for y=x^2-2 do the following.
a. sketch a graph of a equation.
b. indenitfy the vertex.
c. compare the graph of y=f(x) to the graph of y=x^2.
I am LOST!!!!!!!!
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OpenStudy (anonymous):
this is easy as it can get
OpenStudy (anonymous):
Geometric figure= parabola
OpenStudy (anonymous):
or quadratic function
OpenStudy (anonymous):
OpenStudy (anonymous):
y=x^2-2 has a shifts down by 2 because of a verticle shift
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OpenStudy (anonymous):
Helloooooooooooooooooooooo I cant do nothing else until you say something
OpenStudy (anonymous):
i'm trying to figure it out.
OpenStudy (anonymous):
i just dont get the graphing part. sigh
OpenStudy (anonymous):
The graph is right there
OpenStudy (anonymous):
so the point is on 2? right
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OpenStudy (anonymous):
Yea
OpenStudy (anonymous):
so the points are 1,2 or 2,1?
OpenStudy (anonymous):
It decreases by 2 a normal parabola would be on the 0
OpenStudy (anonymous):
2, 0?
OpenStudy (anonymous):
Yea bay
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OpenStudy (anonymous):
i hate graphing!!!!
OpenStudy (anonymous):
so you gone give me best response or something
OpenStudy (anonymous):
i'ma give you best everything in a minute!
OpenStudy (anonymous):
so what are my points? ah! 2,0?
OpenStudy (anonymous):
:)
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OpenStudy (anonymous):
yea and put this just incase -x^2+y+2 = 0
OpenStudy (anonymous):
ok. thanks!!!
OpenStudy (anonymous):
i better get it right! if not i'ma find you! Haaaaaaaaaaaaa