平方完成とy=ax²+bx+cのグラフ

数研出版｜数学Ⅰ[104-901] p.86 練習12

\({\small (1)}~\)
\(\begin{eqnarray}~~~y&=&x^2-6x+4\\[3pt]~~~&=&(x^2-6x+9)-9+4\\[3pt]~~~&=&(x-3)^2-5\end{eqnarray}\)

---

よって、頂点 \((3~,~-5)\) 、軸の方程式 \(x=3\)

---

\(y\) 切片が \(4\) 、下に凸のグラフより、

---

　\(\begin{array}{c|ccccc}
x & \cdots & 0 & \cdots & 3 & \cdots \\
\hline
y & \searrow & 4 & \searrow & -5 & \nearrow
\end{array}\)

---

 
 

\({\small (2)}~\)
\(\begin{eqnarray}~y&=&2x^2+4x+3\\[3pt]~&=&2(x^2+2x)+3\\[3pt]~~~&=&2(x^2+2x+1-1)+3\\[3pt]~~~&=&2(x^2+2x+1)+2 \cdot (-1)+3\\[3pt]~~~&=&2(x+1)^2-2+3\\[3pt]~~~&=&2(x+1)^2+1\end{eqnarray}\)

---

よって、頂点 \((-1~,~1)\) 、軸の方程式 \(x=-1\)

---

\(y\) 切片が \(3\) 、下に凸のグラフより、

---

　\(\begin{array}{c|ccccc}
x & \cdots & -1 & \cdots & 0 & \cdots \\
\hline
y & \searrow & 1 & \nearrow & 3 & \nearrow
\end{array}\)

---

 
 

\({\small (3)}~\)
\(\begin{eqnarray}~y&=&-2x^2+8x-4\\[3pt]~&=&-2(x^2-4x)-4\\[3pt]~~~&=&-2(x^2-4x+4-4)-4\\[3pt]~~~&=&-2(x^2-4x+4)-2 \cdot (-4)-4\\[3pt]~~~&=&-2(x-2)^2+8-4\\[3pt]~~~&=&-2(x-2)^2+4\end{eqnarray}\)

---

よって、頂点 \((2~,~4)\) 、軸の方程式 \(x=2\)

---

\(y\) 切片が \(-4\) 、上に凸のグラフより、

---

　\(\begin{array}{c|ccccc}
x & \cdots & 0 & \cdots & 2 & \cdots \\
\hline
y & \nearrow & -4 & \nearrow & 4 & \searrow
\end{array}\)

---

 
 

\({\small (4)}~\)
\(\begin{eqnarray}~y&=&-3x^2-6x-5\\[3pt]~&=&-3(x^2+2x)-5\\[3pt]~~~&=&-3(x^2+2x+1-1)-5\\[3pt]~~~&=&-3(x^2+2x+1)-3 \cdot (-1)-5\\[3pt]~~~&=&-3(x+1)^2+3-5\\[3pt]~~~&=&-3(x+1)^2-2\end{eqnarray}\)

---

よって、頂点 \((-1~,~-2)\) 、軸の方程式 \(x=-1\)

---

\(y\) 切片が \(-5\) 、上に凸のグラフより、

---

　\(\begin{array}{c|ccccc}
x & \cdots & -1 & \cdots & 0 & \cdots \\
\hline
y & \nearrow & -2 & \searrow & -5 & \searrow
\end{array}\)

---

 
 

\({\small (5)}~\)
\(\begin{eqnarray}~y&=&2x^2-2x+2\\[5pt]~&=&2(x^2-x)+2\\[5pt]~~~&=&2\left(x^2-x+\displaystyle \frac{\,1\,}{\,4\,}-\displaystyle \frac{\,1\,}{\,4\,}\right)+2\\[5pt]~~~&=&2\left(x^2-x+\displaystyle \frac{\,1\,}{\,4\,}\right)+2 \cdot \left(-\displaystyle \frac{\,1\,}{\,4\,}\right)+2\\[5pt]~~~&=&2\left(x-\displaystyle \frac{\,1\,}{\,2\,}\right)^2-\displaystyle \frac{\,1\,}{\,2\,}+2\\[5pt]~~~&=&2\left(x-\displaystyle \frac{\,1\,}{\,2\,}\right)^2+\displaystyle \frac{\,3\,}{\,2\,}\end{eqnarray}\)

---

よって、頂点 \(\left(\displaystyle \frac{\,1\,}{\,2\,}~,~\displaystyle \frac{\,3\,}{\,2\,}\right)\) 、軸の方程式 \(x=\displaystyle \frac{\,1\,}{\,2\,}\)

---

\(y\) 切片が \(2\) 、下に凸のグラフより、

---

　\(\begin{array}{c|ccccc}
x & \cdots & 0 & \cdots & \displaystyle \frac{\,1\,}{\,2\,} & \cdots \\[5pt]
\hline
y & \searrow & 2 & \searrow & \displaystyle \frac{\,3\,}{\,2\,} & \nearrow
\end{array}\)

---

 
 

\({\small (6)}~\)
\(\begin{eqnarray}~y&=&-\displaystyle \frac{\,1\,}{\,2\,}x^2-3x\\[5pt]~&=&-\displaystyle \frac{\,1\,}{\,2\,}(x^2+6x)\\[5pt]~~~&=&-\displaystyle \frac{\,1\,}{\,2\,}(x^2+6x+9-9)\\[5pt]~~~&=&-\displaystyle \frac{\,1\,}{\,2\,}(x^2+6x+9)-\displaystyle \frac{\,1\,}{\,2\,} \cdot (-9)\\[5pt]~~~&=&-\displaystyle \frac{\,1\,}{\,2\,}(x+3)^2+\displaystyle \frac{\,9\,}{\,2\,}\end{eqnarray}\)

---

よって、頂点 \(\left(-3~,~\displaystyle \frac{\,9\,}{\,2\,}\right)\) 、軸の方程式 \(x=-3\)

---

\(y\) 切片が \(0\) 、上に凸のグラフより、

---

　\(\begin{array}{c|ccccc}
x & \cdots & -3 & \cdots & 0 & \cdots \\[5pt]
\hline
y & \nearrow & \displaystyle \frac{\,9\,}{\,2\,} & \searrow & 0 & \searrow
\end{array}\)

