$} -3\\
& \scriptsize\qquad\qquad\qquad\qquad\quad\color{magenta}\Uarr\ -1 \times \left( -4 \right)\\
&= \colorbox{white}{$-$} \left( x +2 \right)^2 +1\\
よって\\
& \scriptsize\color{deepskyblue}x + 2=0\ を解いて\ x = \colorbox{lightcyan}{$\displaystyle-2$}\\
& 軸\ x = \colorbox{lightcyan}{$\displaystyle-2$}\\
& 頂点\ \left( \colorbox{lightcyan}{$\displaystyle-2$},\ 1 \right)
\end{align*}
↓この問題へのリンクはこちら（右クリックで保存）
【解答】
\begin{align*}
& \scriptsize\qquad\color{red}\Darr\bf\ ２次の係数が１以外！\\
y &= \colorbox{mistyrose}{$\displaystyle-$} x^2 -6 x -4\\
& \scriptsize\qquad\qquad\quad\color{red}\Darr\ -6 \div \left( -1 \right) = \colorbox{mistyrose}{$\displaystyle+6$}\\
&= \colorbox{white}{$-$} \left( x^2 \colorbox{mistyrose}{$\displaystyle+6$} x \right) -4\\
& \scriptsize\qquad\qquad\quad\color{orange}\Darr\ +6 \times \dfrac12 = \colorbox{bisque}{$\displaystyle+3$}\\
&= \colorbox{white}{$-$}\left\{ \left( x \colorbox{bisque}{$\displaystyle+3$} \right)^2 \colorbox{palegreen}{$\displaystyle - 9$} \right\} -4\\
& \scriptsize\qquad\qquad\qquad\qquad\quad\color{green}\Uarr\ 2乗した\ 9\ を引く\\
&= \colorbox{white}{$-$} \left( x +3 \right)^2 \colorbox{violet}{$\displaystyle+9$} -4\\
& \scriptsize\qquad\qquad\qquad\qquad\quad\color{magenta}\Uarr\ -1 \times \left( -9 \right)\\
&= \colorbox{white}{$-$} \left( x +3 \right)^2 +5\\
よって\\
& \scriptsize\color{deepskyblue}x + 3=0\ を解いて\ x = \colorbox{lightcyan}{$\displaystyle-3$}\\
& 軸\ x = \colorbox{lightcyan}{$\displaystyle-3$}\\
& 頂点\ \left( \colorbox{lightcyan}{$\displaystyle-3$},\ 5 \right)
\end{align*}
↓この問題へのリンクはこちら（右クリックで保存）
【解答】
\begin{align*}
& \scriptsize\qquad\color{red}\Darr\bf\ ２次の係数が１以外！\\
y &= \colorbox{mistyrose}{$\displaystyle2$} x^2 -8 x +3\\
& \scriptsize\qquad\qquad\quad\color{red}\Darr\ -8 \div \left( 2 \right) = \colorbox{mistyrose}{$\displaystyle-4$}\\
&= \colorbox{white}{$2$} \left( x^2 \colorbox{mistyrose}{$\displaystyle-4$} x \right) +3\\
& \scriptsize\qquad\qquad\quad\color{orange}\Darr\ -4 \times \dfrac12 = \colorbox{bisque}{$\displaystyle-2$}\\
&= \colorbox{white}{$2$}\left\{ \left( x \colorbox{bisque}{$\displaystyle-2$} \right)^2 \colorbox{palegreen}{$\displaystyle - 4$} \right\} +3\\
& \scriptsize\qquad\qquad\qquad\qquad\quad\color{green}\Uarr\ 2乗した\ 4\ を引く\\
&= \colorbox{white}{$2$} \left( x -2 \right)^2 \colorbox{violet}{$\displaystyle-8$} +3\\
& \scriptsize\qquad\qquad\qquad\qquad\quad\color{magenta}\Uarr\ 2 \times \left( -4 \right)\\
&= \colorbox{white}{$2$} \left( x -2 \right)^2 -5\\
よって\\
& \scriptsize\color{deepskyblue}x - 2=0\ を解いて\ x = \colorbox{lightcyan}{$\displaystyle2$}\\
& 軸\ x = \colorbox{lightcyan}{$\displaystyle2$}\\
& 頂点\ \left( \colorbox{lightcyan}{$\displaystyle2$},\ -5 \right)
\end{align*}
↓この問題へのリンクはこちら（右クリックで保存）
【解答】
\begin{align*}
& \scriptsize\qquad\color{red}\Darr\bf\ ２次の係数が１以外！\\
y &= \colorbox{mistyrose}{$\displaystyle-3$} x^2 +6 x \color{lightgray}+0\\
& \scriptsize\qquad\qquad\quad\color{red}\Darr\ +6 \div \left( -3 \right) = \colorbox{mistyrose}{$\displaystyle-2$}\\
&= \colorbox{white}{$-3$} \left( x^2 \colorbox{mistyrose}{$\displaystyle-2$} x \right) \color{lightgray}+0\\
& \scriptsize\qquad\qquad\quad\color{orange}\Darr\ -2 \times \dfrac12 = \colorbox{bisque}{$\displaystyle-1$}\\
&= \colorbox{white}{$-3$}\left\{ \left( x \colorbox{bisque}{$\displaystyle-1$} \right)^2 \colorbox{palegreen}{$\displaystyle - 1$} \right\} \color{lightgray}+0\\
& \scriptsize\qquad\qquad\qquad\qquad\quad\color{green}\Uarr\ 2乗した\ 1\ を引く\\
&= \colorbox{white}{$-3$} \left( x -1 \right)^2 \colorbox{violet}{$\displaystyle+3$} \color{lightgray}+0\\
