Question 695848
Something went wrong (got lost in translation) while you were entering the problem. Maybe you had some typos.
Or maybe not everything you typed is shown the way you meant because of the way characters are coded by computers.
Linear equations (in their most simplified form) do not have exponents on the variables, and I see exponents, so maybe you meant non-linear.
Equations involve equal signs, but I see a > sign instead, which would make it an inequality instead of an equation.
 
Maybe you meant you have to solve, by graphing, the system of non-linear inequalities
{{{system(y<x^2+1,y>-x^2+1)}}}
 
{{{y=x^2+1}}} is a non-linear equation. It can also be called a non-linear function.
Because the {{{x}}} is squared it can also be called a quadratic function.
The graph for the quadratic function {{{y=x^2+1}}} is {{{graph(300,180,-5,5,-5,25,x^2+1)}}}
{{{graph(300,180,-5,5,-25,5,y=6,-x^2+1)}}} is the graph of quadratic function {{{y=x^2+1}}}.
Graphing both together, you would have {{{graph(300,300,-5,5,-25,25,x^2+1,-x^2+1)}}}
The graphs touch at the point where {{{x=0}}} and {{{y=1}}}.
 
To graph {{{y<x^2+1}}} you would just
graph {{{y=x^2+1}}} as a dotted line (to show that those points are not part of the solution),
and then you would color or shade the points (below the curve) that have smaller {{{y}}} values, to get something like this:
{{{graph(300,180,-5,5,-5,25,y<x^2+1)}}}
 
For {{{y>-x^2+1}}}, you would graph {{{y=-x^2+1}}} as a dotted line,
and then you would color or shade the points (above the curve) that have larger {{{y}}} values, to get something like this:
{{{graph(300,180,-5,5,-25,5,y>-x^2+1)}}}
 
To graph {{{system(y<x^2+1,y>-x^2+1)}}} you would just graph both curves as dotted lines
(to show that those points are not part of the solution),
and then you would color or shade the points in between,
the ones that are part of the graph of {{{y<x^2+1}}}
and also are part of the graph of {{{y>-x^2+1}}}