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Mathematics 25 Online
OpenStudy (anonymous):

Graph y > (x + 2)^2

OpenStudy (anonymous):

OpenStudy (anonymous):

mathslover (mathslover):

how did u do that?

OpenStudy (anonymous):

i'm not sure about 1 and 2 actually

mathslover (mathslover):

can u tell me how u did that?

OpenStudy (anonymous):

i just looked at the equations and graphs and i guessed.. but i kinda dont get it

mathslover (mathslover):

ok ... so first of all find x intercept that is put y = 0 and then solve for x

OpenStudy (anonymous):

what do we plug in for x?

mathslover (mathslover):

0<(x+2)^2 solve for x...

mathslover (mathslover):

\[\large{0<(x+2)^2}\] \[\large{\sqrt{0}<\sqrt{(x+2)^2}}\] \[\large{0<x+2}\] right?

OpenStudy (anonymous):

right.

mathslover (mathslover):

what is x now?

OpenStudy (anonymous):

-2?

mathslover (mathslover):

what will be the inequality? x>-2 ? or x < -2 ?

OpenStudy (anonymous):

second one ?

mathslover (mathslover):

no.. \[\large{x+2>0}\] \[\large{x+2-2>0-2}\] \[\large{x>-2}\] remember that sign will not change here

OpenStudy (anonymous):

ok..

mathslover (mathslover):

now .. graph that

mathslover (mathslover):

It will look like a parabola

OpenStudy (anonymous):

#3?

mathslover (mathslover):

it seems good to me

mathslover (mathslover):

yes u r correct

mathslover (mathslover):

@ineedhelpplease57 good work

OpenStudy (anonymous):

thanks @mathslover !

mathslover (mathslover):

I appreciate your perfect responses, a medal for you :) keep up the good work here.. best of luck for your journey in openstudy \[\large{\cal{KEEP}\space\mathbb{ IT}\space\mathsf{ UP}}\]

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