Question 480188
The equation 

x = y²

is not a function because both (1,1) and (1,-1) satisfy the
equation, and a function by definition cannot have two 
different y-values for the same x value.

To make a graph of y² = x, make a table of values:

 x | y
 9 | 3
 9 |-3
 4 | 2
 4 |-2
 1 | 1
 1 |-1
 0 | 0

plot the points:

{{{drawing(400,1600/7,-2,12,-4,4, 
circle(9,3,.1),circle(4,2,.1),
circle(9,-3,.1),
circle(4,-2,.1),
circle(1,1,.1),
circle(1,-1,.1),
circle(0,0,.1),
 graph(400,1600/7,-2,12,-4,4) )}}}

Connect them:

{{{drawing(400,1600/7,-2,12,-4,4, 
circle(9,3,.1),circle(4,2,.1),
circle(9,-3,.1),
circle(4,-2,.1),
circle(1,1,.1),
circle(1,-1,.1),
circle(0,0,.1),
 graph(400,1600/7,-2,12,-4,4,sqrt(x)),

graph(400,1600/7,-2,12,-4,4,-sqrt(x)) )}}}





Also notice that it does not pass the vertical line
test.  Below I have drawn three vertical lines which
intersect the curve twice.  This cannot happen if
the graph is that of a function.


{{{drawing(400,1600/7,-2,12,-4,4, 
green(line(2.7,-15,2.7,15), line(5.3,-15,5.3,15),line(8.4,15,8.4,-15)), 
 graph(400,1600/7,-2,12,-4,4,sqrt(x)),
circle(9,3,.1),circle(4,2,.1),
circle(9,-3,.1),
circle(4,-2,.1),
circle(1,1,.1),
circle(1,-1,.1),
circle(0,0,.1),


graph(400,1600/7,-2,12,-4,4,-sqrt(x)) )}}}