【高校数学Ⅰ】2-1-2 二次関数のグラフ｜問題集

頂点:\((0,1)\)
軸:\(x=0\)

頂点:\((0,-3)\)
軸:\(x=0\)

頂点:\((0,2)\)
軸:\(x=0\)

頂点:\((2,0)\)
軸:\(x=2\)

頂点:\((-1,0)\)
軸:\(x=-1\)

頂点:\((-2,0)\)
軸:\(x=-2\)

頂点:\((1,2)\)
軸:\(x=1\)

頂点:\((2,-4)\)
軸:\(x=2\)

頂点:\((-1,2)\)
軸:\(x=-1\)

頂点:\((-2,-1)\)
軸:\(x=-2\)

\(=x^2+8x+16-16\)
\(=(x+4)^2-16\)

\(=x^2-6x+9-9+8\)
\(=(x-3)^2-1\)

\(=2(x^2+2x)+5\)
\(=2(x^2+2x+1-1)+5\)
\(=2(x+1)^2+3\)

\(=3(x^2-2x)-2\)
\(=3(x^2-2x+1-1)-2\)
\(=3(x-1)^2-5\)

\(\displaystyle =x^2+x+\frac{1}{4}-\frac{1}{4}-2\)
\(\displaystyle =\left(x+\frac{1}{2}\right)^2-\frac{9}{4}\)

\(=-2(x^2-3x)+4\)
\(\displaystyle =-2\left(x^2-3x+\frac{9}{4}-\frac{9}{4}\right)+4\)
\(\displaystyle =-2\left(x-\frac{3}{2}\right)^2+\frac{17}{2}\)

\(=2(x^2+4x)+10\)
\(=2(x^2+4x+4-4)+10\)
\(=2(x+2)^2+2\)

\(=3(x^2+2x)-10\)
\(=3(x^2+2x+1-1)-10\)
\(=3(x+1)^2-13\)

\(=-(x^2+4x)-1\)
\(=-(x^2+4x+4-4)-1\)
\(=-(x+2)^2+3\)

\(=-3(x^2+3x)+4\)
\(\displaystyle =-3\left(x^2+3x+\frac{9}{4}-\frac{9}{4}\right)+4\)
\(\displaystyle =-3\left(x+\frac{3}{2}\right)^2+\frac{43}{4}\)

\(=x^2-4x+4-4+3\)
\(=(x-2)^2-1\)
頂点:\((2,-1)\)
軸:\(x=2\)

\(=2(x^2+4x)+3\)
\(=2(x^2+4x+4-4)+3\)
\(=2(x+2)^2-5\)
頂点:\((-2,-5)\)
軸:\(x=-2\)

\(=-3(x^2-2x)+1\)
\(=-3(x^2-2x+1-1)+1\)
\(=-3(x-1)^2+4\)
頂点:\((1,4)\)
軸:\(x=1\)

\(=-(x^2-3x)\)
\(\displaystyle =-\left(x^2-3x+\frac{9}{4}-\frac{9}{4}\right)\)
\(\displaystyle =-\left(x-\frac{3}{2}\right)^2+\frac{9}{4}\)
頂点:\(\displaystyle \left(\frac{3}{2},\frac{9}{4}\right)\)
軸:\(\displaystyle x=\frac{3}{2}\)

\(=-(x^2-2x)+1\)
\(=-(x^2-2x+1-1)+1\)
\(=-(x-1)^2+2\)
頂点:\((1,2)\)
軸:\(x=1\)

\(\displaystyle =-3\left(x^2+\frac{2}{3}x\right)+1\)
\(\displaystyle =-3\left(x^2+\frac{2}{3}x+\frac{1}{9}-\frac{1}{9}\right)+1\)
\(\displaystyle =-3\left(x+\frac{1}{3}\right)^2+\frac{4}{3}\)
頂点:\(\displaystyle \left(-\frac{1}{3},\frac{4}{3}\right)\)
軸:\(\displaystyle x=-\frac{1}{3}\)