与えられた2次関数を平方完成させる問題です。具体的には、以下の10個の関数を平方完成させます。 (1) $y = x^2 + x - 2$ (2) $y = x^2 - 4x - 2$ (3) $y = x^2 - 4x$ (4) $y = x^2 + 3x + 2$ (5) $y = x^2 - 4x - 4$ (6) $y = x^2 - 5x + 5$ (7) $y = x^2 + x - 3$ (8) $y = x^2 - 3x - 4$ (9) $y = x^2 - 3x$ (10) $y = x^2 + 2x + 5$

## 問題の解答

1. 問題の内容

与えられた2次関数を平方完成させる問題です。具体的には、以下の10個の関数を平方完成させます。

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

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

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

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

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

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

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

(8) $y = x^2 - 3x - 4$

(9) $y = x^2 - 3x$

(10) $y = x^2 + 2x + 5$

2. 解き方の手順

平方完成とは、2次関数を $y = a(x - p)^2 + q$ の形に変形することです。手順は以下の通りです。

(1) $x^2$ の係数で括る(今回は全て1なので省略)

(2) $x$ の係数の半分の2乗を足して引く

(3) $(x + p)^2$ の形を作る

(4) 定数項をまとめる

3. 最終的な答え

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

$y = (x + \frac{1}{2})^2 - (\frac{1}{2})^2 - 2$

$y = (x + \frac{1}{2})^2 - \frac{1}{4} - \frac{8}{4}$

$y = (x + \frac{1}{2})^2 - \frac{9}{4}$

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

$y = (x - 2)^2 - (-2)^2 - 2$

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

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

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

$y = (x - 2)^2 - (-2)^2$

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

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

$y = (x + \frac{3}{2})^2 - (\frac{3}{2})^2 + 2$

$y = (x + \frac{3}{2})^2 - \frac{9}{4} + \frac{8}{4}$

$y = (x + \frac{3}{2})^2 - \frac{1}{4}$

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

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

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

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

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

$y = (x - \frac{5}{2})^2 - (-\frac{5}{2})^2 + 5$

$y = (x - \frac{5}{2})^2 - \frac{25}{4} + \frac{20}{4}$

$y = (x - \frac{5}{2})^2 - \frac{5}{4}$

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

$y = (x + \frac{1}{2})^2 - (\frac{1}{2})^2 - 3$

$y = (x + \frac{1}{2})^2 - \frac{1}{4} - \frac{12}{4}$

$y = (x + \frac{1}{2})^2 - \frac{13}{4}$

(8) $y = x^2 - 3x - 4$

$y = (x - \frac{3}{2})^2 - (-\frac{3}{2})^2 - 4$

$y = (x - \frac{3}{2})^2 - \frac{9}{4} - \frac{16}{4}$

$y = (x - \frac{3}{2})^2 - \frac{25}{4}$

(9) $y = x^2 - 3x$

$y = (x - \frac{3}{2})^2 - (-\frac{3}{2})^2$

$y = (x - \frac{3}{2})^2 - \frac{9}{4}$

(10) $y = x^2 + 2x + 5$

$y = (x + 1)^2 - (1)^2 + 5$

$y = (x + 1)^2 - 1 + 5$

$y = (x + 1)^2 + 4$