Learn this.
 
f(x) = a(x-h)² + k has vertex (h,k) and 1 unit right and
left of the vertex has a y-coordinate which is "a" units above or
below the vertex depending on whether a is positive or negative.

So three points on the parabola are (h,k) (h-1,k+a), and (h+1,k+a)

Compare your equation

f(x) = (x+2)² - 3 or

f(x) = 1(x+2)² - 3

to

f(x) = a(x-h)² + k

And you see that a=1, h=-2 and k=-3

So it has vertex (h,k) = (-2,-3).  That was your first equation.

Now to graph it:

two points on each side of the vertex are (h±1,k+a) = (-2±1,-3+1) = "(-3,-2) and (-1,-2)"

We can also get the x-intercepts, that is, the real zeros by setting
F(x) = 0

(x+2)² - 3 = 0
    (x+2)² = 3
       x+2 = 
         x = -2 = 
         x = approximately -3.7 and -.27

S we plot those points: "(-3,-2) and (-1,-2)

Then draw the parabola:

Edwin