【数学I】例題3.1.8：2次関数のグラフの対称移動（One More）★

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% 例題I3.1.8：2次関数のグラフの対称移動（One More）★
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放物線$y=x^2-2x+3$を，次のように移動した放物線の方程式を求めよ．

(1) $x$軸に関して対称移動

(2) $y$軸に関して対称移動

(3) 原点に関して対称移動

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% 例題I3.1.8の解答（One More）★
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(1) $y$を$-y$におき換えて，$-y=x^2-2x+3$

よって，$y=-x^2+2x-3$

(2) $x$を$-x$におき換えて，$y=(-x)^2-2(-x)+3$

よって，$y=x^2+2x+3$

(3) $x$を$-x$，$y$を$-y$におき換えて，$-y=(-x)^2-2(-x)+3$

よって，$y=-x^2-2x-3$

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% 例題I3.1.8の別解
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$y=x^2-2x+3=(x-1)^2+2$より，頂点は点$(1,2)$で下に凸の放物線である．

(1) 点$(1,2)$を$x$軸に関して対称移動すると，$(1,-2)$

移動したグラフは上に凸であるから，$y=-(x-1)^2-2$

(2) 点$(1,2)$を$y$軸に関して対称移動すると，$(-1,2)$

移動したグラフは下に凸であるから，$y=(x+1)^2+2$

(3) 点$(1,2)$を原点に関して対称移動すると，$(-1,-2)$

移動したグラフは上に凸であるから，$y=-(x+1)^2-2$

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% 問題I3.1.8：2次関数のグラフの対称移動（One More）★
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放物線$y=-x^2+4x-5$を，次のように移動した放物線の方程式を求めよ．

(1) $x$軸に関して対称移動

(2) $y$軸に関して対称移動

(3) 原点に関して対称移動

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% 問題I3.1.8の解答（One More）★
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(1) $y$を$-y$におき換えて，$-y=-x^2+4x-5$

よって，$y=x^2-4x+5$

(2) $x$を$-x$におき換えて，$y=-(-x)^2+4(-x)-5$

よって，$y=-x^2-4x-5$

(3) $x$を$-x$，$y$を$-y$におき換えて，$-y=-(-x)^2+4(-x)-5$

よって，$y=x^2+4x+5$

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% 問題I3.1.8の別解
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$y=-x^2+4x-5=-(x-2)^2-1$より，頂点は点$(2,-1)$で上に凸の放物線である．

(1) 点$(2,-1)$を$x$軸に関して対称移動すると，$(2,1)$

移動したグラフは下に凸であるから，$y=(x-2)^2+1$

(2) 点$(2,-1)$を$y$軸に関して対称移動すると，$(-2,-1)$

移動したグラフは上に凸であるから，$y=-(x+2)^2-1$

(3) 点$(2,-1)$を原点に関して対称移動すると，$(-2,1)$

移動したグラフは下に凸であるから，$y=(x+2)^2+1$

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% 例題I3.1.8の側注：2次関数のグラフの対称移動（One More）★
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% 側注1
答えを記すときは，標準形でも一般形でもよい．

% 側注2
$x^2$の係数の符号が変わり，上に凸のグラフとなる．

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% 問題I3.1.8の側注：2次関数のグラフの対称移動（One More）★
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% 側注1
答えを記すときは，標準形でも一般形でもよい．

% 側注2
$x^2$の係数の符号が変わり，下に凸のグラフとなる．

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% 例題I3.1.8の考え方・補足など：2次関数のグラフの対称移動（One More）★
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% 考え方
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関数$y=f(x)$のグラフをそれぞれ$x$軸，$y$軸，原点に関して対称移動すると，次のようになる．$x$軸$\cdots-y=f(x)$

$y$軸$\cdots y=f(-x)$

原点$\cdots-y=f(-x)$

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% 2次関数のグラフの対称移動
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$x$軸に関して対称移動$\cdots y$を$-y$におき換える．

$y$軸に関して対称移動$\cdots x$を$-x$におき換える．

原点に関して対称移動$\cdots x$を$-x$，$y$を$-y$におき換える．