I don't know which is which

y=(x+9)^2-3

I know it is all real numbers, but i don't know if my answer should be
[-3,00)
or
(-9,00)

Can some one explain this?

I do not know what your question is asking for so will describe the function.

This parabola opens up (holds water) because y gets big positive as x gets big positive or negative.
The axis of symmetry, and therefore x location of the vertex, is when the quantity (x+9) is zero. That is when x is -9
For example if you go one space left and right of x = -9, you should get the same y
well if x = -9-1 = -10, then y = 1-3 = -2
and if x = -9 + 1 = -8, then y = 1-3 = -2 sure enough the same.
Now to find the vertex, we already know that (x+9) is zero at x = -9 so we need y there, but it is obviously 0-3 = -3
so the location of the vertex is (-9,-3)

Now about your "all real numbers" comment.

Yes, the domain of x is all real numbers, you can find a y for any old x.
However we can not find any y below the vertex at y = -3 because the whole parabola is above the vertex at (-9,-3)
Therefore the Range of y is y greater than or equal to -3

To determine the solution set for the inequality y = (x+9)^2 - 3, we need to find the values of x that satisfy the inequality.

First, let's simplify the inequality by expanding the squared term:
y = x^2 + 18x + 81 - 3
y = x^2 + 18x + 78

Now, we can set y greater than or equal to zero and solve for x to find the values that satisfy the inequality:
x^2 + 18x + 78 ≥ 0

To solve this quadratic inequality, we need to find the x-intercepts (where y = 0) and determine the sign of the quadratic expression in each interval. This can be done by factoring, completing the square, or using the quadratic formula.

In this particular case, the quadratic does not factor easily. So we'll use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / 2a

For our quadratic equation x^2 + 18x + 78 = 0, we have a = 1, b = 18, and c = 78.

Using the quadratic formula, we find:
x = (-18 ± √(18^2 - 4*1*78)) / 2*1
x = (-18 ± √(324 - 312)) / 2
x = (-18 ± √12) / 2
x = (-18 ± 2√3) / 2
x = -9 ± √3

Therefore, the x-intercepts for this quadratic equation are x = -9 + √3 and x = -9 - √3.

Now we can determine the sign of the quadratic expression in each interval created by these x-intercepts.

Interval 1: x < -9 - √3
For any value of x less than -9 - √3, the quadratic expression will be positive or zero, since it does not cross the x-axis in this interval.

Interval 2: -9 - √3 < x < -9 + √3
For any value of x between -9 - √3 and -9 + √3, the quadratic expression will be negative, as it crosses the x-axis in this interval.

Interval 3: x > -9 + √3
For any value of x greater than -9 + √3, the quadratic expression will be positive or zero, since it does not cross the x-axis in this interval.

Now let's refer back to the original inequality y ≥ 0, where y = (x+9)^2 - 3.

Since y is always greater than or equal to zero (y ≥ 0), the solution set for x will include the intervals where the quadratic expression is positive or zero.

Thus, the solution set for x is (-∞, -9 - √3] ∪ [-9 + √3, +∞).

In interval notation, this can be expressed as:
(-∞, -9 - √3] ∪ [-9 + √3, +∞)

Therefore, neither [-3, ∞) nor (-9, ∞) is correct.