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OpenStudy (anonymous):
The graph of y = x2 is shown below. Using complete sentences, explain how the graph
of y = –2(x + 3)2 + 1 would differ from this parabola?
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OpenStudy (anonymous):
sorry..not a good artist..the vertex is (0,0)
OpenStudy (anonymous):
|dw:1322072661463:dw|
OpenStudy (karatechopper):
r u doing absolute valuee functions/equations?
OpenStudy (anonymous):
yes..I need hlep with this
OpenStudy (karatechopper):
dont worry
i just learned this a couple of days ago i was frustrated too but i can help in this now:)
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OpenStudy (anonymous):
thank you
OpenStudy (karatechopper):
r the parentheses supposed to be the absolute value signs? and is x2 x times 2?
OpenStudy (anonymous):
the graph of y=-2(x+3)^2+1 will have curvature downwards and its vertex is at \[(1/\sqrt{2}-3.0)\]
OpenStudy (karatechopper):
woah...^
OpenStudy (anonymous):
yes, and x2 is x^2
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OpenStudy (anonymous):
thank you guyz
OpenStudy (karatechopper):
ok
OpenStudy (karatechopper):
well i luckey has given answer so good job
OpenStudy (anonymous):
U still tried
OpenStudy (karatechopper):
yea im proud of my self:) ROBIN AWAY!
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OpenStudy (karatechopper):
thnx
myininaya (myininaya):
\[y=-2(x+3)^2+1 \text{ has vertex } (-3,1)\]
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