# Algebra

posted by .

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?

• Algebra -

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)

• Algebra -

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

### Answer This Question

 First Name: School Subject: Answer:

### Related Questions

More Related Questions

Post a New Question
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}