【数学I】例題3.2.12：2次関数の決定2（One More）★★

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% 例題I3.2.12：2次関数の決定2（One More）★★
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次の3点を通るような放物線をグラフとする2次関数を求めよ．

(1) $(-1,8),(2,8),(-3,-12)$

(2) $(2,0),(-4,0),(1,-10)$

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% 例題I3.2.12の解答（One More）★★
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(1) 求める2次関数を$y=ax^2+bx+c$とする．

この関数のグラフが3点$(-1,8)$，$(2,8)$，$(-3,-12)$を通るから，

\[
\{
\begin{array}{l}
8=a-b+c\cdots(\mathrm{i})\\
8=4a+2b+c\cdots(\mathrm{ii})\\
-12=9a-3b+c\cdots(\mathrm{iii})
\end{array}
\]

$(\mathrm{ii})-(\mathrm{i})$より，$0=3a+3b$，すなわち，$a+b=0\cdots(\mathrm{iv})$

$(\mathrm{ii})-(\mathrm{iii})$より，$20=-5a+5b$，すなわち，$-a+b=4\cdots(\mathrm{v})$

$(\mathrm{iv})$，$(\mathrm{v})$を解いて，$a=-2,b=2$

したがって，これを$(\mathrm{i})$に代入すると，$c=12$

よって，求める2次関数は，$y=-2x^2+2x+12$

(2) $x$軸との共有点の座標が$(2,0),(-4,0)$であるから，求める2次関数は，

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

と表される．

この関数のグラフが点$(1,-10)$を通るから，
\[
-10=a(1-2)(1+4)
\]

したがって，$a=2$

よって，求める2次関数は，$y=2(x-2)(x+4)$

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% 問題I3.2.12：2次関数の決定2（One More）★★
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次の3点を通るような放物線をグラフとする2次関数を求めよ．

(1) $(1,7),(3,7),(-2,-8)$

(2) $(-1,0),(4,0),(2,-12)$

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% 問題I3.2.12の解答（One More）★★
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(1) 求める2次関数を$y=ax^2+bx+c$とする．

この関数のグラフが3点$(1,7)$，$(3,7)$，$(-2,-8)$を通るから，

\[
\{
\begin{array}{l}
7=a+b+c\cdots(\mathrm{i})\\
7=9a+3b+c\cdots(\mathrm{ii})\\
-8=4a-2b+c\cdots(\mathrm{iii})
\end{array}
\]

$(\mathrm{ii})-(\mathrm{i})$より，$0=8a+2b$，すなわち，$4a+b=0\cdots(\mathrm{iv})$

$(\mathrm{ii})-(\mathrm{iii})$より，$15=5a+5b$，すなわち，$a+b=3\cdots(\mathrm{v})$

$(\mathrm{iv})$，$(\mathrm{v})$を解いて，$a=-1,b=4$

したがって，これを$(\mathrm{i})$に代入すると，$c=4$

よって，求める2次関数は，$y=-x^2+4x+4$

(2) $x$軸との共有点の座標が$(-1,0),(4,0)$であるから，求める2次関数は，

\[
y=a(x+1)(x-4)
\]

と表される．

この関数のグラフが点$(2,-12)$を通るから，

\[
-12=a(2+1)(2-4)
\]

したがって，$a=2$

よって，求める2次関数は，$y=2(x+1)(x-4)$

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% 例題I3.2.12の側注：2次関数の決定2（One More）★★
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% 側注1
$y=ax^2+bx+c$に(i)は$x=-1,y=8$，(ii)は$x=2,y=8$，(iii)は$x=-3,y=-12$をそれぞれ代入している．

% 側注2
$c$を消去して，$a,b$についての連立方程式$(\mathrm{iv}),(\mathrm{v})$を解く．

% 側注3
$y=ax^2+bx+c$としてもよいが，連立方程式を解く必要があり，手間が掛かる．

% 側注4
$y=(x-2)(x+4)$とするのは誤りである．$a$を忘れないように注意すること．

% 側注5
$y=2x^2+4x-16$でもよい．

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% 問題I3.2.12の側注：2次関数の決定2（One More）★★
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% 側注1
$y=ax^2+bx+c$に(i)は$x=1,y=7$，(ii)は$x=3,y=7$，(iii)は$x=-2,y=-8$をそれぞれ代入している．

% 側注2
$c$を消去して，$a,b$についての連立方程式$(\mathrm{iv}),(\mathrm{v})$を解く．

% 側注3
$y=2x^2-6x-8$でもよい．

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% 例題I3.2.12の考え方・補足など：2次関数の決定2（One More）★★
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% 考え方
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(1) 3点が与えられたときは，一般形$y=ax^2+bx+c$を用いて考えるとよい．通る3点の座標を代入することで，$a,b,c$の連立方程式を作ることができる．

(2) $x$軸との交点が2つ与えられたときは，$y=a(x-\alpha)(x-\beta)$を用いて考えるとよい．一般に，2次関数のグラフが$x$軸と2点$(\alpha,0),(\beta,0)$で交わるとき，$f(\alpha)=0,f(\beta)=0$となることから，$y=a(x-\alpha)(x-\beta)$と表すことができる．

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3点が与えられたときは，$y=ax^2+bx+c$を用いる．

$x$軸との共有点が与えられたときは，$y=a(x-\alpha)(x-\beta)$を用いる．