---

数研出版｜高等学校数学Ⅰ[104-903] p.84 練習11

\({\small (1)}~\)
\(\begin{eqnarray}~~~y&=&x^2-4x+3\\[3pt]~~~&=&(x^2-4x+4)-4+3\\[3pt]~~~&=&(x-2)^2-1\end{eqnarray}\)

---

よって、頂点 \((2~,~-1)\) 、軸の方程式 \(x=2\)

---

\(y\) 切片が \(3\) 、下に凸のグラフより、

---

　\(\begin{array}{c|ccccc}
x & \cdots & 0 & \cdots & 2 & \cdots \\
\hline
y & \searrow & 3 & \searrow & -1 & \nearrow
\end{array}\)

---

 
 

\({\small (2)}~\)
\(\begin{eqnarray}~~~y&=&2x^2+8x+3\\[3pt]~~~&=&2(x^2+4x)+3\\[3pt]~~~&=&2(x^2+4x+4-4)+3\\[3pt]~~~&=&2(x^2+4x+4)+2 \cdot (-4)+3\\[3pt]~~~&=&2(x+2)^2-8+3\\[3pt]~~~&=&2(x+2)^2-5\end{eqnarray}\)

---

よって、頂点 \((-2~,~-5)\) 、軸の方程式 \(x=-2\)

---

\(y\) 切片が \(3\) 、下に凸のグラフより、

---

　\(\begin{array}{c|ccccc}
x & \cdots & -2 & \cdots & 0 & \cdots \\
\hline
y & \searrow & -5 & \nearrow & 3 & \nearrow
\end{array}\)

---

 
 

\({\small (3)}~\)
\(\begin{eqnarray}~~~y&=&-3x^2+6x+1\\[3pt]~~~&=&-3(x^2-2x)+1\\[3pt]~~~&=&-3(x^2-2x+1-1)+1\\[3pt]~~~&=&-3(x^2-2x+1)-3 \cdot (-1)+1\\[3pt]~~~&=&-3(x-1)^2+3+1\\[3pt]~~~&=&-3(x-1)^2+4\end{eqnarray}\)

---

よって、頂点 \((1~,~4)\) 、軸の方程式 \(x=1\)

---

\(y\) 切片が \(1\) 、上に凸のグラフより、

---

　\(\begin{array}{c|ccccc}
x & \cdots & 0 & \cdots & 1 & \cdots \\
\hline
y & \nearrow & 1 & \nearrow & 4 & \searrow
\end{array}\)

---

 
 

\({\small (4)}~\)
\(\begin{eqnarray}~~~y&=&-x^2-3x\\[5pt]~~~&=&-(x^2+3x)\\[5pt]~~~&=&-\left(x^2+3x+\displaystyle \frac{\,9\,}{\,4\,}-\displaystyle \frac{\,9\,}{\,4\,}\right)\\[5pt]~~~&=&-\left(x^2+3x+\displaystyle \frac{\,9\,}{\,4\,}\right)-1 \cdot \left(-\displaystyle \frac{\,9\,}{\,4\,}\right)\\[5pt]~~~&=&-\left(x+\displaystyle \frac{\,3\,}{\,2\,}\right)^2+\displaystyle \frac{\,9\,}{\,4\,}\end{eqnarray}\)

---

よって、頂点 \(\left(-\displaystyle \frac{\,3\,}{\,2\,}~,~\displaystyle \frac{\,9\,}{\,4\,}\right)\) 、軸の方程式 \(x=-\displaystyle \frac{\,3\,}{\,2\,}\)

---

\(y\) 切片が \(0\) 、上に凸のグラフより、

---

　\(\begin{array}{c|ccccc}
x & \cdots & -\displaystyle \frac{\,3\,}{\,2\,} & \cdots & 0 & \cdots \\[5pt]
\hline
y & \nearrow & \displaystyle \frac{\,9\,}{\,4\,} & \searrow & 0 & \searrow
\end{array}\)

---

数研出版｜高等学校数学Ⅰ[104-903] p.88 問題 4

\({\small (1)}~\)
\(\begin{eqnarray}~~~y&=&\displaystyle \frac{\,1\,}{\,2\,}x^2+x+\displaystyle \frac{\,1\,}{\,2\,}\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,2\,}(x^2+2x)+\displaystyle \frac{\,1\,}{\,2\,}\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,2\,}(x^2+2x+1-1)+\displaystyle \frac{\,1\,}{\,2\,}\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,2\,}(x^2+2x+1)+\displaystyle \frac{\,1\,}{\,2\,} \cdot (-1)+\displaystyle \frac{\,1\,}{\,2\,}\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,2\,}(x+1)^2-\displaystyle \frac{\,1\,}{\,2\,}+\displaystyle \frac{\,1\,}{\,2\,}\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,2\,}(x+1)^2\end{eqnarray}\)

---

よって、頂点 \((-1~,~0)\) 、軸の方程式 \(x=-1\)

---

\(y\) 切片が \(\displaystyle \frac{\,1\,}{\,2\,}\) 、下に凸のグラフより、

---

　\(\begin{array}{c|ccccc}
x & \cdots & -1 & \cdots & 0 & \cdots \\[5pt]
\hline
y & \searrow & 0 & \nearrow & \displaystyle \frac{\,1\,}{\,2\,} & \nearrow
\end{array}\)

---

 
 

\({\small (2)}~\)
\(\begin{eqnarray}~~~y&=&-3x^2+3x+\displaystyle \frac{\,1\,}{\,4\,}\\[5pt]~~~&=&-3(x^2-x)+\displaystyle \frac{\,1\,}{\,4\,}\\[5pt]~~~&=&-3\left(x^2-x+\displaystyle \frac{\,1\,}{\,4\,}-\displaystyle \frac{\,1\,}{\,4\,}\right)+\displaystyle \frac{\,1\,}{\,4\,}\\[5pt]~~~&=&-3\left(x^2-x+\displaystyle \frac{\,1\,}{\,4\,}\right)-3 \cdot \left(-\displaystyle \frac{\,1\,}{\,4\,}\right)+\displaystyle \frac{\,1\,}{\,4\,}\\[5pt]~~~&=&-3\left(x-\displaystyle \frac{\,1\,}{\,2\,}\right)^2+\displaystyle \frac{\,3\,}{\,4\,}+\displaystyle \frac{\,1\,}{\,4\,}\\[5pt]~~~&=&-3\left(x-\displaystyle \frac{\,1\,}{\,2\,}\right)^2+1\end{eqnarray}\)

