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Help.

 Oct 31, 2017
 #1
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If we let x = 0, then y =  2(0)^2  = 0...so  (0,0)  will be on the graph

 

This eliminates the middle two graphs

 

If we let x  = 1, then y =  2(1)^2  =  2

 

So...( 1 , 2)   is on the graph

 

So...the first graph is correct.....

 

 

cool cool cool

 Oct 31, 2017

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