【数学I】例題3.1.10：絶対値記号を含む関数のグラフ1（One More）★★

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% 例題I3.1.10：絶対値記号を含む関数のグラフ1（One More）★★
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次の関数のグラフをかけ．

(1) $y=|x-1|$

(2) $y=|x^2+2x-3|$

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% 例題I3.1.10の解答（One More）★★
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(1)

(i) $x-1\geqq 0$，すなわち，$x\geqq 1$のとき

\[
y=x-1
\]

(ii) $x-1<0$，すなわち，$x<1$のとき

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

よって，(i)，(ii)より，グラフは右上の図のようになる．

(2)

$x^2+2x-3=(x+3)(x-1)$より，

$x^2+2x-3\geqq 0$のとき，$x\leqq-3,1\leqq x$

$x^2+2x-3<0$のとき，$-3<x<1$

(i) $x\leqq-3,1\leqq x$のとき

\[
y=|x^2+2x-3|=x^2+2x-3=(x+1)^2-4
\]

(ii) $-3<x<1$のとき

\[
y=|x^2+2x-3|=-x^2-2x+3=-(x+1)^2+4
\]

よって，(i)，(ii)より，グラフは右上の図のようになる．

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% 問題I3.1.10：絶対値記号を含む関数のグラフ1（One More）★★
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次の関数のグラフをかけ．

(1) $y=|2x+3|$

(2) $y=|x^2-4x+3|$

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% 問題I3.1.10の解答（One More）★★
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(1)

(i) $2x+3\geqq 0$，すなわち，$x\geqq-\frac{3}{2}$のとき

\[
y=2x+3
\]

(ii) $2x+3<0$，すなわち，$x<-\frac{3}{2}$のとき

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

よって，(i)，(ii)より，グラフは右上の図のようになる．

(2)

$x^2-4x+3=(x-3)(x-1)$より，

$x^2-4x+3\geqq 0$のとき，$x\leqq 1,3\leqq x$

$x^2-4x+3<0$のとき，$1<x<3$

(i) $x\leqq 1,x\geqq 3$のとき

\[
y=|x^2-4x+3|=x^2-4x+3=(x-2)^2-1
\]

(ii) $1<x<3$のとき

\[
y=|x^2-4x+3|=-(x^2-4x+3)=-(x-2)^2+1
\]

よって，(i)，(ii)より，グラフは右上の図のようになる．

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% 例題I3.1.10の側注：絶対値記号を含む関数のグラフ1（One More）★★
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% 側注1
$x\geqq 1,x<1$の代わりに，$x>1,x\leqq 1$のように，$x=1$はどちらの範囲に含めて記してもよいが，解答のように記すことが多い．

% 側注2
絶対値記号内が$0$になる値を境に場合分けをする．

% 側注3
なお，$y=|x-1|$は，
\[
y=\{\begin{array}{r}x-1(x\geqq 1)\\-x+1(x<1)\end{array}
\]
のように表されることもある．

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% 側注1
絶対値記号内が$0$になる値を境に場合分けをする．

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% 例題I3.1.10の考え方・補足など：絶対値記号を含む関数のグラフ1（One More）★★
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% 考え方
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絶対値記号内が$0$になる値を境に，場合分けをして絶対値記号を外す．

(1) $|x-1|=\{\begin{array}{ll}x-1&(x-1\geqq 0)\\-(x-1)&(x-1<0)\end{array}=\{\begin{array}{ll}x-1&(x\geqq 1)\\-(x-1)&(x<1)\end{array}$より，$x=1$を境に，場合分けをする．

(2) $x^2+2x-3=(x+3)(x-1)$より，$x=-3,1$を境に，場合分けをする．

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(1)において，$y=x-1$のグラフと，$y=-x+1$のグラフは，$x$軸に関して対称なグラフとなっている．$|x-1|\geqq 0$であるから，$y=|x-1|$のグラフは常に$x$軸より上側にある．また，$y=|x-1|$のグラフは，$y=x-1$のグラフの$x$軸より下側の部分を$x$軸に関して対称に折り返したものになっている．

一般に，$y=|f(x)|$のグラフは，$y=f(x)$のグラフを$x$軸の上下に分けて，$y\geqq 0$（上側）にある部分はそのまま，$y<0$（下側）にある部分は$x$軸に関して対称に折り返したものになる．