What is the domain of the function y=×+6-7

I'm assuming the function you're referring to here is

You can think of the domain of a function as a collection of every number it's possible to "feed" into it as an input. Most of the time, finding out what the domain can't be can help us find out what it can.

The important bit for us to look at here is the term. What restrictions are there on this term? Well, we know that, at least when we're dealing with real numbers (the "number line" numbers) we can't take the square root of a negative number. So we know that has to be positive, or in mathematical terms:

which implies that

This tells us the domain is the collection of all numbers greater than or equal to -6. There are a few ways we can write this:

This format is called interval notation. The [-6 tells us that the collection starts at and includes -6. If we wanted to leave -6 out, we would write (-6. The ∞) tells us that the collection contains everything bigger than six, "on to infinity." Since infinity is more of a concept than a number, we always put that ) after it, since it's impossible for us to actually include.

The second way we could write the solution is

This is called set-builder notation; the first half of the curly braces tells us which collection we're pulling our x's from, and the second half gives us the restrictions on what we can pull exactly. You read as "x is in the set of real numbers" which just means we're pulling it from the collection of "number line" numbers, and tells us we can only pull the real numbers that are greater than or equal to -6.

d. vertex

step-by-step explanation:

for the friend request.

when a figure, such as triangle (xyz) is reflected over the line y = x to create x'y'z' the x-coordinates and y-coordinates of the original image (pre-image) will be reversed for the image. (e.g., (x, y) > (y, x)) this means that the pre-image position of the x- and y-coordinates switch places which in turn provides a mirror like image of the original xyz.  

i notice that when i draw a line segment from x to x', that the points are parallel from each other -- they are exactly across from each other and do not intersect with any other points.  

yes, think i would see the same characteristic if you drew the line segment connecting y with the reflecting line and then y' with the reflecting line because of the essence of reflections. because it is a mirror image, each point was effected the same way which leads me to conclude that they same would apply to y and y'.