【数学I】例題3.3.12：2次関数のグラフと係数の符号（One More）★★

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% 例題I3.3.12：2次関数のグラフと係数の符号（One More）★★
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2次関数$y=ax^2+bx+c$のグラフが右の図のようなとき，次の値の符号を調べよ．

(1) $a$

(2) $b$

(3) $c$

(4) $b^2-4ac$

(5) $a+b+c$

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% 例題I3.3.12の解答（One More）★★
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(1) グラフは下に凸であるから，$a>0$

(2) 軸は直線$x=-\frac{b}{2a}$

軸が$y$軸より右側にあるから，$-\frac{b}{2a}>0$

よって，(1)より，$a>0$であるから，$b<0$

(3) $y$軸との共有点は点$(0,c)$

よって，グラフは$y$軸と$y>0$の部分で交わっているから，$c>0$

(4) 2次方程式$ax^2+bx+c=0$の判別式を$D$とすると，$D=b^2-4ac$

$x$軸と異なる2点で交わっているから，$D>0$

よって，$b^2-4ac>0$

(5) $x=1$のとき，$y=a\cdot 1^2+b\cdot 1+c=a+b+c$

よって，グラフより，$x=1$のとき，$y<0$であるから，$a+b+c<0$

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% 問題I3.3.12：2次関数のグラフと係数の符号（One More）★★
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2次関数$y=ax^2+bx+c$のグラフが右の図のようなとき，次の値の符号を調べよ．ただし，$a<0$とする．

(1) $a$

(2) $b$

(3) $c$

(4) $b^2-4ac$

(5) $a+b+c$

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% 問題I3.3.12の解答（One More）★★
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(1) グラフは上に凸であるから，$a<0$

(2) 軸は直線$x=-\frac{b}{2a}$

軸が$y$軸より右側にあるから，$-\frac{b}{2a}>0$

よって，$a<0$であるから，$b>0$

(3) $y$軸との共有点は点$(0,c)$

よって，グラフは$y$軸と$y>0$の部分で交わっているから，$c>0$

(4) 2次方程式$ax^2+bx+c=0$の判別式を$D$とすると，$D=b^2-4ac$

$x$軸と異なる2点で交わっているから，$D>0$

よって，$b^2-4ac>0$

(5) $x=1$のとき，$y=a\cdot 1^2+b\cdot 1+c=a+b+c$

よって，グラフより，$x=1$のとき，$y>0$であるから，$a+b+c>0$

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% 例題I3.3.12の側注：2次関数のグラフと係数の符号（One More）★★
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% 側注1
$y=ax^2+bx+c=a(x+\frac{b}{2a})^2-\frac{b^2-4ac}{4a}$

% 側注2
$-\frac{b}{2a}>0$から，$\frac{b}{a}<0$より，$a$と$b$は異符号である．

% 側注3
$y$軸との共有点は，$y=ax^2+bx+c$に$x=0$を代入すると，

\[
y=a\cdot 0^2+b\cdot 0+c=c
\]
よって，点$(0,c)$である．

% 側注4
$D=b^2-4ac$

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% 問題I3.3.12の側注：2次関数のグラフと係数の符号（One More）★★
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% 側注1
$y=ax^2+bx+c=a(x+\frac{b}{2a})^2-\frac{b^2-4ac}{4a}$

% 側注2
$-\frac{b}{2a}>0$から，$\frac{b}{a}<0$より，$a$と$b$は異符号である．

% 側注3
$D=b^2-4ac$

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% 例題I3.3.12の考え方・補足など：2次関数のグラフと係数の符号（One More）★★
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% 考え方
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(1) $a$の符号$\cdots$上に凸（$a<0$）か，下に凸（$a>0$）かを調べる．

(2) $b$の符号$\cdots$軸は直線$x=-\frac{b}{2a}$であり，その位置を調べる．このとき，$a$の符号に注意すること．

(3) $c$の符号$\cdots$グラフと$y$軸との共有点を調べる．$x=0$のとき，$y=c$となる．

(4) $b^2-4ac$の符号$\cdots$判別式$D=b^2-4ac$より，グラフと$x$軸との共有点の個数を調べる．なお，頂点$(-\frac{b}{2a},-\frac{b^2-4ac}{4a})$から符号を調べてもよい．

(5) $a+b+c$の符号$\cdots$ $x=1$のときの$y$の値を調べる．$x=1$のとき，$y=a+b+c$となる．