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(1)y=x²に点Pのx座標-2を代入して
y=2x+bにx=-2、y=4を代入して
b=8
直線と放物線の式を連立して
x²=2x+8
x²-2x-8=0
(x-4)(x+2)=0
よって点Qの座標は(4、16)⭕

ちさと

分かりやすくありがとうございます!!

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