【数学I】例題3.1.9：2次関数の平行移動と対称移動（One More）★★★

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% 例題I3.1.9：2次関数の平行移動と対称移動（One More）★★★
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放物線$y=ax^2+bx+c\cdots(\mathrm{i})$を$x$軸方向に$1$，$y$軸方向に$5$だけ平行移動し，さらに$y$軸に関して対称移動すると，放物線$y=2x^2+8x+3\cdots(\mathrm{ii})$になった．このとき，定数$a,b,c$の値を求めよ．

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% 例題I3.1.9の解答（One More）★★★
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(ii) を$y$軸に関して対称移動した放物線の方程式は，

\[
y=2(-x)^2+8(-x)+3
\]

すなわち，$y=2x^2-8x+3\cdots(\mathrm{iii})$

(iii) を$x$軸方向に$-1$，$y$軸方向に$-5$だけ平行移動するから，

\[
y+5=2(x+1)^2-8(x+1)+3
\]

すなわち，$y=2x^2-4x-8$

よって，$y=2x^2-4x-8$が(i)と一致するから，係数を比較すると，

\[
a=2,b=-4,c=-8
\]

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% 例題I3.1.9の別解
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(i)のグラフを$x$軸方向に$1$，$y$軸方向に$5$だけ平行移動した放物線の方程式は，$y-5=a(x-1)^2+b(x-1)+c$

したがって，$y=ax^2+(b-2a)x+a-b+c+5$

$y$軸に関して対称移動すると，

\[
y=a(-x)^2+(b-2a)(-x)+a-b+c+5
\]

ゆえに，$y=ax^2+(2a-b)x+a-b+c+5$

したがって，$y=ax^2+(2a-b)x+a-b+c+5$が(ii)と一致するから，係数を比較すると，$a=2,2a-b=8,a-b+c+5=3$

よって，これを解いて，$a=2,b=-4,c=-8$

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% 問題I3.1.9：2次関数の平行移動と対称移動（One More）★★★
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放物線$y=ax^2+bx+c\cdots(\mathrm{i})$を$x$軸方向に$2$，$y$軸方向に$-3$だけ平行移動し，さらに$y$軸に関して対称移動すると，放物線$y=-x^2+6x+4\cdots(\mathrm{ii})$になった．このとき，定数$a,b,c$の値を求めよ．

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% 問題I3.1.9の解答（One More）★★★
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(ii) を$y$軸に関して対称移動した放物線の方程式は，

\[
y=-(-x)^2+6(-x)+4
\]

すなわち，$y=-x^2-6x+4\cdots(\mathrm{iii})$

(iii) を$x$軸方向に$-2$，$y$軸方向に$3$だけ平行移動するから，

\[
y-3=-(x+2)^2-6(x+2)+4
\]

すなわち，$y=-x^2-10x-9$

よって，$y=-x^2-10x-9$が(i)と一致するから，係数を比較すると，

\[
a=-1,b=-10,c=-9
\]

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% 問題I3.1.9の別解
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(i) のグラフを$x$軸方向に$2$，$y$軸方向に$-3$だけ平行移動した放物線の方程式は，$y+3=a(x-2)^2+b(x-2)+c$

したがって，$y=ax^2+(b-4a)x+(4a-2b+c-3)$

$y$軸に関して対称移動すると，

\[
y=a(-x)^2+(b-4a)(-x)+(4a-2b+c-3)
\]

ゆえに，$y=ax^2+(4a-b)x+(4a-2b+c-3)$

したがって，$y=ax^2+(4a-b)x+(4a-2b+c-3)$が(ii)と一致するから，係数を比較すると，$a=-1,4a-b=6,4a-2b+c-3=4$

よって，これを解いて，$a=-1,b=-10,c=-9$

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% 例題I3.1.9の側注：2次関数の平行移動と対称移動（One More）★★★
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% 側注1
与えられた平行移動の逆向きの平行移動を考える．「$y$軸に関して対称移動」の逆の移動は，そのまま，「$y$軸に関して対称移動」である．

% 側注2
標準形にして，頂点の移動に注目して解いてもよい．

% 側注3
$x$を$x-(-1)$，すなわち，$x+1$，$y$を$y-(-5)$，すなわち，$y+5$におき換える．

% 側注4
$y=ax^2+bx+c\cdots(\mathrm{i})$と係数を比較する．

% 側注5
先に$y=ax^2+bx+c$からグラフの移動を考えた解法である．頂点の移動に注目して解くと，(i)の頂点は
\[
(-\frac{b}{2a},-\frac{b^2-4ac}{4a})
\]
となり，計算に更に手間が掛かる．

% 側注6
$y=2x^2+8x+3\cdots(\mathrm{ii})$と係数を比較する．

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% 問題I3.1.9の側注：2次関数の平行移動と対称移動（One More）★★★
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% 側注1
標準形にして，頂点の移動に注目して解いてもよい．

% 側注2
$y=ax^2+bx+c\cdots(\mathrm{i})$と係数を比較する．

% 側注3
$y=-x^2+6x+4\cdots(\mathrm{ii})$と係数を比較する．

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% 例題I3.1.9の考え方・補足など：2次関数の平行移動と対称移動（One More）★★★
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% 考え方
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グラフが複数回の移動をする問題では，移動の順序に注意する．なお，先に$y=ax^2+bx+c$からグラフの移動を考えてもよいが，計算に手間が掛かる（別解）．そこで，先に$y=2x^2+8x+3$からグラフの移動を考えるとよい．このとき，与えられた平行移動の逆向きに平行移動したものを考えることになるので，符号に注意すること．

\[
(\mathrm{i})y=ax^2+bx+c\overset{\begin{array}{l}x\text{軸方向に}1\\y\text{軸方向に}5\end{array}}{\underset{\begin{array}{l}x\text{軸方向に}-1\\y\text{軸方向に}-5\end{array}}{\longleftrightarrow}}(\mathrm{iii})\overset{y\text{軸に関して対称移動}}{\underset{y\text{軸に関して対称移動}}{\longleftrightarrow}}(\mathrm{ii})y=2x^2+8x+3
\]