【数学I】例題3.2.11：2次関数の決定1（One More）★★

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% 例題I3.2.11：2次関数の決定1（One More）★★
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次の条件を満たすような放物線をグラフとする2次関数を求めよ．

(1) 頂点が点$(-3,2)$で，点$(0,5)$を通る．

(2) 軸が直線$x=-1$で，2点$(0,-6),(2,2)$を通る．

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% 例題I3.2.11の解答（One More）★★
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(1) 頂点が点$(-3,2)$であるから，求める2次関数は，

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

と表される．

この関数のグラフが点$(0,5)$を通るから，

\[
5=a(0+3)^2+2
\]

したがって，$a=\frac{1}{3}$

よって，求める2次関数は，$y=\frac{1}{3}(x+3)^2+2$

(2) 軸が直線$x=-1$であるから，求める2次関数は，

\[
y=a(x+1)^2+q
\]

と表される．

この関数のグラフが2点$(0,-6),(2,2)$を通るから，

\[
-6=a(0+1)^2+q,2=a(2+1)^2+q
\]

したがって，$a+q=-6,9a+q=2$

これを解いて，$a=1,q=-7$

よって，求める2次関数は，$y=(x+1)^2-7$

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% 問題I3.2.11：2次関数の決定1（One More）★★
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次の条件を満たすような放物線をグラフとする2次関数を求めよ．

(1) 頂点が点$(-2,3)$で，点$(1,6)$を通る．

(2) 軸が直線$x=2$で，2点$(0,-4),(3,5)$を通る．

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% 問題I3.2.11の解答（One More）★★
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(1) 頂点が点$(-2,3)$であるから，求める2次関数は，

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

と表される．

この関数のグラフが点$(1,6)$を通るから，

\[
6=a(1+2)^2+3
\]

したがって，$a=\frac{1}{3}$

よって，求める2次関数は，$y=\frac{1}{3}(x+2)^2+3$

(2) 軸が直線$x=2$であるから，求める2次関数は，

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

と表される．

この関数のグラフが2点$(0,-4),(3,5)$を通るから，

\[
-4=a(0-2)^2+q,5=a(3-2)^2+q
\]

したがって，$4a+q=-4,a+q=5$

これを解いて，$a=-3,q=8$

よって，求める2次関数は，$y=-3(x-2)^2+8$

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% 例題I3.2.11の側注：2次関数の決定1（One More）★★
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% 側注1
頂点が与えられているから，標準形を用いる．

% 側注2
$y=a(x+3)^2+2$に，$x=0,y=5$を代入する．

% 側注3
$y=\frac{1}{3}x^2+2x+5$でもよい．

% 側注4
軸が与えられているから，標準形を用いる．

% 側注5
$x=0,y=-6$と$x=2,y=2$を，それぞれ$y=a(x+1)^2+q$に代入する．

% 側注6
連立方程式を解く．

% 側注7
$y=x^2+2x-6$でもよい．

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% 問題I3.2.11の側注：2次関数の決定1（One More）★★
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% 側注1
頂点が与えられているので，標準形を用いる．

% 側注2
$y=a(x+2)^2+3$に，$x=1,y=6$を代入する．

% 側注3
軸が与えられているので，標準形を用いる．

% 側注4
$x=0,y=-4$と$x=3,y=5$を，それぞれ$y=a(x-2)^2+q$に代入する．

% 側注5
連立方程式を解く．

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% 例題I3.2.11の考え方・補足など：2次関数の決定1（One More）★★
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頂点$(p,q)$や軸$x=p$に関する条件が与えられたときは，標準形$y=a(x-p)^2+q$を用いて考えるとよい（計算が楽になることが見込まれる）．

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頂点$(p,q)$や軸$x=p$が与えられたときは，標準形$y=a(x-p)^2+q$を用いる．