X= x^2 + 4x + 4 axis of symmetry: vertex: domain: range: and i also need the graph points show your work and how you got it.

You mean Y=x^2 + 4x + 4, this is quadratic function (convex parabola) which  standard form is y=ax^2+bx+c 

1. First we will find zero of the function, y=0 or as you like  intersection
    with x axis.
    y=0 =>. x^2+4x+4=0 => this is the square of the binomial  ( x + 2)^2 = 0
    => x + 2 = 0 => x= -2 In this case it's a touch point with x axis.
 2. Intersection with y axis  => If x=0 => y=4  (0, 4)
 3. The coordinate of vertex are x= - b/2a = - 4/2 = -2 and y=(4ac-b^2)/4a
      y= (4* 1*4 - 4^2)/ 4 = (16-16)/4= 0   Vmin ( -2, 0) this point is at same              time zero of the function. 
 4.  Domain:. Dx=R (field of real numbers)
 5.  Range:  y belong to interval ( - infinite, 0) or R-
 6.  Function's sign y>0 => x belong to intervals (-infinite, -2) U ( -2, +infinite)
 7.  Function's growth or decline 
      for x belong interval ( -infinite, -2) => function decline
      for x belong interval ( -2, +infinite) => function is growing

      Thats all

21, i'm pretty sure it's right lol