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Algebra 23 Online
OpenStudy (anonymous):

The vertex is (0,3) and point (2,5) lies on graph.

OpenStudy (anonymous):

what are you trying to find?

OpenStudy (anonymous):

the equation of the line

OpenStudy (anonymous):

hehe jst borrowed ds account from my frnd :)

OpenStudy (anonymous):

in y=mx + B format?

OpenStudy (anonymous):

yes..

OpenStudy (anonymous):

so to find slope, you want to do y2-y1/x2-x1 since your point is the 2nd point, its (x2, y2) slope is 5-3/2-0=1

OpenStudy (anonymous):

also, for point b, use the y intercept for your vertex

OpenStudy (anonymous):

your equation looks like y=x+1

OpenStudy (anonymous):

the slope is 1

OpenStudy (anonymous):

yea

OpenStudy (anonymous):

isnt it?

OpenStudy (anonymous):

yes it is. and in the equation y=mx + b, m is where you put the slope

OpenStudy (anonymous):

ok...got it

OpenStudy (anonymous):

and i will use (y2-y1) = m(x2-x1) ... ???

OpenStudy (anonymous):

no.

OpenStudy (anonymous):

it's a parabola?!

OpenStudy (anonymous):

(y2-y1)/(x2-x1)

OpenStudy (anonymous):

no its a simple equation If you want, try graphing it out and do rise over run

OpenStudy (anonymous):

|dw:1376134648702:dw| Graph would look something like this

OpenStudy (anonymous):

but theres a vertex.. hehe

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