【数学I】例題3.1.5：2次関数のグラフ2（One More）★

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% 例題I3.1.5：2次関数のグラフ2（One More）★
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次の2次関数のグラフをかき，その軸と頂点を求めよ．

(1) $y=3x^2-6x+2$

(2) $y=-2x^2+4x-5$

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% 例題I3.1.5の解答（One More）★
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(1)

\begin{align*}
y&=3x^2-6x+2\\
&=3(x^2-2x)+2\\
&=3\{(x-1)^2-1^2\}+2\\
&=3(x-1)^2-3\cdot 1^2+2\\
&=3(x-1)^2-1
\end{align*}

よって，グラフは右の図のようになる．

軸は直線$x=1$

頂点は点$(1,-1)$

(2)

\begin{align*}
y&=-2x^2+4x-5\\
&=-2(x^2-2x)-5\\
&=-2\{(x-1)^2-1^2\}-5\\
&=-2(x-1)^2+2\cdot 1^2-5\\
&=-2(x-1)^2-3
\end{align*}

よって，グラフは右の図のようになる．

軸は直線$x=1$

頂点は点$(1,-3)$

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% 問題I3.1.5：2次関数のグラフ2（One More）★
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次の2次関数のグラフをかき，その軸と頂点を求めよ．

(1) $y=2x^2+8x+3$

(2) $y=-x^2+6x-8$

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% 問題I3.1.5の解答（One More）★
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(1)

\begin{align*}
y&=2x^2+8x+3\\
&=2(x^2+4x)+3\\
&=2\{(x+2)^2-2^2\}+3\\
&=2(x+2)^2-8+3\\
&=2(x+2)^2-5
\end{align*}

よって，グラフは右の図のようになる．

軸は直線$x=-2$

頂点は点$(-2,-5)$

(2)

\begin{align*}
y&=-x^2+6x-8\\
&=-(x^2-6x)-8\\
&=-\{(x-3)^2-3^2\}-8\\
&=-(x-3)^2+9-8\\
&=-(x-3)^2+1
\end{align*}

よって，グラフは右の図のようになる．

軸は直線$x=3$

頂点は点$(3,1)$

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% 例題I3.1.5の側注：2次関数のグラフ2（One More）★
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% 側注1
$3$でくくり出す．

% 側注2
$x$の係数の半分を2乗する．

% 側注3
$y=a(x-p)^2+q$の形であるから，軸や頂点を求めることができ，グラフがかける．

% 側注4
$-2$でくくり出す．

% 側注5
$x$の係数の半分を2乗する．

% 側注6
$-2$を分配して展開する．

% 側注7
なお，平方完成した後は，展開すればもとの式に戻るか検算することができる．

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% 問題I3.1.5の側注：2次関数のグラフ2（One More）★
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% 側注1
$2$でくくり出す．

% 側注2
$x$の係数の半分を2乗する．

% 側注3
$y=a(x-p)^2+q$の形であるから，軸や頂点を求めることができ，グラフがかける．

% 側注4
$-1$でくくり出す．

% 側注5
$x$の係数の半分を2乗する．

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% 例題I3.1.5の考え方・補足など：2次関数のグラフ2（One More）★
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% 考え方
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一般形$y=ax^2+bx+c$の右辺を標準形$y=a(x-p)^2+q$の形に平方完成して，グラフをかく．このとき，$x^2+mx=(x+\frac{m}{2})^2-(\frac{m}{2})^2$の利用を考えるとよい．

[width=5.5cm]$x^2+mx=(x+\underbrace{\frac{m}{2}}_{m\text{の半分}})^2-(\underbrace{\frac{m}{2}}_{m\text{の半分}})^2$