---

よって、頂点 \(\left(\displaystyle \frac{\,1\,}{\,2\,}~,~1\right)\) 、軸の方程式 \(x=\displaystyle \frac{\,1\,}{\,2\,}\)

---

\(y\) 切片が \(\displaystyle \frac{\,1\,}{\,4\,}\) 、上に凸のグラフより、

---

　\(\begin{array}{c|ccccc}
x & \cdots & 0 & \cdots & \displaystyle \frac{\,1\,}{\,2\,} & \cdots \\[5pt]
\hline
y & \nearrow & \displaystyle \frac{\,1\,}{\,4\,} & \nearrow & 1 & \searrow
\end{array}\)

---

 
 

\({\small (3)}~\)
\(\begin{eqnarray}~~~y&=&(x-1)(x-5)\\[3pt]~~~&=&x^2-6x+5\\[3pt]~~~&=&(x^2-6x+9)-9+5\\[3pt]~~~&=&(x-3)^2-4\end{eqnarray}\)

---

よって、頂点 \((3~,~-4)\) 、軸の方程式 \(x=3\)

---

\(y\) 切片が \(5\) 、下に凸のグラフより、

---

　\(\begin{array}{c|ccccc}
x & \cdots & 0 & \cdots & 3 & \cdots \\
\hline
y & \searrow & 5 & \searrow & -4 & \nearrow
\end{array}\)

---

 
 

\({\small (4)}~\)
\(\begin{eqnarray}~~~y&=&(2x-1)(x+3)\\[5pt]~~~&=&2x^2+5x-3\\[5pt]~~~&=&2\left(x^2+\displaystyle \frac{\,5\,}{\,2\,}x\right)-3\\[5pt]~~~&=&2\left(x^2+\displaystyle \frac{\,5\,}{\,2\,}x+\displaystyle \frac{\,25\,}{\,16\,}-\displaystyle \frac{\,25\,}{\,16\,}\right)-3\\[5pt]~~~&=&2\left(x^2+\displaystyle \frac{\,5\,}{\,2\,}x+\displaystyle \frac{\,25\,}{\,16\,}\right)+2 \cdot \left(-\displaystyle \frac{\,25\,}{\,16\,}\right)-3\\[5pt]~~~&=&2\left(x+\displaystyle \frac{\,5\,}{\,4\,}\right)^2-\displaystyle \frac{\,25\,}{\,8\,}-3\\[5pt]~~~&=&2\left(x+\displaystyle \frac{\,5\,}{\,4\,}\right)^2-\displaystyle \frac{\,49\,}{\,8\,}\end{eqnarray}\)

---

よって、頂点 \(\left(-\displaystyle \frac{\,5\,}{\,4\,}~,~-\displaystyle \frac{\,49\,}{\,8\,}\right)\) 、軸の方程式 \(x=-\displaystyle \frac{\,5\,}{\,4\,}\)

---

\(y\) 切片が \(-3\) 、下に凸のグラフより、

---

　\(\begin{array}{c|ccccc}
x & \cdots & -\displaystyle \frac{\,5\,}{\,4\,} & \cdots & 0 & \cdots \\[5pt]
\hline
y & \searrow & -\displaystyle \frac{\,49\,}{\,8\,} & \nearrow & -3 & \nearrow
\end{array}\)

---

数研出版｜新編数学Ⅰ[104-904] p.92 練習12

\({\small (1)}~\)
\(\begin{eqnarray}~~~y&=&x^2-6x+5\\[3pt]~~~&=&(x^2-6x+9)-9+5\\[3pt]~~~&=&(x-3)^2-4\end{eqnarray}\)

---

よって、頂点 \((3~,~-4)\) 、軸の方程式 \(x=3\)

---

\(y\) 切片が \(5\) 、下に凸のグラフより、

---

　\(\begin{array}{c|ccccc}
x & \cdots & 0 & \cdots & 3 & \cdots \\
\hline
y & \searrow & 5 & \searrow & -4 & \nearrow
\end{array}\)

---

 
 

\({\small (2)}~\)
\(\begin{eqnarray}~~~y&=&2x^2+8x+3\\[3pt]~~~&=&2(x^2+4x)+3\\[3pt]~~~&=&2(x^2+4x+4-4)+3\\[3pt]~~~&=&2(x^2+4x+4)+2 \cdot (-4)+3\\[3pt]~~~&=&2(x+2)^2-8+3\\[3pt]~~~&=&2(x+2)^2-5\end{eqnarray}\)

---

よって、頂点 \((-2~,~-5)\) 、軸の方程式 \(x=-2\)

---

\(y\) 切片が \(3\) 、下に凸のグラフより、

---

　\(\begin{array}{c|ccccc}
x & \cdots & -2 & \cdots & 0 & \cdots \\
\hline
y & \searrow & -5 & \nearrow & 3 & \nearrow
\end{array}\)

---

 
 

\({\small (3)}~\)
\(\begin{eqnarray}~~~y&=&-3x^2+6x+1\\[3pt]~~~&=&-3(x^2-2x)+1\\[3pt]~~~&=&-3(x^2-2x+1-1)+1\\[3pt]~~~&=&-3(x^2-2x+1)-3 \cdot (-1)+1\\[3pt]~~~&=&-3(x-1)^2+3+1\\[3pt]~~~&=&-3(x-1)^2+4\end{eqnarray}\)

---

よって、頂点 \((1~,~4)\) 、軸の方程式 \(x=1\)

---

\(y\) 切片が \(1\) 、上に凸のグラフより、

---

　\(\begin{array}{c|ccccc}
x & \cdots & 0 & \cdots & 1 & \cdots \\
\hline
y & \nearrow & 1 & \nearrow & 4 & \searrow
\end{array}\)

---

 
 

\({\small (4)}~\)
\(\begin{eqnarray}~~~y&=&-x^2-4x+2\\[3pt]~~~&=&-(x^2+4x)+2\\[3pt]~~~&=&-(x^2+4x+4-4)+2\\[3pt]~~~&=&-(x^2+4x+4)-1 \cdot (-4)+2\\[3pt]~~~&=&-(x+2)^2+4+2\\[3pt]~~~&=&-(x+2)^2+6\end{eqnarray}\)

