# Confusion with the vertical line test of functions

Recently i saw the following article on wikipedia to see whether a graph is a graph of a function or not.

http://en.wikipedia.org/wiki/Vertical_line_test

It states that To use the vertical line test,"draw a line parallel to the y-axis for any chosen value of x. If the vertical line you drew intersects the graph more than once for any value of x then the graph is not the graph of a function. If, alternatively, a vertical line intersects the graph no more than once, no matter where the vertical line is placed, then the graph is the graph of a function." So in case of a parabola with $y^2=4ax$ a line $\parallel$ to the $y-axis$ will cut it into two points. So $y^2=4ax$ is not a function?

Parabola not a function? http://www5a.wolframalpha.com/Calculate/MSP/MSP180921i11501eh29c4ab00005i6e19ih12i0if69?MSPStoreType=image/gif&s=52&w=279.&h=302.&cdf=RangeControl

Wikipedia writes the same: "As an example, a sideways parabola (one whose directrix is a vertical line) is not the graph of a function because some vertical lines will intersect the parabola twice."

$Thanks$
Everything you say is correct. $y^2 = 4ax$ is not a function of $x$, since a given (positive) value of $x$ corresponds to two possible values of $y$. But if you turn your head on its side, you see that the graph is the graph of a function of $y$ - any given value of $y$ corresponds to exactly one value of $x$.
If we want to make light on @rogerl 's post we may consider two ordered pairs: $$(1, 2\sqrt{a}), ~~(1, -2\sqrt{a})$$ Both are placed on the curve of $y^2=4ax, ~~(x>0)$. You cut the curve at $x=1$ with two different $y$. This violets the definition of a function. Indeed, this is a relation instead. Moreover, as you can find $y$ with respect to $x$, $y^2=4ax,~~(x\ge 0)$ is an implicit function: $$F(x,y)=y^2-4ax=0$$
• @neil_roy: If you have this formula, no it is not a function. Of course if you want it to be a function so you should restrict the $y$. And then the above last plot will have one branch of positive values of $y$ or just the negative ones. Note that $x^2=4ay$ is a well-known function. – mrs Aug 13 '14 at 19:15