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Mathematics 8 Online
OpenStudy (anonymous):

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!!!!!!!!

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

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

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

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):

:)

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

OpenStudy (anonymous):

lol

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