& \scriptsize\qquad\qquad\qquad\qquad\quad\color{magenta}\Uarr\ -3 \times \left( -1 \right)\\
&= \colorbox{white}{$-3$} \left( x -1 \right)^2 +3\\
よって\\
& \scriptsize\color{deepskyblue}x - 1=0\ を解いて\ x = \colorbox{lightcyan}{$\displaystyle1$}\\
& 軸\ x = \colorbox{lightcyan}{$\displaystyle1$}\\
& 頂点\ \left( \colorbox{lightcyan}{$\displaystyle1$},\ 3 \right)
\end{align*}
↓この問題へのリンクはこちら（右クリックで保存）
【解答】
\begin{align*}
& \scriptsize\qquad\color{red}\Darr\bf\ ２次の係数が１以外！\\
y &= \colorbox{mistyrose}{$\displaystyle-2$} x^2 +12 x +2\\
& \scriptsize\qquad\qquad\quad\color{red}\Darr\ +12 \div \left( -2 \right) = \colorbox{mistyrose}{$\displaystyle-6$}\\
&= \colorbox{white}{$-2$} \left( x^2 \colorbox{mistyrose}{$\displaystyle-6$} x \right) +2\\
& \scriptsize\qquad\qquad\quad\color{orange}\Darr\ -6 \times \dfrac12 = \colorbox{bisque}{$\displaystyle-3$}\\
&= \colorbox{white}{$-2$}\left\{ \left( x \colorbox{bisque}{$\displaystyle-3$} \right)^2 \colorbox{palegreen}{$\displaystyle - 9$} \right\} +2\\
& \scriptsize\qquad\qquad\qquad\qquad\quad\color{green}\Uarr\ 2乗した\ 9\ を引く\\
&= \colorbox{white}{$-2$} \left( x -3 \right)^2 \colorbox{violet}{$\displaystyle+18$} +2\\
& \scriptsize\qquad\qquad\qquad\qquad\quad\color{magenta}\Uarr\ -2 \times \left( -9 \right)\\
&= \colorbox{white}{$-2$} \left( x -3 \right)^2 +20\\
よって\\
& \scriptsize\color{deepskyblue}x - 3=0\ を解いて\ x = \colorbox{lightcyan}{$\displaystyle3$}\\
& 軸\ x = \colorbox{lightcyan}{$\displaystyle3$}\\
& 頂点\ \left( \colorbox{lightcyan}{$\displaystyle3$},\ 20 \right)
\end{align*}
↓この問題へのリンクはこちら（右クリックで保存）
【解答】
\begin{align*}
& \scriptsize\qquad\color{red}\Darr\bf\ ２次の係数が１以外！\\
y &= \colorbox{mistyrose}{$\displaystyle-$} x^2 +6 x +5\\
& \scriptsize\qquad\qquad\quad\color{red}\Darr\ +6 \div \left( -1 \right) = \colorbox{mistyrose}{$\displaystyle-6$}\\
&= \colorbox{white}{$-$} \left( x^2 \colorbox{mistyrose}{$\displaystyle-6$} x \right) +5\\
& \scriptsize\qquad\qquad\quad\color{orange}\Darr\ -6 \times \dfrac12 = \colorbox{bisque}{$\displaystyle-3$}\\
&= \colorbox{white}{$-$}\left\{ \left( x \colorbox{bisque}{$\displaystyle-3$} \right)^2 \colorbox{palegreen}{$\displaystyle - 9$} \right\} +5\\
& \scriptsize\qquad\qquad\qquad\qquad\quad\color{green}\Uarr\ 2乗した\ 9\ を引く\\
&= \colorbox{white}{$-$} \left( x -3 \right)^2 \colorbox{violet}{$\displaystyle+9$} +5\\
& \scriptsize\qquad\qquad\qquad\qquad\quad\color{magenta}\Uarr\ -1 \times \left( -9 \right)\\
&= \colorbox{white}{$-$} \left( x -3 \right)^2 +14\\
よって\\
& \scriptsize\color{deepskyblue}x - 3=0\ を解いて\ x = \colorbox{lightcyan}{$\displaystyle3$}\\
& 軸\ x = \colorbox{lightcyan}{$\displaystyle3$}\\
& 頂点\ \left( \colorbox{lightcyan}{$\displaystyle3$},\ 14 \right)
\end{align*}
↓この問題へのリンクはこちら（右クリックで保存）
【解答】
\begin{align*}
& \scriptsize\qquad\color{red}\Darr\bf\ ２次の係数が１以外！\\
y &= \colorbox{mistyrose}{$\displaystyle2$} x^2 +4 x +1\\
& \scriptsize\qquad\qquad\quad\color{red}\Darr\ +4 \div \left( 2 \right) = \colorbox{mistyrose}{$\displaystyle+2$}\\
&= \colorbox{white}{$2$} \left( x^2 \colorbox{mistyrose}{$\displaystyle+2$} x \right) +1\\
& \scriptsize\qquad\qquad\quad\color{orange}\Darr\ +2 \times \dfrac12 = \colorbox{bisque}{$\displaystyle+1$}\\
&= \colorbox{white}{$2$}\left\{ \left( x \colorbox{bisque}{$\displaystyle+1$} \right)^2 \colorbox{palegreen}{$\displaystyle - 1$} \right\} +1\\
& \scriptsize\qquad\qquad\qquad\qquad\quad\color{green}\Uarr\ 2乗した\ 1\ を引く\\
&= \colorbox{white}{$2$} \left( x +1 \right)^2 \colorbox{violet}{$\displaystyle-2$} +1\\
& \scriptsize\qquad\qquad\qquad\qquad\quad\color{magenta}\Uarr\ 2 \times \left( -1 \right)\\
