【数学I】例題3.1.6：2次関数のグラフの平行移動1（One More）★★

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% 例題I3.1.6：2次関数のグラフの平行移動1（One More）★★
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放物線$y=-x^2+8x-12$を$x$軸方向に$-1$，$y$軸方向に$2$だけ平行移動して得られる放物線の方程式を求めよ．

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% 例題I3.1.6の解答（One More）★★
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放物線$y=-x^2+8x-12$の$x$を$x-(-1)$，すなわち，$x+1$，$y$を$y-2$におき換えると，

\[
y-2=-(x+1)^2+8(x+1)-12
\]

よって，求める放物線の方程式は，$y=-x^2+6x-3$

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% 例題I3.1.6の別解
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\begin{align*}
y=&-(x^2-8x)-12\\
=&-\{(x-4)^2-4^2\}-12\\
=&-(x-4)^2+16-12\\
=&-(x-4)^2+4
\end{align*}

したがって，もとの放物線$y=-x^2+8x-12$の頂点は点$(4,4)$である．

この頂点を平行移動すると，点$(4-1,4+2)$

すなわち，点$(3,6)$になる．

よって，求める放物線の方程式は，$y=-(x-3)^2+6$

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% 問題I3.1.6：2次関数のグラフの平行移動1（One More）★★
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放物線$y=x^2-6x+5$を$x$軸方向に$2$，$y$軸方向に$-3$だけ平行移動して得られる放物線の方程式を求めよ．

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% 問題I3.1.6の解答（One More）★★
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放物線$y=x^2-6x+5$の$x$を$x-2$，$y$を$y+3$におき換えると，

\[
y+3=(x-2)^2-6(x-2)+5
\]

よって，求める放物線の方程式は，$y=x^2-10x+18$

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% 問題I3.1.6の別解
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\begin{align*}
y&=x^2-6x+5\\
&=(x^2-6x)+5\\
&=(x-3)^2-9+5\\
&=(x-3)^2-4
\end{align*}

したがって，もとの放物線$y=x^2-6x+5$の頂点は点$(3,-4)$である．

この頂点を平行移動すると，点$(3+2,-4-3)$

すなわち，点$(5,-7)$になる．

よって，求める放物線の方程式は，$y=(x-5)^2-7$

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% 例題I3.1.6の側注：2次関数のグラフの平行移動1（One More）★★
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% 側注1
平方完成する．

% 側注2
符号に注意すること．

% 側注3
$x+(-1),y+2$におき換えるのは誤りである．符号に注意すること．

% 側注4
移項して整理する．

% 側注5
$x^2$の係数は変わらない．なお，$y=-x^2+6x-3$でもよい．

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% 問題I3.1.6の側注：2次関数のグラフの平行移動1（One More）★★
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% 側注1
平方完成する．

% 側注2
符号に注意すること．

% 側注3
$x-2,y+3$におき換える．

% 側注4
移項して整理する．

% 側注5
$x^2$の係数は変わらない．$y=x^2-10x+18$としてもよい．

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% 例題I3.1.6の考え方・補足など：2次関数のグラフの平行移動1（One More）★★
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% 考え方
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放物線$y=ax^2+bx+c$を$x$軸方向に$p,y$軸方向に$q$だけ平行移動して得られる放物線の方程式は，

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

これは，$y=ax^2+bx+c$の$x$を$x-p$，$y$を$y-q$におき換えたものになっている．

なお，別解のように放物線の方程式を標準形に変形して，頂点の移動に注目して解いてもよい．このとき，平行移動しても$x^2$の係数は変わらないので注意すること．

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% Notes
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放物線は平行移動しても，$x^2$の係数は変わらない．

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% 注意
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別解において，頂点である点$(3,6)$から放物線の方程式を求めるとき，$y=(x-3)^2+6$などとするのは誤りである．平行移動しても$x^2$の係数は変わらないので，$y=-(x-3)^2+6$とするように注意すること．