I"m a little confused by the rule: If you attract a upright line that intersects the graph at an ext than 1 suggest then it is not a function.

You are watching: Is a circle graph a function

Because then a circle choose $y^2 + x^2 = 1$ is no a function?

And indeed if ns rewrite it together $f(x) = \sqrt(1 - x^2)$ climate wolfram alpha doesn"t draw a circle. I guess I"m absent the intuition as to why this is though?



The definition of a duty is for this reason important.In addition to the above, the snapshot below (taken from: What is a function) may help.

(the left hand side is her X and the best hand side is the worth Y)



A function is a preeminence that assigns uniquely to a member the domain set, a member the the photo set.The an essential word is "uniquely".So if you entrust say 2 and -2 to number 1, climate you have actually a rule, but not a function.That is the logic behind the vertical heat test. If you attract a vertical line and it intersects the graph that the duty in two distinctive points, climate you can see the it method I have assigned both of these points come the point where my vertical line the cross 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=+$\sqrt1-x^2$) and the bottom fifty percent is the negative square root (y=-$\sqrt1-x^2$).


Functions should be well-defined as part of their definition, so because that a provided input there deserve to only be one output.

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

However creating $y^2+x^2=1$ as a function of $x$ alone can not be done, together $x=\dfrac12$ has actually two options ($y=\pm\sqrt\dfrac34$).


If you want to have a duty that "draws" a circle through radius $r$ and center $P = (x_0, y_0)$ top top the cartesian plane, you have the right to use the role $f : <0, 2\pi> \rightarrow \gaianation.netbbR \times \gaianation.netbbR$ characterized by $$f(\varphi) = (x_0 + r \cos \varphi, y_0 + r \sin \varphi)$$But, the course, this is no a role from $\gaianation.netbbR$ come $\gaianation.netbbR$.

Also, girlfriend can specify a curve in the aircraft by method of one equation of two variables $x$ and also $y$. If you have actually a (continuous) role $f : A\subseteq \gaianation.netbbR\rightarrow \gaianation.netbbR$, friend can gain an equation $y = f(x)$ native it, which specifies a curve.But friend cannot constantly transform one equation containing 2 variables come an tantamount equation $y = f(x)$. The equation $x^2 + y^2 = r^2, r\in\gaianation.netbbR$ is an instance of this fact.

A role $f(x_1, \ldots, x_n)$ has actually the property, the for one collection of worths $(v_1, \ldots, v_n)$ there is at most one result. If friend compare, her $f(0) = 1$, however there are 2 worths for $y$ s.t. $y^2 + x^2 = 1 \mid x = 0$, namely $\ 1, -1 \$.

The standard definition of a role $f$ is that it takes one worth $f(x)$ because that each $x$ (where it is defined).

In particular, the square root is a single valued role - because that a real number $x$, the square source is provided by $\sqrtx^2 =|x|$.

In your example, when solving for $y$ in the one equation $y^2+x^2=1$ there space two possibilities $$y=\sqrt1-x^2\qquad \textor\qquad y=-\sqrt1-x^2$$which room two different functions and the union of their graphs is the circle.

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

An equivalent explicit meaning of $y$ is:

$y=\pm \sqrt1-x^2$ , with problem $x\in <-1,1> $

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