---

よって、頂点 \((-2~,~6)\) 、軸の方程式 \(x=-2\)

---

\(y\) 切片が \(2\) 、上に凸のグラフより、

---

　\(\begin{array}{c|ccccc}
x & \cdots & -2 & \cdots & 0 & \cdots \\
\hline
y & \nearrow & 6 & \searrow & 2 & \searrow
\end{array}\)

---

 
 

\({\small (5)}~\)
\(\begin{eqnarray}~~~y&=&2x^2-6x-1\\[5pt]~~~&=&2(x^2-3x)-1\\[5pt]~~~&=&2\left(x^2-3x+\displaystyle \frac{\,9\,}{\,4\,}-\displaystyle \frac{\,9\,}{\,4\,}\right)-1\\[5pt]~~~&=&2\left(x^2-3x+\displaystyle \frac{\,9\,}{\,4\,}\right)+2 \cdot \left(-\displaystyle \frac{\,9\,}{\,4\,}\right)-1\\[5pt]~~~&=&2\left(x-\displaystyle \frac{\,3\,}{\,2\,}\right)^2-\displaystyle \frac{\,9\,}{\,2\,}-1\\[5pt]~~~&=&2\left(x-\displaystyle \frac{\,3\,}{\,2\,}\right)^2-\displaystyle \frac{\,11\,}{\,2\,}\end{eqnarray}\)

---

よって、頂点 \(\left(\displaystyle \frac{\,3\,}{\,2\,}~,~-\displaystyle \frac{\,11\,}{\,2\,}\right)\) 、軸の方程式 \(x=\displaystyle \frac{\,3\,}{\,2\,}\)

---

\(y\) 切片が \(-1\) 、下に凸のグラフより、

---

　\(\begin{array}{c|ccccc}
x & \cdots & 0 & \cdots & \displaystyle \frac{\,3\,}{\,2\,} & \cdots \\[5pt]
\hline
y & \searrow & -1 & \searrow & -\displaystyle \frac{\,11\,}{\,2\,} & \nearrow
\end{array}\)

---

 
 

\({\small (6)}~\)
\(\begin{eqnarray}~~~y&=&-x^2+3x\\[5pt]~~~&=&-(x^2-3x)\\[5pt]~~~&=&-\left(x^2-3x+\displaystyle \frac{\,9\,}{\,4\,}-\displaystyle \frac{\,9\,}{\,4\,}\right)\\[5pt]~~~&=&-\left(x^2-3x+\displaystyle \frac{\,9\,}{\,4\,}\right)-1 \cdot \left(-\displaystyle \frac{\,9\,}{\,4\,}\right)\\[5pt]~~~&=&-\left(x-\displaystyle \frac{\,3\,}{\,2\,}\right)^2+\displaystyle \frac{\,9\,}{\,4\,}\end{eqnarray}\)

---

よって、頂点 \(\left(\displaystyle \frac{\,3\,}{\,2\,}~,~\displaystyle \frac{\,9\,}{\,4\,}\right)\) 、軸の方程式 \(x=\displaystyle \frac{\,3\,}{\,2\,}\)

---

\(y\) 切片が \(0\) 、上に凸のグラフより、

---

　\(\begin{array}{c|ccccc}
x & \cdots & 0 & \cdots & \displaystyle \frac{\,3\,}{\,2\,} & \cdots \\[5pt]
\hline
y & \nearrow & 0 & \nearrow & \displaystyle \frac{\,9\,}{\,4\,} & \searrow
\end{array}\)

---

数研出版｜新編数学Ⅰ[104-904] p.96 補充問題 2

\({\small (1)}~\)
\(\begin{eqnarray}~~~y&=&2x^2-4x+2\\[3pt]~~~&=&2(x^2-2x)+2\\[3pt]~~~&=&2(x^2-2x+1-1)+2\\[3pt]~~~&=&2(x^2-2x+1)+2 \cdot (-1)+2\\[3pt]~~~&=&2(x-1)^2-2+2\\[3pt]~~~&=&2(x-1)^2\end{eqnarray}\)

---

よって、頂点 \((1~,~0)\) 、軸の方程式 \(x=1\)

---

\(y\) 切片が \(2\) 、下に凸のグラフより、

---

　\(\begin{array}{c|ccccc}
x & \cdots & 0 & \cdots & 1 & \cdots \\
\hline
y & \searrow & 2 & \searrow & 0 & \nearrow
\end{array}\)

---

 
 

\({\small (2)}~\)
\(\begin{eqnarray}~~~y&=&-\displaystyle \frac{\,1\,}{\,2\,}x^2+x-1\\[5pt]~~~&=&-\displaystyle \frac{\,1\,}{\,2\,}(x^2-2x)-1\\[5pt]~~~&=&-\displaystyle \frac{\,1\,}{\,2\,}(x^2-2x+1-1)-1\\[5pt]~~~&=&-\displaystyle \frac{\,1\,}{\,2\,}(x^2-2x+1)-\displaystyle \frac{\,1\,}{\,2\,} \cdot (-1)-1\\[5pt]~~~&=&-\displaystyle \frac{\,1\,}{\,2\,}(x-1)^2+\displaystyle \frac{\,1\,}{\,2\,}-1\\[5pt]~~~&=&-\displaystyle \frac{\,1\,}{\,2\,}(x-1)^2-\displaystyle \frac{\,1\,}{\,2\,}\end{eqnarray}\)

---

よって、頂点 \(\left(1~,~-\displaystyle \frac{\,1\,}{\,2\,}\right)\) 、軸の方程式 \(x=1\)

---

\(y\) 切片が \(-1\) 、上に凸のグラフより、

---

　\(\begin{array}{c|ccccc}
x & \cdots & 0 & \cdots & 1 & \cdots \\[5pt]
\hline
y & \nearrow & -1 & \nearrow & -\displaystyle \frac{\,1\,}{\,2\,} & \searrow
\end{array}\)

---

 
 