&= \colorbox{white}{$2$} \left( x +1 \right)^2 -1\\
よって\\
& \scriptsize\color{deepskyblue}x + 1=0\ を解いて\ x = \colorbox{lightcyan}{$\displaystyle-1$}\\
& 軸\ x = \colorbox{lightcyan}{$\displaystyle-1$}\\
& 頂点\ \left( \colorbox{lightcyan}{$\displaystyle-1$},\ -1 \right)
\end{align*}
↓この問題へのリンクはこちら（右クリックで保存）
【解答】
\begin{align*}
& \scriptsize\qquad\color{red}\Darr\bf\ ２次の係数が１以外！\\
y &= \colorbox{mistyrose}{$\displaystyle-$} x^2 -2 x +3\\
& \scriptsize\qquad\qquad\quad\color{red}\Darr\ -2 \div \left( -1 \right) = \colorbox{mistyrose}{$\displaystyle+2$}\\
&= \colorbox{white}{$-$} \left( x^2 \colorbox{mistyrose}{$\displaystyle+2$} x \right) +3\\
& \scriptsize\qquad\qquad\quad\color{orange}\Darr\ +2 \times \dfrac12 = \colorbox{bisque}{$\displaystyle+1$}\\
&= \colorbox{white}{$-$}\left\{ \left( x \colorbox{bisque}{$\displaystyle+1$} \right)^2 \colorbox{palegreen}{$\displaystyle - 1$} \right\} +3\\
& \scriptsize\qquad\qquad\qquad\qquad\quad\color{green}\Uarr\ 2乗した\ 1\ を引く\\
&= \colorbox{white}{$-$} \left( x +1 \right)^2 \colorbox{violet}{$\displaystyle+1$} +3\\
& \scriptsize\qquad\qquad\qquad\qquad\quad\color{magenta}\Uarr\ -1 \times \left( -1 \right)\\
&= \colorbox{white}{$-$} \left( x +1 \right)^2 +4\\
よって\\
& \scriptsize\color{deepskyblue}x + 1=0\ を解いて\ x = \colorbox{lightcyan}{$\displaystyle-1$}\\
& 軸\ x = \colorbox{lightcyan}{$\displaystyle-1$}\\
& 頂点\ \left( \colorbox{lightcyan}{$\displaystyle-1$},\ 4 \right)
\end{align*}
↓この問題へのリンクはこちら（右クリックで保存）
【解答】
\begin{align*}
& \scriptsize\qquad\color{red}\Darr\bf\ ２次の係数が１以外！\\
y &= \colorbox{mistyrose}{$\displaystyle-$} x^2 +8 x \color{lightgray}+0\\
& \scriptsize\qquad\qquad\quad\color{red}\Darr\ +8 \div \left( -1 \right) = \colorbox{mistyrose}{$\displaystyle-8$}\\
&= \colorbox{white}{$-$} \left( x^2 \colorbox{mistyrose}{$\displaystyle-8$} x \right) \color{lightgray}+0\\
& \scriptsize\qquad\qquad\quad\color{orange}\Darr\ -8 \times \dfrac12 = \colorbox{bisque}{$\displaystyle-4$}\\
&= \colorbox{white}{$-$}\left\{ \left( x \colorbox{bisque}{$\displaystyle-4$} \right)^2 \colorbox{palegreen}{$\displaystyle - 16$} \right\} \color{lightgray}+0\\
& \scriptsize\qquad\qquad\qquad\qquad\quad\color{green}\Uarr\ 2乗した\ 16\ を引く\\
&= \colorbox{white}{$-$} \left( x -4 \right)^2 \colorbox{violet}{$\displaystyle+16$} \color{lightgray}+0\\
& \scriptsize\qquad\qquad\qquad\qquad\quad\color{magenta}\Uarr\ -1 \times \left( -16 \right)\\
&= \colorbox{white}{$-$} \left( x -4 \right)^2 +16\\
よって\\
& \scriptsize\color{deepskyblue}x - 4=0\ を解いて\ x = \colorbox{lightcyan}{$\displaystyle4$}\\
& 軸\ x = \colorbox{lightcyan}{$\displaystyle4$}\\
& 頂点\ \left( \colorbox{lightcyan}{$\displaystyle4$},\ 16 \right)
\end{align*}
↓この問題へのリンクはこちら（右クリックで保存）
【解答】
\begin{align*}
& \scriptsize\qquad\color{red}\Darr\bf\ ２次の係数が１以外！\\
y &= \colorbox{mistyrose}{$\displaystyle2$} x^2 -12 x +19\\
& \scriptsize\qquad\qquad\quad\color{red}\Darr\ -12 \div \left( 2 \right) = \colorbox{mistyrose}{$\displaystyle-6$}\\
&= \colorbox{white}{$2$} \left( x^2 \colorbox{mistyrose}{$\displaystyle-6$} x \right) +19\\
& \scriptsize\qquad\qquad\quad\color{orange}\Darr\ -6 \times \dfrac12 = \colorbox{bisque}{$\displaystyle-3$}\\
&= \colorbox{white}{$2$}\left\{ \left( x \colorbox{bisque}{$\displaystyle-3$} \right)^2 \colorbox{palegreen}{$\displaystyle - 9$} \right\} +19\\
& \scriptsize\qquad\qquad\qquad\qquad\quad\color{green}\Uarr\ 2乗した\ 9\ を引く\\
&= \colorbox{white}{$2$} \left( x -3 \right)^2 \colorbox{violet}{$\displaystyle-18$} +19\\
& \scriptsize\qquad\qquad\qquad\qquad\quad\color{magenta}\Uarr\ 2 \times \left( -9 \right)\\
