# Help w Functions: math progr. says -x+y^2=2 is function

#### jennie

##### New member
I was doing this math program, and I got problems about whether a certain equation was a function. -x+y^2=2 was considered a function, and y=sqrt(x+1) was also considered a function. I don't get how. When solving square roots, don't we have to consider the positive or negative solution? Thanks in advance.

#### Alfredo Dawlabany

##### New member
-x+y^2=2 is not a function but it is a conic which is a union of two functions
conic= parabola
hyperbola
ellipse
circle
the equation:-x+y^2=2 is a parabola(graph below)

-x+y^2=2 can be written as y=squr(x+2) and y=-squr(x+2)
for y=squr(x+2) the graph is like that

for y=-squr(x+2) the graph is like that

therefore as you notice that these 2 graphs are the union of the 1st
these graphs are called functions while the 1st one is a conic
a function is not a function when a vertical line cuts it at 2 points

#### Alfredo Dawlabany

##### New member

I was doing this math program, and I got problems about whether a certain equation was a function. -x+y^2=2 was considered a function, and y=sqrt(x+1) was also considered a function. I don't get how. When solving square roots, don't we have to consider the positive or negative solution? Thanks in advance.
the equation -x+y^2=2 is not a function
it is a conic(parabola,hyperbola,ellipse,circle)
-x+y^2=2 can be written as y=squ(x+2) and y=-squ(x+2)
-x+y^2=2 is the union of y=squ(x+2) and y=-squ(x+2)
example by graph:
> -x+y^2=2

> y=squ(x+2)

> y=-squ(x+2)

as you see that the 1st graph is the union of the 2nd and the 3rd one
remark:the 2nd and the 3rd equations are function but the first is not
an equation is not a function when a vertical line cuts its graph at two points

#### JeffM

##### Elite Member
I was doing this math program, and I got problems about whether a certain equation was a function. -x+y^2=2 was considered a function, and y=sqrt(x+1) was also considered a function. I don't get how. When solving square roots, don't we have to consider the positive or negative solution? Thanks in advance.
The square root function is always considered to be non-negative.

$$\displaystyle (-\ 2)^2 = 4 \implies (-\ 2) = -\ \sqrt{4} \ne \sqrt{4} = 2.$$

in other words

$$\displaystyle x^2 - 4 = 0$$ has two roots, the square root and its additive inverse.

EDIT: Your first example can be shown as a valid function only as f = x. Do you see why?

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#### Otis

##### Full Member
When solving square roots, don't we have to consider the positive or negative solution?
Yes, if you solve y^2 = 2 - x for y, by taking the square root of each side, then you do need to consider both the square root of 2-x and its opposite. This generates two functions, as explained below.

In the relationship between x and y given by the equation

x + y^2 = 2

it's true that y is not a function of x. Rather, as JeffM posted, x is a function of y.

However, if we restrict the values of y to be non-negative, then y is a function of x.

Alternatively, if we restrict y to be negative, then y becomes a different function of x.

y1(x) = sqrt(2 - x)

y2(x) = - sqrt(2 - x)

The domain for each of these functions is all numbers less than or equal to 2.

The range for y1(x) is all non-negative numbers.

The range for y2(x) is all negative numbers and zero.

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#### stapel

##### Super Moderator
Staff member
I was doing this math program, and I got problems about whether a certain equation was a function. -x+y^2=2 was considered a function...
We can only speak to the general definition of functions, using the common orientation of variables. We can't speak to whatever math program you're using, nor how it defines the variables; especially since we don't know what you've entered, how the program is set up, or what is meant by "is considered a function".

If you could provide all of the necessary information, we may be able to assist. But, with the current information, I can only suggest that perhaps the software is fluid with the variables (because "-x + y^2 = 2" is a function, of x in terms of y).

and y=sqrt(x+1) was also considered a function. I don't get how. When solving square roots, don't we have to consider the positive or negative solution?
Are you saying that you'd started out with "y^2 = x + 1", and then took one of the square roots? Because, if you'd started with "y = sqrt[x + 1]", then there is nothing to "solve"; you've already been given a function (of y in terms of x).