Question 161526
given the algebraic relation x=y^2+1, determine the domain, range, and whether or not it is a funtion.   
<pre><b>

I think you are a little confused. Before you enter it in 
your calculator you must solve the equation for y:

{{{x=y^2+1}}}
{{{y^2+1=x}}}
{{{y^2=x-1}}}
{{{y}}}=±{{{sqrt(x-1)}}}

This means the equation consists of two function graphs,

One function graph is the graph of this function, using
the positive square root:

{{{y}}}={{{sqrt(x-1)}}}

{{{drawing(400,400,-10,10,-10,10,

graph(400,400,-10,10,-10,10,sqrt(x-1)) )}}}

The other function graph is the graph of this function,
using the negative square root:

{{{y=-sqrt(x-1)}}}

{{{drawing(400,400,-10,10,-10,10,

graph(400,400,-10,10,-10,10,-sqrt(x-1)) )}}}

But the original equation

{{{x=y^2+1}}}

consiste of BOTH of these at the same time on the
same set of axes, like this:

{{{drawing(400,400,-10,10,-10,10,
graph(400,400,-10,10,-10,10,sqrt(x-1)) ,
graph(400,400,-10,10,-10,10,-sqrt(x-1)) )}}}

and this is not a function because it does not
pass the vertical line test.  It is a parabola
that opens to the right.

Since the x-values of the graph start where x=1,
including that point, and the graph goes to 
infinity on the right the domain is {{{matrix(1,5,"[",1,",",infinity,")")}}}

Since the y-values go both upward and downward
from 0 both to positive and negative infinity,
the range is   {{{matrix(1,5,"(",-infinity,",",infinity,")")}}}

No, it is not a function.

Edwin</pre>