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quadratic functions: axis of symmetry and vertex formula for this equation y= x^2 + 2 how do you find and graph the vertex? If you plug it into the formula, -b/2a, which is 0/2*1 you get x=0, but what about the y?
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x=0 means symmetric by the y-axis
so it's exactly on the y axis??
\[f(x)=0^2+2=2\]
yes
but what about the point? how do you know what point it is on?
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it's supposed to be 0,2
i already answered?
is that a formula?
I just put x=0 into a function
OOOOH!!!! so you just plugged in x for x^2+2?
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i mean 0
yep!
sweet thanks!
you vetex(0,2)
this is an awesome site i just want to say. Before this, i would be struggling and watching so many videos and reading so many math sites just to figure out something like this
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