\({\small (3)}~\)
\(\begin{eqnarray}~~~y&=&(x-1)(x-2)\\[5pt]~~~&=&x^2-3x+2\\[5pt]~~~&=&\left(x^2-3x+\displaystyle \frac{\,9\,}{\,4\,}\right)-\displaystyle \frac{\,9\,}{\,4\,}+2\\[5pt]~~~&=&\left(x-\displaystyle \frac{\,3\,}{\,2\,}\right)^2-\displaystyle \frac{\,1\,}{\,4\,}\end{eqnarray}\)

---

よって、頂点 \(\left(\displaystyle \frac{\,3\,}{\,2\,}~,~-\displaystyle \frac{\,1\,}{\,4\,}\right)\) 、軸の方程式 \(x=\displaystyle \frac{\,3\,}{\,2\,}\)

---

\(y\) 切片が \(2\) 、下に凸のグラフより、

---

　\(\begin{array}{c|ccccc}
x & \cdots & 0 & \cdots & \displaystyle \frac{\,3\,}{\,2\,} & \cdots \\[5pt]
\hline
y & \searrow & 2 & \searrow & -\displaystyle \frac{\,1\,}{\,4\,} & \nearrow
\end{array}\)

---

 
 

\({\small (4)}~\)
\(\begin{eqnarray}~~~y&=&(2x-1)(x+3)\\[5pt]~~~&=&2x^2+5x-3\\[5pt]~~~&=&2\left(x^2+\displaystyle \frac{\,5\,}{\,2\,}x\right)-3\\[5pt]~~~&=&2\left(x^2+\displaystyle \frac{\,5\,}{\,2\,}x+\displaystyle \frac{\,25\,}{\,16\,}-\displaystyle \frac{\,25\,}{\,16\,}\right)-3\\[5pt]~~~&=&2\left(x^2+\displaystyle \frac{\,5\,}{\,2\,}x+\displaystyle \frac{\,25\,}{\,16\,}\right)+2 \cdot \left(-\displaystyle \frac{\,25\,}{\,16\,}\right)-3\\[5pt]~~~&=&2\left(x+\displaystyle \frac{\,5\,}{\,4\,}\right)^2-\displaystyle \frac{\,25\,}{\,8\,}-3\\[5pt]~~~&=&2\left(x+\displaystyle \frac{\,5\,}{\,4\,}\right)^2-\displaystyle \frac{\,49\,}{\,8\,}\end{eqnarray}\)

---

よって、頂点 \(\left(-\displaystyle \frac{\,5\,}{\,4\,}~,~-\displaystyle \frac{\,49\,}{\,8\,}\right)\) 、軸の方程式 \(x=-\displaystyle \frac{\,5\,}{\,4\,}\)

---

\(y\) 切片が \(-3\) 、下に凸のグラフより、

---

　\(\begin{array}{c|ccccc}
x & \cdots & -\displaystyle \frac{\,5\,}{\,4\,} & \cdots & 0 & \cdots \\[5pt]
\hline
y & \searrow & -\displaystyle \frac{\,49\,}{\,8\,} & \nearrow & -3 & \nearrow
\end{array}\)

---

東京書籍｜Advanced数学Ⅰ[002-901] p.84 問12

\({\small (1)}~\)
\(\begin{eqnarray}~~~y&=&x^2+4x+5\\[3pt]~~~&=&(x^2+4x+4)-4+5\\[3pt]~~~&=&(x+2)^2+1\end{eqnarray}\)

 
 

\({\small (2)}~\)
\(\begin{eqnarray}~~~y&=&3x^2-6x+2\\[3pt]~~~&=&3(x^2-2x)+2\\[3pt]~~~&=&3(x^2-2x+1-1)+2\\[3pt]~~~&=&3(x^2-2x+1)+3 \cdot (-1)+2\\[3pt]~~~&=&3(x-1)^2-3+2\\[3pt]~~~&=&3(x-1)^2-1\end{eqnarray}\)

 
 

\({\small (3)}~\)
\(\begin{eqnarray}~~~y&=&-x^2+6x+1\\[3pt]~~~&=&-(x^2-6x)+1\\[3pt]~~~&=&-(x^2-6x+9-9)+1\\[3pt]~~~&=&-(x^2-6x+9)-1 \cdot (-9)+1\\[3pt]~~~&=&-(x-3)^2+9+1\\[3pt]~~~&=&-(x-3)^2+10\end{eqnarray}\)

 
 

\({\small (4)}~\)
\(\begin{eqnarray}~~~y&=&\displaystyle \frac{\,1\,}{\,2\,}x^2+4x+6\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,2\,}(x^2+8x)+6\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,2\,}(x^2+8x+16-16)+6\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,2\,}(x^2+8x+16)+\displaystyle \frac{\,1\,}{\,2\,} \cdot (-16)+6\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,2\,}(x+4)^2-8+6\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,2\,}(x+4)^2-2\end{eqnarray}\)

 
 

\({\small (5)}~\)
\(\begin{eqnarray}~~~y&=&x^2+3x+4\\[5pt]~~~&=&\left(x^2+3x+\displaystyle \frac{\,9\,}{\,4\,}\right)-\displaystyle \frac{\,9\,}{\,4\,}+4\\[5pt]~~~&=&\left(x+\displaystyle \frac{\,3\,}{\,2\,}\right)^2+\displaystyle \frac{\,7\,}{\,4\,}\end{eqnarray}\)

 
 

\({\small (6)}~\)
\(\begin{eqnarray}~~~y&=&-2x^2+2x+3\\[5pt]~~~&=&-2(x^2-x)+3\\[5pt]~~~&=&-2\left(x^2-x+\displaystyle \frac{\,1\,}{\,4\,}-\displaystyle \frac{\,1\,}{\,4\,}\right)+3\\[5pt]~~~&=&-2\left(x^2-x+\displaystyle \frac{\,1\,}{\,4\,}\right)-2 \cdot \left(-\displaystyle \frac{\,1\,}{\,4\,}\right)+3\\[5pt]~~~&=&-2\left(x-\displaystyle \frac{\,1\,}{\,2\,}\right)^2+\displaystyle \frac{\,1\,}{\,2\,}+3\\[5pt]~~~&=&-2\left(x-\displaystyle \frac{\,1\,}{\,2\,}\right)^2+\displaystyle \frac{\,7\,}{\,2\,}\end{eqnarray}\)

