1.次の二次関数のグラフをかき、頂点と軸を答えなさい。
(1)\(y=x^2+1\)
頂点:\((0,1)\)
軸:\(x=0\)
(2)\(y=2x^2-3\)
頂点:\((0,-3)\)
軸:\(x=0\)
(3)\(y=-x^2+2\)
頂点:\((0,2)\)
軸:\(x=0\)
(4)\(y=(x-2)^2\)
頂点:\((2,0)\)
軸:\(x=2\)
(5)\(y=2(x+1)^2\)
頂点:\((-1,0)\)
軸:\(x=-1\)
(6)\(y=-2(x+2)^2\)
頂点:\((-2,0)\)
軸:\(x=-2\)
(7)\(y=(x-1)^2+2\)
頂点:\((1,2)\)
軸:\(x=1\)
(8)\(y=2(x-2)^2-4\)
頂点:\((2,-4)\)
軸:\(x=2\)
(9)\(y=-2(x+1)^2+2\)
頂点:\((-1,2)\)
軸:\(x=-1\)
(10)\(\displaystyle y=-\frac{1}{2}(x+2)^2-1\)
頂点:\((-2,-1)\)
軸:\(x=-2\)
2.次の二次式を平方完成しなさい。
(1)\(y=x^2+8x\)
\begin{align} y &= x^2+8x \\ &= x^2+8x+16-16 \\ &= (x+4)^2-16 \end{align}
(2)\(y=x^2-6x+8\)
\begin{align} y &= x^2-6x+8 \\ &= x^2-6x+9-9+8 \\ &= (x-3)^2-1 \end{align}
(3)\(2x^2+4x+5\)
\begin{align} y &= 2x^2+4x+5 \\ &= 2(x^2+2x)+5 \\ &= 2(x^2+2x+1-1)+5 \\ &= 2(x+1)^2+3 \end{align}
(4)\(y=3x^2-6x-2\)
\begin{align} y &= 3x^2-6x-2 \\ &= 3(x^2-2x)-2 \\ &= 3(x^2-2x+1-1)-2 \\ &= 3(x-1)^2-5 \end{align}
(5)\(y=x^2+x-2\)
\begin{align} y &= x^2+x-2 \\ &= x^2+x+\frac{1}{4}-\frac{1}{4}-2 \\ &= \left(x+\frac{1}{2}\right)^2-\frac{9}{4} \end{align}
(6)\(y=-2x^2+6x+4\)
\begin{align} y &= -2x^2+6x+4 \\ &= -2(x^2-3x)+4 \\ &= -2\left(x^2-3x+\frac{9}{4}-\frac{9}{4}\right)+4 \\ &= -2\left(x-\frac{3}{2}\right)^2+\frac{17}{2} \end{align}
(7)\(y=2x^2+8x+10\)
\begin{align} y &= 2x^2+8x+10 \\ &= 2(x^2+4x)+10 \\ &= 2(x^2+4x+4-4)+10 \\ &= 2(x+2)^2+2 \end{align}
(8)\(y=3x^2+6x-10\)
\begin{align} y &= 3x^2+6x-10 \\ &= 3(x^2+2x)-10 \\ &= 3(x^2+2x+1-1)-10 \\ &= 3(x+1)^2-13 \end{align}
(9)\(y=-x^2-4x-1\)
\begin{align} y &= -x^2-4x-1 \\ &= -(x^2+4x)-1 \\ &= -(x^2+4x+4-4)-1 \\ &= -(x+2)^2+3 \end{align}
(10)\(y=-3x^2-9x+4\)
\begin{align} y &= -3x^2-9x+4 \\ &= -3(x^2+3x)+4 \\ &= -3\left(x^2+3x+\frac{9}{4}-\frac{9}{4}\right)+4 \\ &= -3\left(x+\frac{3}{2}\right)^2+\frac{43}{4} \end{align}
3.次の二次関数のグラフをかき、頂点と軸を答えなさい。
(1)\(y=x^2-4x+3\)
\begin{align}
y &= x^2-4x+3 \\
&= x^2-4x+4-4+3 \\
&= (x-2)^2-1
\end{align}
頂点:\((2,-1)\)
軸:\(x=2\)
(2)\(y=2x^2+8x+3\)
\begin{align}
y &= 2x^2+8x+3 \\
&= 2(x^2+4x)+3 \\
&= 2(x^2+4x+4-4)+3 \\
&= 2(x+2)^2-5
\end{align}
頂点:\((-2,-5)\)
軸:\(x=-2\)
(3)\(y=-3x^2+6x+1\)
\begin{align}
y &= -3x^2+6x+1 \\
&= -3(x^2-2x)+1 \\
&= -3(x^2-2x+1-1)+1 \\
&= -3(x-1)^2+4
\end{align}
頂点:\((1,4)\)
軸:\(x=1\)
(4)\(y=-x^2+3x\)
\begin{align}
y &= -x^2+3x \\
&= -(x^2-3x) \\
&= -\left(x^2-3x+\frac{9}{4}-\frac{9}{4}\right) \\
&= -\left(x-\frac{3}{2}\right)^2+\frac{9}{4}
\end{align}
頂点:\(\displaystyle \left(\frac{3}{2},\frac{9}{4}\right)\)
軸:\(\displaystyle x=\frac{3}{2}\)
(5)\(y=-x^2+2x+1\)
\begin{align}
y &= -x^2+2x+1 \\
&= -(x^2-2x)+1 \\
&= -(x^2-2x+1-1)+1 \\
&= -(x-1)^2+2
\end{align}
頂点:\((1,2)\)
軸:\(x=1\)
(6)\(y=-3x^2-2x+1\)
\begin{align}
y &= -3x^2-2x+1 \\
&= -3\left(x^2+\frac{2}{3}x\right)+1 \\
&= -3\left(x^2+\frac{2}{3}x+\frac{1}{9}-\frac{1}{9}\right)+1 \\
&= -3\left(x+\frac{1}{3}\right)^2+\frac{4}{3}
\end{align}
頂点:\(\displaystyle \left(-\frac{1}{3},\frac{4}{3}\right)\)
軸:\(\displaystyle x=-\frac{1}{3}\)