&= \colorbox{white}{$2$} \left( x -3 \right)^2 +1\\
よって\\
& \scriptsize\color{deepskyblue}x - 3=0\ を解いて\ x = \colorbox{lightcyan}{$\displaystyle3$}\\
& 軸\ x = \colorbox{lightcyan}{$\displaystyle3$}\\
& 頂点\ \left( \colorbox{lightcyan}{$\displaystyle3$},\ 1 \right)
\end{align*}
↓この問題へのリンクはこちら（右クリックで保存）
【解答】
\begin{align*}
& \scriptsize\qquad\color{red}\Darr\bf\ ２次の係数が１以外！\\
y &= \colorbox{mistyrose}{$\displaystyle2$} x^2 -8 x +4\\
& \scriptsize\qquad\qquad\quad\color{red}\Darr\ -8 \div \left( 2 \right) = \colorbox{mistyrose}{$\displaystyle-4$}\\
&= \colorbox{white}{$2$} \left( x^2 \colorbox{mistyrose}{$\displaystyle-4$} x \right) +4\\
& \scriptsize\qquad\qquad\quad\color{orange}\Darr\ -4 \times \dfrac12 = \colorbox{bisque}{$\displaystyle-2$}\\
&= \colorbox{white}{$2$}\left\{ \left( x \colorbox{bisque}{$\displaystyle-2$} \right)^2 \colorbox{palegreen}{$\displaystyle - 4$} \right\} +4\\
& \scriptsize\qquad\qquad\qquad\qquad\quad\color{green}\Uarr\ 2乗した\ 4\ を引く\\
&= \colorbox{white}{$2$} \left( x -2 \right)^2 \colorbox{violet}{$\displaystyle-8$} +4\\
& \scriptsize\qquad\qquad\qquad\qquad\quad\color{magenta}\Uarr\ 2 \times \left( -4 \right)\\
&= \colorbox{white}{$2$} \left( x -2 \right)^2 -4\\
よって\\
& \scriptsize\color{deepskyblue}x - 2=0\ を解いて\ x = \colorbox{lightcyan}{$\displaystyle2$}\\
& 軸\ x = \colorbox{lightcyan}{$\displaystyle2$}\\
& 頂点\ \left( \colorbox{lightcyan}{$\displaystyle2$},\ -4 \right)
\end{align*}
↓この問題へのリンクはこちら（右クリックで保存）
【解答】
\begin{align*}
& \scriptsize\qquad\color{red}\Darr\bf\ ２次の係数が１以外！\\
y &= \colorbox{mistyrose}{$\displaystyle-2$} x^2 -4 x \color{lightgray}+0\\
& \scriptsize\qquad\qquad\quad\color{red}\Darr\ -4 \div \left( -2 \right) = \colorbox{mistyrose}{$\displaystyle+2$}\\
&= \colorbox{white}{$-2$} \left( x^2 \colorbox{mistyrose}{$\displaystyle+2$} x \right) \color{lightgray}+0\\
& \scriptsize\qquad\qquad\quad\color{orange}\Darr\ +2 \times \dfrac12 = \colorbox{bisque}{$\displaystyle+1$}\\
&= \colorbox{white}{$-2$}\left\{ \left( x \colorbox{bisque}{$\displaystyle+1$} \right)^2 \colorbox{palegreen}{$\displaystyle - 1$} \right\} \color{lightgray}+0\\
& \scriptsize\qquad\qquad\qquad\qquad\quad\color{green}\Uarr\ 2乗した\ 1\ を引く\\
&= \colorbox{white}{$-2$} \left( x +1 \right)^2 \colorbox{violet}{$\displaystyle+2$} \color{lightgray}+0\\
& \scriptsize\qquad\qquad\qquad\qquad\quad\color{magenta}\Uarr\ -2 \times \left( -1 \right)\\
&= \colorbox{white}{$-2$} \left( x +1 \right)^2 +2\\
よって\\
& \scriptsize\color{deepskyblue}x + 1=0\ を解いて\ x = \colorbox{lightcyan}{$\displaystyle-1$}\\
& 軸\ x = \colorbox{lightcyan}{$\displaystyle-1$}\\
& 頂点\ \left( \colorbox{lightcyan}{$\displaystyle-1$},\ 2 \right)
\end{align*}
↓この問題へのリンクはこちら（右クリックで保存）
【解答】
\begin{align*}
& \scriptsize\qquad\color{red}\Darr\bf\ ２次の係数が１以外！\\
y &= \colorbox{mistyrose}{$\displaystyle2$} x^2 +4 x +5\\
& \scriptsize\qquad\qquad\quad\color{red}\Darr\ +4 \div \left( 2 \right) = \colorbox{mistyrose}{$\displaystyle+2$}\\
&= \colorbox{white}{$2$} \left( x^2 \colorbox{mistyrose}{$\displaystyle+2$} x \right) +5\\
& \scriptsize\qquad\qquad\quad\color{orange}\Darr\ +2 \times \dfrac12 = \colorbox{bisque}{$\displaystyle+1$}\\
&= \colorbox{white}{$2$}\left\{ \left( x \colorbox{bisque}{$\displaystyle+1$} \right)^2 \colorbox{palegreen}{$\displaystyle - 1$} \right\} +5\\
& \scriptsize\qquad\qquad\qquad\qquad\quad\color{green}\Uarr\ 2乗した\ 1\ を引く\\
&= \colorbox{white}{$2$} \left( x +1 \right)^2 \colorbox{violet}{$\displaystyle-2$} +5\\
& \scriptsize\qquad\qquad\qquad\qquad\quad\color{magenta}\Uarr\ 2 \times \left( -1 \right)\\
&= \colorbox{white}{$2$} \left( x +1 \right)^2 +3\\