東京書籍｜Advanced数学Ⅰ[002-901] p.85 問13

\({\small (1)}~\)
\(\begin{eqnarray}~~~y&=&2x^2+12x+8\\[3pt]~~~&=&2(x^2+6x)+8\\[3pt]~~~&=&2(x^2+6x+9-9)+8\\[3pt]~~~&=&2(x^2+6x+9)+2 \cdot (-9)+8\\[3pt]~~~&=&2(x+3)^2-18+8\\[3pt]~~~&=&2(x+3)^2-10\end{eqnarray}\)

---

よって、頂点 \((-3~,~-10)\) 、軸の方程式 \(x=-3\)

---

\(y\) 切片が \(8\) 、下に凸のグラフより、

---

　\(\begin{array}{c|ccccc}
x & \cdots & -3 & \cdots & 0 & \cdots \\
\hline
y & \searrow & -10 & \nearrow & 8 & \nearrow
\end{array}\)

---

 
 

\({\small (2)}~\)
\(\begin{eqnarray}~~~y&=&\displaystyle \frac{\,1\,}{\,2\,}x^2-2x-1\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,2\,}(x^2-4x)-1\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,2\,}(x^2-4x+4-4)-1\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,2\,}(x^2-4x+4)+\displaystyle \frac{\,1\,}{\,2\,} \cdot (-4)-1\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,2\,}(x-2)^2-2-1\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,2\,}(x-2)^2-3\end{eqnarray}\)

---

よって、頂点 \((2~,~-3)\) 、軸の方程式 \(x=2\)

---

\(y\) 切片が \(-1\) 、下に凸のグラフより、

---

　\(\begin{array}{c|ccccc}
x & \cdots & 0 & \cdots & 2 & \cdots \\
\hline
y & \searrow & -1 & \searrow & -3 & \nearrow
\end{array}\)

---

 
 

\({\small (3)}~\)
\(\begin{eqnarray}~~~y&=&-\displaystyle \frac{\,1\,}{\,2\,}x^2+3x\\[5pt]~~~&=&-\displaystyle \frac{\,1\,}{\,2\,}(x^2-6x)\\[5pt]~~~&=&-\displaystyle \frac{\,1\,}{\,2\,}(x^2-6x+9-9)\\[5pt]~~~&=&-\displaystyle \frac{\,1\,}{\,2\,}(x^2-6x+9)-\displaystyle \frac{\,1\,}{\,2\,} \cdot (-9)\\[5pt]~~~&=&-\displaystyle \frac{\,1\,}{\,2\,}(x-3)^2+\displaystyle \frac{\,9\,}{\,2\,}\end{eqnarray}\)

---

よって、頂点 \(\left(3~,~\displaystyle \frac{\,9\,}{\,2\,}\right)\) 、軸の方程式 \(x=3\)

---

\(y\) 切片が \(0\) 、上に凸のグラフより、

---

　\(\begin{array}{c|ccccc}
x & \cdots & 0 & \cdots & 3 & \cdots \\[5pt]
\hline
y & \nearrow & 0 & \nearrow & \displaystyle \frac{\,9\,}{\,2\,} & \searrow
\end{array}\)

---

 
 

\({\small (4)}~\)
\(\begin{eqnarray}~~~y&=&-x^2-x+1\\[5pt]~~~&=&-(x^2+x)+1\\[5pt]~~~&=&-\left(x^2+x+\displaystyle \frac{\,1\,}{\,4\,}-\displaystyle \frac{\,1\,}{\,4\,}\right)+1\\[5pt]~~~&=&-\left(x^2+x+\displaystyle \frac{\,1\,}{\,4\,}\right)-1 \cdot \left(-\displaystyle \frac{\,1\,}{\,4\,}\right)+1\\[5pt]~~~&=&-\left(x+\displaystyle \frac{\,1\,}{\,2\,}\right)^2+\displaystyle \frac{\,1\,}{\,4\,}+1\\[5pt]~~~&=&-\left(x+\displaystyle \frac{\,1\,}{\,2\,}\right)^2+\displaystyle \frac{\,5\,}{\,4\,}\end{eqnarray}\)

---

よって、頂点 \(\left(-\displaystyle \frac{\,1\,}{\,2\,}~,~\displaystyle \frac{\,5\,}{\,4\,}\right)\) 、軸の方程式 \(x=-\displaystyle \frac{\,1\,}{\,2\,}\)

---

\(y\) 切片が \(1\) 、上に凸のグラフより、

---

　\(\begin{array}{c|ccccc}
x & \cdots & -\displaystyle \frac{\,1\,}{\,2\,} & \cdots & 0 & \cdots \\[5pt]
\hline
y & \nearrow & \displaystyle \frac{\,5\,}{\,4\,} & \searrow & 1 & \searrow
\end{array}\)

---

東京書籍｜Advanced数学Ⅰ[002-901] p.98 問題 2

\({\small (1)}~\)
\(\begin{eqnarray}~~~y&=&-x^2+6x-5\\[3pt]~~~&=&-(x^2-6x)-5\\[3pt]~~~&=&-(x^2-6x+9-9)-5\\[3pt]~~~&=&-(x^2-6x+9)-1 \cdot (-9)-5\\[3pt]~~~&=&-(x-3)^2+9-5\\[3pt]~~~&=&-(x-3)^2+4\end{eqnarray}\)

---

よって、頂点 \((3~,~4)\) 、軸の方程式 \(x=3\)

---

\(y\) 切片が \(-5\) 、上に凸のグラフより、

---

　\(\begin{array}{c|ccccc}
x & \cdots & 0 & \cdots & 3 & \cdots \\
\hline
y & \nearrow & -5 & \nearrow & 4 & \searrow
\end{array}\)

---

 
 

\({\small (2)}~\)
\(\begin{eqnarray}~~~y&=&2(x-1)(x-3)\\[3pt]~~~&=&2x^2-8x+6\\[3pt]~~~&=&2(x^2-4x)+6\\[3pt]~~~&=&2(x^2-4x+4-4)+6\\[3pt]~~~&=&2(x^2-4x+4)+2 \cdot (-4)+6\\[3pt]~~~&=&2(x-2)^2-8+6\\[3pt]~~~&=&2(x-2)^2-2\end{eqnarray}\)

---

よって、頂点 \((2~,~-2)\) 、軸の方程式 \(x=2\)

