Mathematics
Senior High

放物線x^+(6a+2)+3a+4の頂点の座標を求めよ。

この問題が解けません。因数分解の途中式を詳しく書いていただけるとありがたいです🙇

二次関数

Answers

x²+(6a+2)x+3a+4
=x²+2(3a+1)x+3a+4
=(x+3a+1)²-(3a+1)²+3a+4
=(x+3a+1)²-9a²-3a+3なので頂点は(-3a-1,-9a²-3a+3)ですね

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