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What is the coordinate of the vertex for the following prabola: (x-10)^2 = 12(y+2)
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What is the coordinate of the vertex for the following prabola: (x-10)^2 = 12(y+2)

User Merwok
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Answer:

Hi,

Explanation:


(x-10)^2=12(y+2)\\y=((x-10)^2)/(12)-2\\y'=0 == > 2(x-10)/12=0 == > x=10 == > y=-2\\\\Vertex\ is\ (10,-2)\\

User MojoTosh
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