与えられた二次関数を $y=a(x-p)^2+q$ の形に変形し、空欄を埋める問題です。問題は全部で4つあります。 (1) $y = x^2 - 2x$ (2) $y = x^2 + 4x + 1$ (3) $y = 2x^2 + 8x$ (4) $y = -x^2 + 8x + 2$

1. 問題の内容

与えられた二次関数を $y=a(x-p)^2+q$ の形に変形し、空欄を埋める問題です。問題は全部で4つあります。

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

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

(3) $y = 2x^2 + 8x$

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

2. 解き方の手順

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

$y = x^2 - 2 \times 1 \times x$

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

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

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

$y = x^2 + 2 \times 2 \times x + 1$

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

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

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

(3) $y = 2x^2 + 8x$

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

$y = 2(x^2 + 2 \times 2 \times x)$

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

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

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

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

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

$y = -(x^2 - 2 \times 4 \times x) + 2$

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

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

$y = -(x-4)^2 + 16 + 2$

$y = -(x-4)^2 + 18$

3. 最終的な答え

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

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

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

(4) $y = -(x-4)^2 + 18$