
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*}