よって\\
& \scriptsize\color{deepskyblue}x + 1=0\ を解いて\ x = \colorbox{lightcyan}{$\displaystyle-1$}\\
& 軸\ x = \colorbox{lightcyan}{$\displaystyle-1$}\\
& 頂点\ \left( \colorbox{lightcyan}{$\displaystyle-1$},\ 3 \right)
\end{align*}
↓この問題へのリンクはこちら（右クリックで保存）
【解答】
\begin{align*}
& \scriptsize\qquad\color{red}\Darr\bf\ ２次の係数が１以外！\\
y &= \colorbox{mistyrose}{$\displaystyle3$} x^2 +6 x +2\\
& \scriptsize\qquad\qquad\quad\color{red}\Darr\ +6 \div \left( 3 \right) = \colorbox{mistyrose}{$\displaystyle+2$}\\
&= \colorbox{white}{$3$} \left( x^2 \colorbox{mistyrose}{$\displaystyle+2$} x \right) +2\\
& \scriptsize\qquad\qquad\quad\color{orange}\Darr\ +2 \times \dfrac12 = \colorbox{bisque}{$\displaystyle+1$}\\
&= \colorbox{white}{$3$}\left\{ \left( x \colorbox{bisque}{$\displaystyle+1$} \right)^2 \colorbox{palegreen}{$\displaystyle - 1$} \right\} +2\\
& \scriptsize\qquad\qquad\qquad\qquad\quad\color{green}\Uarr\ 2乗した\ 1\ を引く\\
&= \colorbox{white}{$3$} \left( x +1 \right)^2 \colorbox{violet}{$\displaystyle-3$} +2\\
& \scriptsize\qquad\qquad\qquad\qquad\quad\color{magenta}\Uarr\ 3 \times \left( -1 \right)\\
&= \colorbox{white}{$3$} \left( x +1 \right)^2 -1\\
よって\\
& \scriptsize\color{deepskyblue}x + 1=0\ を解いて\ x = \colorbox{lightcyan}{$\displaystyle-1$}\\
& 軸\ x = \colorbox{lightcyan}{$\displaystyle-1$}\\
& 頂点\ \left( \colorbox{lightcyan}{$\displaystyle-1$},\ -1 \right)
\end{align*}
↓この問題へのリンクはこちら（右クリックで保存）
【解答】
\begin{align*}
& \scriptsize\qquad\color{red}\Darr\bf\ ２次の係数が１以外！\\
y &= \colorbox{mistyrose}{$\displaystyle2$} x^2 -8 x -3\\
& \scriptsize\qquad\qquad\quad\color{red}\Darr\ -8 \div \left( 2 \right) = \colorbox{mistyrose}{$\displaystyle-4$}\\
&= \colorbox{white}{$2$} \left( x^2 \colorbox{mistyrose}{$\displaystyle-4$} x \right) -3\\
& \scriptsize\qquad\qquad\quad\color{orange}\Darr\ -4 \times \dfrac12 = \colorbox{bisque}{$\displaystyle-2$}\\
&= \colorbox{white}{$2$}\left\{ \left( x \colorbox{bisque}{$\displaystyle-2$} \right)^2 \colorbox{palegreen}{$\displaystyle - 4$} \right\} -3\\
& \scriptsize\qquad\qquad\qquad\qquad\quad\color{green}\Uarr\ 2乗した\ 4\ を引く\\
&= \colorbox{white}{$2$} \left( x -2 \right)^2 \colorbox{violet}{$\displaystyle-8$} -3\\
& \scriptsize\qquad\qquad\qquad\qquad\quad\color{magenta}\Uarr\ 2 \times \left( -4 \right)\\
&= \colorbox{white}{$2$} \left( x -2 \right)^2 -11\\
よって\\
& \scriptsize\color{deepskyblue}x - 2=0\ を解いて\ x = \colorbox{lightcyan}{$\displaystyle2$}\\
& 軸\ x = \colorbox{lightcyan}{$\displaystyle2$}\\
& 頂点\ \left( \colorbox{lightcyan}{$\displaystyle2$},\ -11 \right)
\end{align*}
↓この問題へのリンクはこちら（右クリックで保存）
【解答】
\begin{align*}
& \scriptsize\qquad\color{red}\Darr\bf\ ２次の係数が１以外！\\
y &= \colorbox{mistyrose}{$\displaystyle-2$} x^2 -4 x +1\\
& \scriptsize\qquad\qquad\quad\color{red}\Darr\ -4 \div \left( -2 \right) = \colorbox{mistyrose}{$\displaystyle+2$}\\
&= \colorbox{white}{$-2$} \left( x^2 \colorbox{mistyrose}{$\displaystyle+2$} x \right) +1\\
& \scriptsize\qquad\qquad\quad\color{orange}\Darr\ +2 \times \dfrac12 = \colorbox{bisque}{$\displaystyle+1$}\\
&= \colorbox{white}{$-2$}\left\{ \left( x \colorbox{bisque}{$\displaystyle+1$} \right)^2 \colorbox{palegreen}{$\displaystyle - 1$} \right\} +1\\
& \scriptsize\qquad\qquad\qquad\qquad\quad\color{green}\Uarr\ 2乗した\ 1\ を引く\\
&= \colorbox{white}{$-2$} \left( x +1 \right)^2 \colorbox{violet}{$\displaystyle+2$} +1\\
& \scriptsize\qquad\qquad\qquad\qquad\quad\color{magenta}\Uarr\ -2 \times \left( -1 \right)\\
&= \colorbox{white}{$-2$} \left( x +1 \right)^2 +3\\
よって\\
& \scriptsize\color{deepskyblue}x + 1=0\ を解いて\ x = \colorbox{lightcyan}{$\displaystyle-1$}\\
