- Thread starter jennie
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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

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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

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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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.When solving square roots, don't we have to consider the positive or negative solution?

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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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".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...

If you could provide

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).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?

Please reply with clarification. Thank you!