[Shurephile]

x=y^2+2, create table with three points and graph. I'm choosing -1,0, and 1 for the heck of it. I'm not used to seeing x on that side of the equal sign so I have a bit of writers block. 

 

[VNandor]

I'm not sure if I understand correctly what the task is because English is not my native language.
However it is easy to make the X appear on the other side of the equation
 
You substract X from both sides : 0=y^2+2-X
You substract y^2+2 from both sides: -y^2-2=-X
You multiply both sides by -1: y^2+2=X
 
and that should be it.

 

[Shurephile]

  I'm not sure if I understand correctly what the task is because English is not my native language.
However it is easy to make the X appear on the other side of the equation
 
You substract X from both sides : 0=y^2+2-X
You substract y^2+2 from both sides: -y^2-2=-X
You multiply both sides by -1: y^2+2=X
 
and that should be it.

 
Thank you, trying to solve y^2+2=-1 and y^2+2=0 and than I have to graph. Should be a half-moon pointed to the right. 

 

[VNandor]

Ah I think I get what your task is. You have to draw a graph in a coordinate system.
Remember that in a coordinate system the  the horizontal axis represents X and the vertical axis represents Y.
So you want your equation to look like Y=square root(X-2). From that you can see can that must be bigger than 2 otherwise you can't solve the equation or more like the solution is that there is no such X. (If we don't take imaginary/complex numers into account, but you probably don't have to.)
After that You choose X to be 3 6 and 11 so you are going to get 1 2 and 3 in the Y axis respectively.

 

[algebrat]

Remember that when you take the square root of both sides you get + and - the square root.... So you have 2 functions y=√(x-2) and y=-√(x-2). So to plot the equation y^2+2=x you'll have to plot both those equations above.

 

[VNandor]

 Remember that when you take the square root of both sides you get + and - the square root.... So you have 2 functions y=√(x-2) and y=-√(x-2). So to plot the equation y^2+2=x you'll have to plot both those equations above.

From a strictly algebric standpoint the first statement is true and I totally forgot that. However you shouldn't make a plot which assigns two Ys for one X because a graph always should have only one Y assigned to a given X.

 

[Shurephile]

So looking at the graph it comes out linear, should it be a half moon? Also regarding relations, how can you tell if one is a function? Any cleavers ones? 

 

[VNandor]

This is how a square root graph should look like. Yours should start at (2;0) the first number is the x axis and the second is the y axis because your Equation is not Y= square root(X) but Y=Square root(X-2)

 

[algebrat]

Question to you then:
What makes y=√(x-2) the answer over y=-√(x-2)?

 

[Shurephile]

 

This is how a square root graph should look like. Yours should start at (1;2) the first number is the x axis and the second is the y axis.

 
Yes that looks right, my book shows it also going into the fourth quadrant. As in inverted. 

 

[VNandor]

Oh does your book show a graph that does have things going on below the x axis as well? If so then disregard what I have said about not caring about the minus part of the equation.

 

[algebrat]

A function is a relation that is one to one. Such that if f(x)=y then for every x there is exactly one y value. When graphed you can use ther vertical line test. If the line passes through 2 or more points on the graph at any location then it is not a function.

 

[VNandor]

 A function is a relation that is one to one. Such that if f(x)=y then for every x there is exactly one y value. When graphed you can use ther vertical line test. If the line passes through 2 or more points on the graph at any location then it is not a function.

As far as I know the y=√x is supposed to be a function that is why i said to disregard the minus part of the equation. However his math book seems to say otherwise.

 

[Shurephile]

So I can invert it to finish it? My question was to described the equation by picking any three points, describing the shape and domain and range. So we didn't pick any negative numbers, so it doesn't technically have to go into the fourth quadrant. My original equation is x=y^2+2. 

 

[algebrat]

Y^2+2=x is not a function in terms of x, but y=√x is a function. The 4th quadrant is still positive for x so those values are still valid values. So my points would be (66,8), (66,-8), (102,10), (102,-10), (123,11), (123,-11) as an example.