& 軸\ x = \colorbox{lightcyan}{$\displaystyle-1$}\\
& 頂点\ \left( \colorbox{lightcyan}{$\displaystyle-1$},\ 3 \right)
\end{align*}
↓この問題へのリンクはこちら（右クリックで保存）
【解答】
\begin{align*}
& \scriptsize\qquad\color{red}\Darr\bf\ ２次の係数が１以外！\\
y &= \colorbox{mistyrose}{$\displaystyle2$} x^2 +8 x +3\\
& \scriptsize\qquad\qquad\quad\color{red}\Darr\ +8 \div \left( 2 \right) = \colorbox{mistyrose}{$\displaystyle+4$}\\
&= \colorbox{white}{$2$} \left( x^2 \colorbox{mistyrose}{$\displaystyle+4$} x \right) +3\\
& \scriptsize\qquad\qquad\quad\color{orange}\Darr\ +4 \times \dfrac12 = \colorbox{bisque}{$\displaystyle+2$}\\
&= \colorbox{white}{$2$}\left\{ \left( x \colorbox{bisque}{$\displaystyle+2$} \right)^2 \colorbox{palegreen}{$\displaystyle - 4$} \right\} +3\\
& \scriptsize\qquad\qquad\qquad\qquad\quad\color{green}\Uarr\ 2乗した\ 4\ を引く\\
&= \colorbox{white}{$2$} \left( x +2 \right)^2 \colorbox{violet}{$\displaystyle-8$} +3\\
& \scriptsize\qquad\qquad\qquad\qquad\quad\color{magenta}\Uarr\ 2 \times \left( -4 \right)\\
&= \colorbox{white}{$2$} \left( x +2 \right)^2 -5\\
よって\\
& \scriptsize\color{deepskyblue}x + 2=0\ を解いて\ x = \colorbox{lightcyan}{$\displaystyle-2$}\\
& 軸\ x = \colorbox{lightcyan}{$\displaystyle-2$}\\
& 頂点\ \left( \colorbox{lightcyan}{$\displaystyle-2$},\ -5 \right)
\end{align*}
↓この問題へのリンクはこちら（右クリックで保存）
【解答】
\begin{align*}
& \scriptsize\qquad\color{red}\Darr\bf\ ２次の係数が１以外！\\
y &= \colorbox{mistyrose}{$\displaystyle-3$} x^2 +6 x +1\\
& \scriptsize\qquad\qquad\quad\color{red}\Darr\ +6 \div \left( -3 \right) = \colorbox{mistyrose}{$\displaystyle-2$}\\
&= \colorbox{white}{$-3$} \left( x^2 \colorbox{mistyrose}{$\displaystyle-2$} x \right) +1\\
& \scriptsize\qquad\qquad\quad\color{orange}\Darr\ -2 \times \dfrac12 = \colorbox{bisque}{$\displaystyle-1$}\\
&= \colorbox{white}{$-3$}\left\{ \left( x \colorbox{bisque}{$\displaystyle-1$} \right)^2 \colorbox{palegreen}{$\displaystyle - 1$} \right\} +1\\
& \scriptsize\qquad\qquad\qquad\qquad\quad\color{green}\Uarr\ 2乗した\ 1\ を引く\\
&= \colorbox{white}{$-3$} \left( x -1 \right)^2 \colorbox{violet}{$\displaystyle+3$} +1\\
& \scriptsize\qquad\qquad\qquad\qquad\quad\color{magenta}\Uarr\ -3 \times \left( -1 \right)\\
&= \colorbox{white}{$-3$} \left( x -1 \right)^2 +4\\
よって\\
& \scriptsize\color{deepskyblue}x - 1=0\ を解いて\ x = \colorbox{lightcyan}{$\displaystyle1$}\\
& 軸\ x = \colorbox{lightcyan}{$\displaystyle1$}\\
& 頂点\ \left( \colorbox{lightcyan}{$\displaystyle1$},\ 4 \right)
\end{align*}
↓この問題へのリンクはこちら（右クリックで保存）
【解答】
\begin{align*}
& \scriptsize\qquad\color{red}\Darr\bf\ ２次の係数が１以外！\\
y &= \colorbox{mistyrose}{$\displaystyle-$} x^2 -4 x +2\\
& \scriptsize\qquad\qquad\quad\color{red}\Darr\ -4 \div \left( -1 \right) = \colorbox{mistyrose}{$\displaystyle+4$}\\
&= \colorbox{white}{$-$} \left( x^2 \colorbox{mistyrose}{$\displaystyle+4$} x \right) +2\\
& \scriptsize\qquad\qquad\quad\color{orange}\Darr\ +4 \times \dfrac12 = \colorbox{bisque}{$\displaystyle+2$}\\
&= \colorbox{white}{$-$}\left\{ \left( x \colorbox{bisque}{$\displaystyle+2$} \right)^2 \colorbox{palegreen}{$\displaystyle - 4$} \right\} +2\\
& \scriptsize\qquad\qquad\qquad\qquad\quad\color{green}\Uarr\ 2乗した\ 4\ を引く\\
&= \colorbox{white}{$-$} \left( x +2 \right)^2 \colorbox{violet}{$\displaystyle+4$} +2\\
& \scriptsize\qquad\qquad\qquad\qquad\quad\color{magenta}\Uarr\ -1 \times \left( -4 \right)\\
&= \colorbox{white}{$-$} \left( x +2 \right)^2 +6\\
よって\\
& \scriptsize\color{deepskyblue}x + 2=0\ を解いて\ x = \colorbox{lightcyan}{$\displaystyle-2$}\\
& 軸\ x = \colorbox{lightcyan}{$\displaystyle-2$}\\
