Question

Need help on this problem

Answer 1

Answer:

Step-by-step explanation:

$2(x-2)^2=8(7+y)\\2(x-2)^2=56+8y \Rightarrow y={1\over{8}}[2(x-2)^2-56]={1\over{4}}(x-2)^2-7\\finally\\y={1\over{4}}(x-2)^2-7\\let ~change~x~and~y\\x={1\over{4}}(y-2)^2-7\\x+7={1\over{4}}(y-2)^2\\4(x+7)=(y-2)^2\\y-2=\pm\sqrt{4(x+7)}\\\\y=2\pm\sqrt{4x+28}$