Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm a little confused by the rule: If you draw a vertical line that intersects the graph at more than 1 point then it is not a function.

Because then a circle like $y^2 + x^2 = 1$ is not a function?

And indeed if I rewrite it as $f(x) = \sqrt(1 - x^2)$ then wolfram alpha doesn't draw a circle. I guess I'm missing the intuition as to why this is though?

share|cite|improve this question
Now you got several similar answers - I hope it helps! – AD. Oct 27 '11 at 7:32
up vote 15 down vote accepted

The definition of a function is so important. In addition to the above, the picture below (taken from: What is a function) may help.

(the left hand side is your X and the right hand side is the value Y)

What is a function

share|cite|improve this answer

A function is a rule that assigns uniquely to a member of domain set, a member of the image set. The key word is "uniquely". So if you assign say 2 as well as -2 to number 1, then you have a rule, but not a function. That is the logic behind the vertical line test. If you draw a vertical line and it intersects the graph of the function in two distinct points, then you can see that it means I have assigned both of these points to the point where my vertical line crosses the x-axis. An example of this is the circle.

However a semi-circle is a legit function-the upper half is the positive square root (y=+$\sqrt{1-x^2}$) and the bottom half is the negative square root (y=-$\sqrt{1-x^2}$).

share|cite|improve this answer

Functions need to be well-defined as part of their definition, so for a given input there can only be one output.

$f(x,y)=x^2+y^2-1$ is a function of two variables, and the set of points for which this function gets $0$ is the unit circle.

However writing $y^2+x^2=1$ as a function of $x$ alone cannot be done, as $x=\dfrac12$ has two solutions ($y=\pm\sqrt{\dfrac34}$).

share|cite|improve this answer

If you want have a function that "draws" a circle with radius $r$ and center $P = (x_0, y_0)$ on the cartesian plane, you can use the function $f : [0, 2\pi] \rightarrow \mathbb{R} \times \mathbb{R}$ defined by $$f(\varphi) = (x_0 + r \cos \varphi, y_0 + r \sin \varphi)$$ But, of course, this is not a function from $\mathbb{R}$ to $\mathbb{R}$.

Also, you can define a curve in the plane by means of an equation of two variables $x$ and $y$. If you have a (continuous) function $f : A\subseteq \mathbb{R}\rightarrow \mathbb{R}$, you can get an equation $y = f(x)$ from it, which defines a curve. But you cannot always transform an equation containing two variables to an equivalent equation $y = f(x)$. The equation $x^2 + y^2 = r^2, r\in\mathbb{R}$ is an example of this fact.

share|cite|improve this answer
+1 for “…from $\mathbb{R}$ to $\mathbb{R}$”; the OP’s equation is indeed a function, just one from reals to pairs of reals, like your parametric version. – Jon Purdy Oct 27 '11 at 14:13
I would not say that an equation is function: an equation can define a function. Both an equation and a function can define a curve on the plane, but some equations do not have an associated function. – Giorgio Oct 27 '11 at 15:00
Fair enough. It seems to me that the OP is saying $\pm y$ intuitively ought to represent the pair $(+y, -y)$. – Jon Purdy Oct 27 '11 at 16:14
Ah, ok, you mean $\pm y = (+y, -y)$, so you would have a function $f:\mathbb{R}\rightarrow \mathbb{R}\times\mathbb{R}$. – Giorgio Oct 27 '11 at 16:16

A function $f(x_1, \ldots, x_n)$ has the property, that for one set of values $(v_1, \ldots, v_n)$ there is at most one result. If you compare, your $f(0) = 1$, but there are 2 values for $y$ s.t. $y^2 + x^2 = 1 \mid x = 0$, namely $\{ 1, -1 \}$.

share|cite|improve this answer

The standard definition of a function $f$ is that it takes one value $f(x)$ for each $x$ (where it is defined).

In particular, the square root is a single valued function - for a real number $x$, the square root is given by $\sqrt{x^2} =|x|$.

In your example, when solving for $y$ in the circle equation $y^2+x^2=1$ there are two possibilities $$y=\sqrt{1-x^2}\qquad \text{or}\qquad y=-\sqrt{1-x^2}$$ which are two different functions and the union of their graphs is the circle.

share|cite|improve this answer

$y^2+x^2=1$ is implicit definition of $y$

An equivalent explicit definition of $y$ is:

$y=\pm \sqrt{1-x^2}$ , with condition $x\in [-1,1] $

share|cite|improve this answer
Yes, but $y = \pm \sqrt{1 - x^2}$ is not a function. – Giorgio Oct 27 '11 at 15:03
@Giorgio,$\pm\sqrt{1-x^2}$ means $y=+\sqrt{1-x^2} \lor y=-\sqrt{1-x^2}$ – pedja Oct 27 '11 at 19:27

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.