与えられた10個の二次関数の式を $y = ax^2 + bx + c$ の形に変形し、それぞれの $a, b, c$ の値を求める。

## 数学の問題の解答

1. 問題の内容

与えられた10個の二次関数の式を $y = ax^2 + bx + c$ の形に変形し、それぞれの $a, b, c$ の値を求める。

2. 解き方の手順

各関数について、以下の手順で $y = ax^2 + bx + c$ の形に変形し、$a, b, c$ の値を特定する。

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

$y = x^2 - 6x + 9$

$a = 1, b = -6, c = 9$

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

$y = x^2 - 10x + 25 - 10$

$y = x^2 - 10x + 15$

$a = 1, b = -10, c = 15$

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

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

$y = 2x^2 - 4x + 2$

$a = 2, b = -4, c = 2$

(4) $y = (x - \frac{3}{2})^2 - \frac{5}{4}$

$y = x^2 - 3x + \frac{9}{4} - \frac{5}{4}$

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

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

$a = 1, b = -3, c = 1$

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

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

$y = 3x^2 + 6x + 3 - 2$

$y = 3x^2 + 6x + 1$

$a = 3, b = 6, c = 1$

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

$y = -(x^2 - 6x + 9) + 1$

$y = -x^2 + 6x - 9 + 1$

$y = -x^2 + 6x - 8$

$a = -1, b = 6, c = -8$

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

$y = -(x^2 - 3x + \frac{9}{4}) + \frac{1}{4}$

$y = -x^2 + 3x - \frac{9}{4} + \frac{1}{4}$

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

$y = -x^2 + 3x - 2$

$a = -1, b = 3, c = -2$

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

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

$y = 2x^2 + 8x + 8$

$a = 2, b = 8, c = 8$

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

$y = -2(x^2 + 6x + 9) - 1$

$y = -2x^2 - 12x - 18 - 1$

$y = -2x^2 - 12x - 19$

$a = -2, b = -12, c = -19$

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

$y = -2(x^2 - 6x + 9) - 1$

$y = -2x^2 + 12x - 18 - 1$

$y = -2x^2 + 12x - 19$

$a = -2, b = 12, c = -19$

3. 最終的な答え

(1) 1,-6,9

(2) 1,-10,15

(3) 2,-4,2

(4) 1,-3,1

(5) 3,6,1

(6) -1,6,-8

(7) -1,3,-2

(8) 2,8,8

(9) -2,-12,-19

(10) -2,12,-19