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Mathematics 16 Online
OpenStudy (anonymous):

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?

OpenStudy (anonymous):

x=0 means symmetric by the y-axis

OpenStudy (anonymous):

so it's exactly on the y axis??

OpenStudy (anonymous):

\[f(x)=0^2+2=2\]

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

but what about the point? how do you know what point it is on?

OpenStudy (anonymous):

it's supposed to be 0,2

OpenStudy (anonymous):

i already answered?

OpenStudy (anonymous):

is that a formula?

OpenStudy (anonymous):

I just put x=0 into a function

OpenStudy (anonymous):

OOOOH!!!! so you just plugged in x for x^2+2?

OpenStudy (anonymous):

i mean 0

OpenStudy (anonymous):

yep!

OpenStudy (anonymous):

sweet thanks!

OpenStudy (anonymous):

you vetex(0,2)

OpenStudy (anonymous):

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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