Solveit

End point: -b/2a

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

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

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

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 

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

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

(x+2y)^2≤-3

(x+2y)^2+3≤0

there is no solution may be you forget to wrote smth.

i think: 3

i think:

x^3-1 because ^3 it is cub and the others are parabola

-3=3/7n

multiply both sides by 7

-3*7=3/7n*7

-21=3n

n=-7

4.7

23.000x10/100=2.3

x percent of something it is mean= 

\(x*smth\over 100\)

or 

smth*0.01x