& 頂点\ \left( \colorbox{lightcyan}{$\displaystyle-2$},\ 6 \right)
\end{align*}
↓この問題へのリンクはこちら（右クリックで保存）
【解答】
\begin{align*}
& \scriptsize\qquad\color{red}\Darr\bf\ ２次の係数が１以外！\\
y &= \colorbox{mistyrose}{$\displaystyle2$} x^2 -4 x \color{lightgray}+0\\
& \scriptsize\qquad\qquad\quad\color{red}\Darr\ -4 \div \left( 2 \right) = \colorbox{mistyrose}{$\displaystyle-2$}\\
&= \colorbox{white}{$2$} \left( x^2 \colorbox{mistyrose}{$\displaystyle-2$} x \right) \color{lightgray}+0\\
& \scriptsize\qquad\qquad\quad\color{orange}\Darr\ -2 \times \dfrac12 = \colorbox{bisque}{$\displaystyle-1$}\\
&= \colorbox{white}{$2$}\left\{ \left( x \colorbox{bisque}{$\displaystyle-1$} \right)^2 \colorbox{palegreen}{$\displaystyle - 1$} \right\} \color{lightgray}+0\\
& \scriptsize\qquad\qquad\qquad\qquad\quad\color{green}\Uarr\ 2乗した\ 1\ を引く\\
&= \colorbox{white}{$2$} \left( x -1 \right)^2 \colorbox{violet}{$\displaystyle-2$} \color{lightgray}+0\\
& \scriptsize\qquad\qquad\qquad\qquad\quad\color{magenta}\Uarr\ 2 \times \left( -1 \right)\\
&= \colorbox{white}{$2$} \left( x -1 \right)^2 -2\\
よって\\
& \scriptsize\color{deepskyblue}x - 1=0\ を解いて\ x = \colorbox{lightcyan}{$\displaystyle1$}\\
& 軸\ x = \colorbox{lightcyan}{$\displaystyle1$}\\
& 頂点\ \left( \colorbox{lightcyan}{$\displaystyle1$},\ -2 \right)
\end{align*}
↓この問題へのリンクはこちら（右クリックで保存）
【解答】
\begin{align*}
& \scriptsize\qquad\color{red}\Darr\bf\ ２次の係数が１以外！\\
y &= \colorbox{mistyrose}{$\displaystyle2$} x^2 +4 x -3\\
& \scriptsize\qquad\qquad\quad\color{red}\Darr\ +4 \div \left( 2 \right) = \colorbox{mistyrose}{$\displaystyle+2$}\\
&= \colorbox{white}{$2$} \left( x^2 \colorbox{mistyrose}{$\displaystyle+2$} x \right) -3\\
& \scriptsize\qquad\qquad\quad\color{orange}\Darr\ +2 \times \dfrac12 = \colorbox{bisque}{$\displaystyle+1$}\\
&= \colorbox{white}{$2$}\left\{ \left( x \colorbox{bisque}{$\displaystyle+1$} \right)^2 \colorbox{palegreen}{$\displaystyle - 1$} \right\} -3\\
& \scriptsize\qquad\qquad\qquad\qquad\quad\color{green}\Uarr\ 2乗した\ 1\ を引く\\
&= \colorbox{white}{$2$} \left( x +1 \right)^2 \colorbox{violet}{$\displaystyle-2$} -3\\
& \scriptsize\qquad\qquad\qquad\qquad\quad\color{magenta}\Uarr\ 2 \times \left( -1 \right)\\
&= \colorbox{white}{$2$} \left( x +1 \right)^2 -5\\
よって\\
& \scriptsize\color{deepskyblue}x + 1=0\ を解いて\ x = \colorbox{lightcyan}{$\displaystyle-1$}\\
& 軸\ x = \colorbox{lightcyan}{$\displaystyle-1$}\\
& 頂点\ \left( \colorbox{lightcyan}{$\displaystyle-1$},\ -5 \right)
\end{align*}
↓この問題へのリンクはこちら（右クリックで保存）
【解答】
\begin{align*}
& \scriptsize\qquad\color{red}\Darr\bf\ ２次の係数が１以外！\\
y &= \colorbox{mistyrose}{$\displaystyle2$} x^2 -4 x +2\\
& \scriptsize\qquad\qquad\quad\color{red}\Darr\ -4 \div \left( 2 \right) = \colorbox{mistyrose}{$\displaystyle-2$}\\
&= \colorbox{white}{$2$} \left( x^2 \colorbox{mistyrose}{$\displaystyle-2$} x \right) +2\\
& \scriptsize\qquad\qquad\quad\color{orange}\Darr\ -2 \times \dfrac12 = \colorbox{bisque}{$\displaystyle-1$}\\
&= \colorbox{white}{$2$}\left\{ \left( x \colorbox{bisque}{$\displaystyle-1$} \right)^2 \colorbox{palegreen}{$\displaystyle - 1$} \right\} +2\\
& \scriptsize\qquad\qquad\qquad\qquad\quad\color{green}\Uarr\ 2乗した\ 1\ を引く\\
&= \colorbox{white}{$2$} \left( x -1 \right)^2 \colorbox{violet}{$\displaystyle-2$} +2\\
& \scriptsize\qquad\qquad\qquad\qquad\quad\color{magenta}\Uarr\ 2 \times \left( -1 \right)\\
&= \colorbox{white}{$2$} \left( x -1 \right)^2 \\
よって\\
& \scriptsize\color{deepskyblue}x - 1=0\ を解いて\ x = \colorbox{lightcyan}{$\displaystyle1$}\\
& 軸\ x = \colorbox{lightcyan}{$\displaystyle1$}\\
& 頂点\ \left( \colorbox{lightcyan}{$\displaystyle1$},\ 0 \right)
