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

Sagot :

caylus

Answer:

Hi,

Step-by-step explanation:

[tex](x-10)^2=12(y+2)\\y=\dfrac{(x-10)^2}{12}-2\\y'=0 == > 2(x-10)/12=0 == > x=10 == > y=-2\\\\Vertex\ is\ (10,-2)\\[/tex]

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