---

\(y\) 切片が \(6\) 、下に凸のグラフより、

---

　\(\begin{array}{c|ccccc}
x & \cdots & 0 & \cdots & 2 & \cdots \\
\hline
y & \searrow & 6 & \searrow & -2 & \nearrow
\end{array}\)

---

東京書籍｜Standard数学Ⅰ[002-902] p.84 問9

\({\small (1)}~\)
\(\begin{eqnarray}~~~y&=&x^2-2x\\[3pt]~~~&=&(x^2-2x+1)-1\\[3pt]~~~&=&(x-1)^2-1\end{eqnarray}\)

 
 

\({\small (2)}~\)
\(\begin{eqnarray}~~~y&=&x^2+6x-2\\[3pt]~~~&=&(x^2+6x+9)-9-2\\[3pt]~~~&=&(x+3)^2-11\end{eqnarray}\)

 
 

\({\small (3)}~\)
\(\begin{eqnarray}~~~y&=&x^2+3x+4\\[5pt]~~~&=&\left(x^2+3x+\displaystyle \frac{\,9\,}{\,4\,}\right)-\displaystyle \frac{\,9\,}{\,4\,}+4\\[5pt]~~~&=&\left(x+\displaystyle \frac{\,3\,}{\,2\,}\right)^2+\displaystyle \frac{\,7\,}{\,4\,}\end{eqnarray}\)

 
 

\({\small (4)}~\)
\(\begin{eqnarray}~~~y&=&x^2-9x+21\\[5pt]~~~&=&\left(x^2-9x+\displaystyle \frac{\,81\,}{\,4\,}\right)-\displaystyle \frac{\,81\,}{\,4\,}+21\\[5pt]~~~&=&\left(x-\displaystyle \frac{\,9\,}{\,2\,}\right)^2+\displaystyle \frac{\,3\,}{\,4\,}\end{eqnarray}\)

東京書籍｜Standard数学Ⅰ[002-902] p.85 問10

\({\small (1)}~\)
\(\begin{eqnarray}~~~y&=&2x^2+8x+3\\[3pt]~~~&=&2(x^2+4x)+3\\[3pt]~~~&=&2(x^2+4x+4-4)+3\\[3pt]~~~&=&2(x^2+4x+4)+2 \cdot (-4)+3\\[3pt]~~~&=&2(x+2)^2-8+3\\[3pt]~~~&=&2(x+2)^2-5\end{eqnarray}\)

 
 

\({\small (2)}~\)
\(\begin{eqnarray}~~~y&=&3x^2-12x-2\\[3pt]~~~&=&3(x^2-4x)-2\\[3pt]~~~&=&3(x^2-4x+4-4)-2\\[3pt]~~~&=&3(x^2-4x+4)+3 \cdot (-4)-2\\[3pt]~~~&=&3(x-2)^2-12-2\\[3pt]~~~&=&3(x-2)^2-14\end{eqnarray}\)

 
 

\({\small (3)}~\)
\(\begin{eqnarray}~~~y&=&-x^2+10x+7\\[3pt]~~~&=&-(x^2-10x)+7\\[3pt]~~~&=&-(x^2-10x+25-25)+7\\[3pt]~~~&=&-(x^2-10x+25)-1 \cdot (-25)+7\\[3pt]~~~&=&-(x-5)^2+25+7\\[3pt]~~~&=&-(x-5)^2+32\end{eqnarray}\)

 
 

\({\small (4)}~\)
\(\begin{eqnarray}~~~y&=&-2x^2-6x-5\\[5pt]~~~&=&-2(x^2+3x)-5\\[5pt]~~~&=&-2\left(x^2+3x+\displaystyle \frac{\,9\,}{\,4\,}-\displaystyle \frac{\,9\,}{\,4\,}\right)-5\\[5pt]~~~&=&-2\left(x^2+3x+\displaystyle \frac{\,9\,}{\,4\,}\right)-2 \cdot \left(-\displaystyle \frac{\,9\,}{\,4\,}\right)-5\\[5pt]~~~&=&-2\left(x+\displaystyle \frac{\,3\,}{\,2\,}\right)^2+\displaystyle \frac{\,9\,}{\,2\,}-5\\[5pt]~~~&=&-2\left(x+\displaystyle \frac{\,3\,}{\,2\,}\right)^2-\displaystyle \frac{\,1\,}{\,2\,}\end{eqnarray}\)

東京書籍｜Standard数学Ⅰ[002-902] p.85 問11

\({\small (1)}~\)
\(\begin{eqnarray}~~~y&=&x^2-4x+3\\[3pt]~~~&=&(x^2-4x+4)-4+3\\[3pt]~~~&=&(x-2)^2-1\end{eqnarray}\)

---

よって、頂点 \((2~,~-1)\) 、軸の方程式 \(x=2\)

---

\(y\) 切片が \(3\) 、下に凸のグラフより、

---

　\(\begin{array}{c|ccccc}
x & \cdots & 0 & \cdots & 2 & \cdots \\
\hline
y & \searrow & 3 & \searrow & -1 & \nearrow
\end{array}\)

---

 
 

\({\small (2)}~\)
\(\begin{eqnarray}~~~y&=&2x^2+4x+3\\[3pt]~~~&=&2(x^2+2x)+3\\[3pt]~~~&=&2(x^2+2x+1-1)+3\\[3pt]~~~&=&2(x^2+2x+1)+2 \cdot (-1)+3\\[3pt]~~~&=&2(x+1)^2-2+3\\[3pt]~~~&=&2(x+1)^2+1\end{eqnarray}\)

---

よって、頂点 \((-1~,~1)\) 、軸の方程式 \(x=-1\)

---

\(y\) 切片が \(3\) 、下に凸のグラフより、

---

　\(\begin{array}{c|ccccc}
x & \cdots & -1 & \cdots & 0 & \cdots \\
\hline
y & \searrow & 1 & \nearrow & 3 & \nearrow
\end{array}\)

---

 
 

