How do you graph the parabola `y = (x + 4)^2 - 3` using vertex, intercepts and additional points?

1. Notice The Form
   Recall that the general forms of the parabola are:
2. `(±y-k)^2=(x-j)` opening to theleft/right
3. `(y-k)=(±x-j)^2` opening up/down

     *where (j,k) is the vertex of the parabola
   

Note that the given follows the form of opening up/down. Additonally, its form follows the opening up parabola since the given has a positive 'x' component. Also we now know that the vertex of the parabola is at (-4,-3).
How?
By rearranging the given, we can see the general form:
• `y+3=(x+4)^2`
• `y-(-3)=(x-(-4))^2`

Hence, the verex is at (-4,-3).

1. Solve For The Intercepts

For the x-intercept, let `x =0`

`y=(0+4)^2-3`
`y=16-3`
`y=13`

Hence, the x-intercept is at (0,13).

Next is the y-intercept. Let `y=0`.

`0=(x+4)^2-3`
`3=(x+4)^2`
`±√3=x+4`
`-4±√3=x`
`x=-4±√3`

Hence, the y-intercepts are at (-4-√3,0) and (-4+√3,0).

1. Pair Up.

Choose any number as your `x` and substitute it into the given equation. Then, pair the chosen `x`'s to their corresponding solution.

Example:
Let `x=1`

`y=(1+4)^2-3`
`y=5^2-3`
`y=25-3`
`y=22`

Thus, the pairing would be: (1,22). And so on.

Hope this helps!