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sorry..not a good artist..the vertex is (0,0)

|dw:1322072661463:dw|

r u doing absolute valuee functions/equations?

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yes..I need hlep with this

thank you

r the parentheses supposed to be the absolute value signs? and is x2 x times 2?

the graph of y=-2(x+3)^2+1 will have curvature downwards and its vertex is at \[(1/\sqrt{2}-3.0)\]

woah...^

yes, and x2 is x^2

thank you guyz

well i luckey has given answer so good job

U still tried

yea im proud of my self:) ROBIN AWAY!

thnx

\[y=-2(x+3)^2+1 \text{ has vertex } (-3,1)\]

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