\end{align*}
↓この問題へのリンクはこちら（右クリックで保存）
【解答】
\begin{align*}
& \scriptsize\qquad\color{red}\Darr\bf\ ２次の係数が１以外！\\
y &= \colorbox{mistyrose}{$\displaystyle2$} x^2 +4 x -1\\
& \scriptsize\qquad\qquad\quad\color{red}\Darr\ +4 \div \left( 2 \right) = \colorbox{mistyrose}{$\displaystyle+2$}\\
&= \colorbox{white}{$2$} \left( x^2 \colorbox{mistyrose}{$\displaystyle+2$} x \right) -1\\
& \scriptsize\qquad\qquad\quad\color{orange}\Darr\ +2 \times \dfrac12 = \colorbox{bisque}{$\displaystyle+1$}\\
&= \colorbox{white}{$2$}\left\{ \left( x \colorbox{bisque}{$\displaystyle+1$} \right)^2 \colorbox{palegreen}{$\displaystyle - 1$} \right\} -1\\
& \scriptsize\qquad\qquad\qquad\qquad\quad\color{green}\Uarr\ 2乗した\ 1\ を引く\\
&= \colorbox{white}{$2$} \left( x +1 \right)^2 \colorbox{violet}{$\displaystyle-2$} -1\\
& \scriptsize\qquad\qquad\qquad\qquad\quad\color{magenta}\Uarr\ 2 \times \left( -1 \right)\\
&= \colorbox{white}{$2$} \left( x +1 \right)^2 -3\\
よって\\
& \scriptsize\color{deepskyblue}x + 1=0\ を解いて\ x = \colorbox{lightcyan}{$\displaystyle-1$}\\
& 軸\ x = \colorbox{lightcyan}{$\displaystyle-1$}\\
& 頂点\ \left( \colorbox{lightcyan}{$\displaystyle-1$},\ -3 \right)
\end{align*}
↓この問題へのリンクはこちら（右クリックで保存）
【解答】
\begin{align*}
& \scriptsize\qquad\color{red}\Darr\bf\ ２次の係数が１以外！\\
y &= \colorbox{mistyrose}{$\displaystyle-2$} x^2 +8 x \color{lightgray}+0\\
& \scriptsize\qquad\qquad\quad\color{red}\Darr\ +8 \div \left( -2 \right) = \colorbox{mistyrose}{$\displaystyle-4$}\\
&= \colorbox{white}{$-2$} \left( x^2 \colorbox{mistyrose}{$\displaystyle-4$} x \right) \color{lightgray}+0\\
& \scriptsize\qquad\qquad\quad\color{orange}\Darr\ -4 \times \dfrac12 = \colorbox{bisque}{$\displaystyle-2$}\\
&= \colorbox{white}{$-2$}\left\{ \left( x \colorbox{bisque}{$\displaystyle-2$} \right)^2 \colorbox{palegreen}{$\displaystyle - 4$} \right\} \color{lightgray}+0\\
& \scriptsize\qquad\qquad\qquad\qquad\quad\color{green}\Uarr\ 2乗した\ 4\ を引く\\
&= \colorbox{white}{$-2$} \left( x -2 \right)^2 \colorbox{violet}{$\displaystyle+8$} \color{lightgray}+0\\
& \scriptsize\qquad\qquad\qquad\qquad\quad\color{magenta}\Uarr\ -2 \times \left( -4 \right)\\
&= \colorbox{white}{$-2$} \left( x -2 \right)^2 +8\\
よって\\
& \scriptsize\color{deepskyblue}x - 2=0\ を解いて\ x = \colorbox{lightcyan}{$\displaystyle2$}\\
& 軸\ x = \colorbox{lightcyan}{$\displaystyle2$}\\
& 頂点\ \left( \colorbox{lightcyan}{$\displaystyle2$},\ 8 \right)
\end{align*}
↓この問題へのリンクはこちら（右クリックで保存）
【解答】
\begin{align*}
& \scriptsize\qquad\color{red}\Darr\bf\ ２次の係数が１以外！\\
y &= \colorbox{mistyrose}{$\displaystyle-2$} x^2 +12 x -10\\
& \scriptsize\qquad\qquad\quad\color{red}\Darr\ +12 \div \left( -2 \right) = \colorbox{mistyrose}{$\displaystyle-6$}\\
&= \colorbox{white}{$-2$} \left( x^2 \colorbox{mistyrose}{$\displaystyle-6$} x \right) -10\\
& \scriptsize\qquad\qquad\quad\color{orange}\Darr\ -6 \times \dfrac12 = \colorbox{bisque}{$\displaystyle-3$}\\
&= \colorbox{white}{$-2$}\left\{ \left( x \colorbox{bisque}{$\displaystyle-3$} \right)^2 \colorbox{palegreen}{$\displaystyle - 9$} \right\} -10\\
& \scriptsize\qquad\qquad\qquad\qquad\quad\color{green}\Uarr\ 2乗した\ 9\ を引く\\
&= \colorbox{white}{$-2$} \left( x -3 \right)^2 \colorbox{violet}{$\displaystyle+18$} -10\\
& \scriptsize\qquad\qquad\qquad\qquad\quad\color{magenta}\Uarr\ -2 \times \left( -9 \right)\\
&= \colorbox{white}{$-2$} \left( x -3 \right)^2 +8\\
よって\\
& \scriptsize\color{deepskyblue}x - 3=0\ を解いて\ x = \colorbox{lightcyan}{$\displaystyle3$}\\
& 軸\ x = \colorbox{lightcyan}{$\displaystyle3$}\\
& 頂点\ \left( \colorbox{lightcyan}{$\displaystyle3$},\ 8 \right)
\end{align*}