\({\small (3)}~\)
\(\begin{eqnarray}~~~y&=&-2x^2-6x-3\\[5pt]~~~&=&-2(x^2+3x)-3\\[5pt]~~~&=&-2\left(x^2+3x+\displaystyle \frac{\,9\,}{\,4\,}-\displaystyle \frac{\,9\,}{\,4\,}\right)-3\\[5pt]~~~&=&-2\left(x^2+3x+\displaystyle \frac{\,9\,}{\,4\,}\right)-2 \cdot \left(-\displaystyle \frac{\,9\,}{\,4\,}\right)-3\\[5pt]~~~&=&-2\left(x+\displaystyle \frac{\,3\,}{\,2\,}\right)^2+\displaystyle \frac{\,9\,}{\,2\,}-3\\[5pt]~~~&=&-2\left(x+\displaystyle \frac{\,3\,}{\,2\,}\right)^2+\displaystyle \frac{\,3\,}{\,2\,}\end{eqnarray}\)

---

よって、頂点 \(\left(-\displaystyle \frac{\,3\,}{\,2\,}~,~\displaystyle \frac{\,3\,}{\,2\,}\right)\) 、軸の方程式 \(x=-\displaystyle \frac{\,3\,}{\,2\,}\)

---

\(y\) 切片が \(-3\) 、上に凸のグラフより、

---

　\(\begin{array}{c|ccccc}
x & \cdots & -\displaystyle \frac{\,3\,}{\,2\,} & \cdots & 0 & \cdots \\[5pt]
\hline
y & \nearrow & \displaystyle \frac{\,3\,}{\,2\,} & \searrow & -3 & \searrow
\end{array}\)

---

 
 

\({\small (4)}~\)
\(\begin{eqnarray}~~~y&=&\displaystyle \frac{\,1\,}{\,2\,}x^2+2x+5\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,2\,}(x^2+4x)+5\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,2\,}(x^2+4x+4-4)+5\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,2\,}(x^2+4x+4)+\displaystyle \frac{\,1\,}{\,2\,} \cdot (-4)+5\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,2\,}(x+2)^2-2+5\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,2\,}(x+2)^2+3\end{eqnarray}\)

---

よって、頂点 \((-2~,~3)\) 、軸の方程式 \(x=-2\)

---

\(y\) 切片が \(5\) 、下に凸のグラフより、

---

　\(\begin{array}{c|ccccc}
x & \cdots & -2 & \cdots & 0 & \cdots \\
\hline
y & \searrow & 3 & \nearrow & 5 & \nearrow
\end{array}\)

---

東京書籍｜Standard数学Ⅰ[002-902] p.100 Training 2

\({\small (1)}~\)
\(\begin{eqnarray}~~~y&=&2x^2-12x+16\\[3pt]~~~&=&2(x^2-6x)+16\\[3pt]~~~&=&2(x^2-6x+9-9)+16\\[3pt]~~~&=&2(x^2-6x+9)+2 \cdot (-9)+16\\[3pt]~~~&=&2(x-3)^2-18+16\\[3pt]~~~&=&2(x-3)^2-2\end{eqnarray}\)

---

よって、頂点 \((3~,~-2)\) 、軸の方程式 \(x=3\)

---

\(y\) 切片が \(16\) 、下に凸のグラフより、

---

　\(\begin{array}{c|ccccc}
x & \cdots & 0 & \cdots & 3 & \cdots \\
\hline
y & \searrow & 16 & \searrow & -2 & \nearrow
\end{array}\)

---

 
 

\({\small (2)}~\)
\(\begin{eqnarray}~~~y&=&-x^2+8x-15\\[3pt]~~~&=&-(x^2-8x)-15\\[3pt]~~~&=&-(x^2-8x+16-16)-15\\[3pt]~~~&=&-(x^2-8x+16)-1 \cdot (-16)-15\\[3pt]~~~&=&-(x-4)^2+16-15\\[3pt]~~~&=&-(x-4)^2+1\end{eqnarray}\)

---

よって、頂点 \((4~,~1)\) 、軸の方程式 \(x=4\)

---

\(y\) 切片が \(-15\) 、上に凸のグラフより、

---

　\(\begin{array}{c|ccccc}
x & \cdots & 0 & \cdots & 4 & \cdots \\
\hline
y & \nearrow & -15 & \nearrow & 1 & \searrow
\end{array}\)

---

 
 

\({\small (3)}~\)
\(\begin{eqnarray}~~~y&=&-\displaystyle \frac{\,1\,}{\,2\,}x^2-x+\displaystyle \frac{\,3\,}{\,2\,}\\[5pt]~~~&=&-\displaystyle \frac{\,1\,}{\,2\,}(x^2+2x)+\displaystyle \frac{\,3\,}{\,2\,}\\[5pt]~~~&=&-\displaystyle \frac{\,1\,}{\,2\,}(x^2+2x+1-1)+\displaystyle \frac{\,3\,}{\,2\,}\\[5pt]~~~&=&-\displaystyle \frac{\,1\,}{\,2\,}(x^2+2x+1)-\displaystyle \frac{\,1\,}{\,2\,} \cdot (-1)+\displaystyle \frac{\,3\,}{\,2\,}\\[5pt]~~~&=&-\displaystyle \frac{\,1\,}{\,2\,}(x+1)^2+\displaystyle \frac{\,1\,}{\,2\,}+\displaystyle \frac{\,3\,}{\,2\,}\\[5pt]~~~&=&-\displaystyle \frac{\,1\,}{\,2\,}(x+1)^2+2\end{eqnarray}\)

---

よって、頂点 \((-1~,~2)\) 、軸の方程式 \(x=-1\)

---

\(y\) 切片が \(\displaystyle \frac{\,3\,}{\,2\,}\) 、上に凸のグラフより、

---

　\(\begin{array}{c|ccccc}
x & \cdots & -1 & \cdots & 0 & \cdots \\[5pt]
\hline
y & \nearrow & 2 & \searrow & \displaystyle \frac{\,3\,}{\,2\,} & \searrow
\end{array}\)

---

 
 

\({\small (4)}~\)
\(\begin{eqnarray}~~~y&=&(x+2)(x-4)\\[3pt]~~~&=&x^2-2x-8\\[3pt]~~~&=&(x^2-2x+1)-1-8\\[3pt]~~~&=&(x-1)^2-9\end{eqnarray}\)

---

よって、頂点 \((1~,~-9)\) 、軸の方程式 \(x=1\)

---

\(y\) 切片が \(-8\) 、下に凸のグラフより、

---

　\(\begin{array}{c|ccccc}
x & \cdots & 0 & \cdots & 1 & \cdots \\
\hline
y & \searrow & -8 & \searrow & -9 & \nearrow
\end{array}\)

---