System of Quadratic Inequalities.

Sketch the intersection of the given inequalities. 1 y≥x^2 and 2 y≤-x^2+2x+4

I can do this. You get a region of overlap, which is where the solution lies. My question is, this is a bounded region, but would we still say there are infinitely many solutions since, for instance, we can take x=0, x=0.1, x=0.01, x=0.001, x=0.0001 etc...that is, we can just move to another point in the region that is such a small difference over? Or since the region is bounded, is the solution set finite?

Also,

can you get a system of quadratic inequalities with all real numbers as the solution set? I know you can if the inequalities are just multiples of one another, but is there another way?

Re: System of Quadratic Inequalities.

sure the solution is not finite as you know any interval (a,b) is not finite for b>a

Quote:

can you get a system of quadratic inequalities with all real numbers as the solution set? I know you can if the inequalities are just multiples of one another, but is there another way?

I think you are talking about x values if so

you can take y < x^2 , and y> -x^2 -1