Last math question! Promise! Maybe.. :)

ohh okay! Great, Thank you, to the both of you! I greatly appreciate it! :)

I'm sorry if some of those questions are stupid. :/

Ah fel dat UR stapid, U gt to get gud, m8. U cant' re-lie on dis cite, srry. ): ):) :)

Y(x+2)2 is that all that they gived to you ?

Solveit, U men gave*, gived iz nono word U reaturd!

Yes, Solviet :)

That's all they gave me :/

its pretty hard to work with just these terms, but we can work around it :)

ok, so if we put the whole factor equal to 0 it'll be simple

and;

Y(x+2)^2=0

you can take the whole thing apart

so (x+2)^2=0 and Y=0

isolate for x 

x=-2 or +2

and for a Y root the x should be equal to 0 so;

replacing X with 0 in the initial formula 

Y=0

thats all i can work with :)

hope this helps

i think

y=(x+2)^2

Domain it is the all values which x can be so there is can be all the numbers 

So your domain is (-infinity,+infinity)

Range is the all values of y since the equetion is squared so the y value cannot be less than 0 let s check:

(x+2)^2=0

x=-2

so your range is: (0,+infinity)

X-intercept it is when Y is 0 :

(x+2)^2=0

x=-2

Y-intercept when X is 0:

(0+2)^2=Y

Y=4

here s your graph:

https://www.desmos.com/calculator/xcxwilih1w

and you can do that by yourself

my explonation so cool that even i didn t understand what did i wrote if you didn t understan smth ask i will try to help 

End point: -b/2a

\(Y=(x+2)^2\)

\(Y=x^2+4x+4\)

\(Y=ax^2+bx+c\)

Thank you. ;)