【数学I】例題3.1.7：2次関数のグラフの平行移動2（One More）★★

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% 例題I3.1.7：2次関数のグラフの平行移動2（One More）★★
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(1) 放物線$y=-x^2+6x-4$は放物線$y=-x^2-2x+3$をどのように平行移動したものか．

(2) $x$軸方向に$3$，$y$軸方向に$2$だけ平行移動すると，放物線$y=3x^2-2x+5$になるような放物線$C$の方程式を求めよ．

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% 例題I3.1.7の解答（One More）★★
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(1) $y=-x^2+6x-4=-(x-3)^2+5$より，頂点は点$(3,5)$

$y=-x^2-2x+3=-(x+1)^2+4$より，頂点は点$(-1,4)$

頂点$(-1,4)$が点$(3,5)$に移されるから，

$x$軸方向に$3-(-1)=4$，

$y$軸方向に$5-4=1$

だけ平行移動している．

よって，$x$軸方向に$4$，$y$軸方向に$1$だけ平行移動したものである．

(2) 放物線$y=3x^2-2x+5$において，

$x$軸方向に$-3$，$y$軸方向に$-2$

だけ平行移動したものが放物線$C$である．

したがって，放物線$y=3x^2-2x+5$の$x$を$x+3$，$y$を$y+2$におき換えて，

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

よって，$y=3x^2+16x+24$

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% 例題I3.1.7の別解
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$y=3x^2-2x+5=3(x-\frac{1}{3})^2+\frac{14}{3}$より，頂点は点$(\frac{1}{3},\frac{14}{3})$

したがって，放物線$C$の頂点は点$(\frac{1}{3}-3,\frac{14}{3}-2)$

すなわち，点$(-\frac{8}{3},\frac{8}{3})$

よって，放物線$C$の方程式は，

\[
y=3(x+\frac{8}{3})^2+\frac{8}{3}=3x^2+16x+24
\]

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% 問題I3.1.7：2次関数のグラフの平行移動2（One More）★★
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(1) 放物線$y=x^2+4x+1$は放物線$y=x^2-2x-3$をどのように平行移動したものか．

(2) $x$軸方向に$-2$，$y$軸方向に$3$だけ平行移動すると，放物線$y=-2x^2+5x-7$になるような放物線$C$の方程式を求めよ．

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% 問題I3.1.7の解答（One More）★★
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(1) $y=x^2+4x+1=(x+2)^2-3$より，頂点は点$(-2,-3)$

$y=x^2-2x-3=(x-1)^2-4$より，頂点は点$(1,-4)$

頂点$(1,-4)$が点$(-2,-3)$に移されるから，

$x$軸方向に$-2-1=-3$

$y$軸方向に$-3-(-4)=1$

だけ平行移動している．

よって，$x$軸方向に$-3$，$y$軸方向に$1$だけ平行移動したものである．

(2) 放物線$y=-2x^2+5x-7$において，

$x$軸方向に$2$，$y$軸方向に$-3$

だけ平行移動したものが放物線$C$である．

したがって，放物線$y=-2x^2+5x-7$の$x$を$x-2$，$y$を$y+3$におき換えて，

\[
y+3=-2(x-2)^2+5(x-2)-7
\]

よって，$y=-2x^2+13x-28$

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% 問題I3.1.7の別解
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$y=-2x^2+5x-7=-2(x-\frac{5}{4})^2-\frac{31}{8}$より，頂点は点$(\frac{5}{4},-\frac{31}{8})$

したがって，放物線$C$の頂点は点$(\frac{5}{4}+2,-\frac{31}{8}-3)$

すなわち，点$(\frac{13}{4},-\frac{55}{8})$

よって，放物線$C$の方程式は，

\[
y=-2(x-\frac{13}{4})^2-\frac{55}{8}=-2x^2+13x-28
\]

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% 例題I3.1.7の側注：2次関数のグラフの平行移動2（One More）★★
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% 側注1
頂点の座標を求める．

% 側注2
移動した後の座標から，移動する前の座標を引くことを考える．

% 側注3
与えられた平行移動の逆向きの移動を考える．「$x$軸方向に$3$，$y$軸方向に$2$だけ平行移動」とするのは誤りであるので注意すること．

% 側注4
頂点の移動に注目した解法である．

% 側注5
平行移動しても，$x^2$の係数は変わらない．

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% 問題I3.1.7の側注：2次関数のグラフの平行移動2（One More）★★
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% 側注1
頂点の座標を求める．

% 側注2
移動した後の座標から，移動する前の座標を引くことを考える．

% 側注3
頂点の移動に注目した解法である．

% 側注4
平行移動しても，$x^2$の係数は変わらない．

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% 例題I3.1.7の考え方・補足など：2次関数のグラフの平行移動2（One More）★★
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% 考え方
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(1) 頂点の移動に注目するとよい．

(2) 放物線$y=3x^2-2x+5$を$x$軸方向に$-3$，$y$軸方向に$-2$だけ平行移動すると，放物線$C$が得られる．求めるグラフは，与えられた平行移動の逆向きに平行移動したものであるので注意すること．

\[
C\overset{\begin{array}{l}x\text{軸方向に}p\\y\text{軸方向に}q\end{array}}{\underset{\begin{array}{l}x\text{軸方向に}-p\\y\text{軸方向に}-q\end{array}}{\longleftrightarrow}}C^{\prime}
\]