与えられた二次関数を平方完成させる問題です。全部で10問あります。

1. 問題の内容

与えられた二次関数を平方完成させる問題です。全部で10問あります。

2. 解き方の手順

平方完成を行うには、一般的に以下の手順を踏みます。

1. $x^2$ の係数で $x^2$ と $x$ の項をくくり出します。

2. 括弧の中を $(x + a)^2 + b$ の形に変形します。具体的には、$x$ の係数の半分を $a$ とし、$a^2$ を足して引きます。

3. 括弧を外し、定数項を整理します。

それぞれの問題について解いていきます。

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

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

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

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

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

$y = -2(x - \frac{3}{4})^2 + \frac{9}{8} + \frac{72}{8}$

$y = -2(x - \frac{3}{4})^2 + \frac{81}{8}$

(2) $y = 4x^2 + 6x - 8$

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

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

$y = 4((x + \frac{3}{4})^2 - \frac{9}{16}) - 8$

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

$y = 4(x + \frac{3}{4})^2 - \frac{9}{4} - \frac{32}{4}$

$y = 4(x + \frac{3}{4})^2 - \frac{41}{4}$

(3) $y = -4x^2 - 8x - 10$

$y = -4(x^2 + 2x) - 10$

$y = -4(x^2 + 2x + 1 - 1) - 10$

$y = -4((x + 1)^2 - 1) - 10$

$y = -4(x + 1)^2 + 4 - 10$

$y = -4(x + 1)^2 - 6$

(4) $y = -x^2 - 10x - 6$

$y = -(x^2 + 10x) - 6$

$y = -(x^2 + 10x + 25 - 25) - 6$

$y = -((x + 5)^2 - 25) - 6$

$y = -(x + 5)^2 + 25 - 6$

$y = -(x + 5)^2 + 19$

(5) $y = 3x^2 - 5x - 3$

$y = 3(x^2 - \frac{5}{3}x) - 3$

$y = 3(x^2 - \frac{5}{3}x + (\frac{5}{6})^2 - (\frac{5}{6})^2) - 3$

$y = 3((x - \frac{5}{6})^2 - \frac{25}{36}) - 3$

$y = 3(x - \frac{5}{6})^2 - \frac{25}{12} - 3$

$y = 3(x - \frac{5}{6})^2 - \frac{25}{12} - \frac{36}{12}$

$y = 3(x - \frac{5}{6})^2 - \frac{61}{12}$

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

$y = (x^2 + 2x + 1) + 3 - 1$

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

(7) $y = -4x^2 + 6x + 3$

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

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

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

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

$y = -4(x - \frac{3}{4})^2 + \frac{9}{4} + \frac{12}{4}$

$y = -4(x - \frac{3}{4})^2 + \frac{21}{4}$

(8) $y = -5x^2 - 5x - 7$

$y = -5(x^2 + x) - 7$

$y = -5(x^2 + x + (\frac{1}{2})^2 - (\frac{1}{2})^2) - 7$

$y = -5((x + \frac{1}{2})^2 - \frac{1}{4}) - 7$

$y = -5(x + \frac{1}{2})^2 + \frac{5}{4} - 7$

$y = -5(x + \frac{1}{2})^2 + \frac{5}{4} - \frac{28}{4}$

$y = -5(x + \frac{1}{2})^2 - \frac{23}{4}$

(9) $y = 2x^2 - x - 10$

$y = 2(x^2 - \frac{1}{2}x) - 10$

$y = 2(x^2 - \frac{1}{2}x + (\frac{1}{4})^2 - (\frac{1}{4})^2) - 10$

$y = 2((x - \frac{1}{4})^2 - \frac{1}{16}) - 10$

$y = 2(x - \frac{1}{4})^2 - \frac{1}{8} - 10$

$y = 2(x - \frac{1}{4})^2 - \frac{1}{8} - \frac{80}{8}$

$y = 2(x - \frac{1}{4})^2 - \frac{81}{8}$

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

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

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

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

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

$y = 2(x - \frac{5}{2})^2 - \frac{25}{2} - \frac{10}{2}$

$y = 2(x - \frac{5}{2})^2 - \frac{35}{2}$

3. 最終的な答え

(1) $y = -2(x - \frac{3}{4})^2 + \frac{81}{8}$

(2) $y = 4(x + \frac{3}{4})^2 - \frac{41}{4}$

(3) $y = -4(x + 1)^2 - 6$

(4) $y = -(x + 5)^2 + 19$

(5) $y = 3(x - \frac{5}{6})^2 - \frac{61}{12}$

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

(7) $y = -4(x - \frac{3}{4})^2 + \frac{21}{4}$

(8) $y = -5(x + \frac{1}{2})^2 - \frac{23}{4}$

(9) $y = 2(x - \frac{1}{4})^2 - \frac{81}{8}$

(10) $y = 2(x - \frac{5}{2})^2 - \frac{35}{2}$