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(1)

(    )

例題

5

３

(1) (2)

解

(1)

(2)

のグラフの特徴 y = a(x − p)

2

例

y = 3(x − 5)

2

＞ 第２章 ２次関数 ＞ 第１節 ２次関数  ＞ 第２講：２次関数 数

 I

２次関数  y = a(x − p) 2 のグラフ

１

３

a > 0

頂点は(   ，  )

a < 0

２ 軸は，

x = p

軸

y = 2 x

2

x y

O

y = 3x

2 x

 軸方向に

y = ax2をx 軸方向に

p だけ平行移動した

y = 2(x − 3)

2

3

もとの２次関数を

y = 3x

2 とするとき，

次の２次関数が x 軸方向にどれだけ平行移 動したか。また，その頂点も答えなさい。

y = 3(x + 2)

2

y = 3(x

−

5)

2

5

頂点は， (5, 0)

y = 3x

2 x

 軸方向に y = 3(x + 2)

2

−

2

頂点は， (− 2, 0)

x y

O

y = 3x2 y = 3(x −5)2 5

−2 x

y

O

y = 3x2 y = 3(x + 2)2

x = 3

日付(        月         日        曜日  )   名前 (       ）

p 0

のとき，(   )に凸 のとき，(   )に凸

下

上

参照

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