Parabola: Finding (x,y) Pairs Without Crossing Curve

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Find all the pairs (x, y)R^2, through which does not cross any curve : y = - x^2 + (4-2p)x + p^2 . Finding pairs (x,y) are the co-ordinate points . Thanks for help
 
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Points don't cross things...What exactly do you mean? Show us an attempt at a solution by the way, or else we can't offer any help.
 
Maybe this help
There are possible answers: Finding pairs (x,y) are the co-ordinate points
a) lyings below straight line y = -2x + 1;
b) lyings inside the circle x^2 + (y -3)^2= 9 ;
c) lyings outside the circle x^2 + (y - 3)^2 = 9;
d) lyings below the parabola y = -2 x^2 + 4x;
e) lyings on the parabola y = - x^2 + 4x;
f) lyings above parabola y = x^2 + 2x;
g) lyings below parabola y = x^2 + 2x;
h) lyings above parabola y = -2 x^2 + 4x;
i) Every point (x, y) lies on some of these curves;
j) all answers are false
I don't have any idea to solve this task.
 
Yes this points are laying on the graph y= f(x)
"inside" means that this points are in the circle or outside
below ,above the parabola.
 
Okay. Now, you know that the formula for a circle is [itex]x^2 + (y -3)^2= 9[/itex] because the are points exactly 3 units from the center point (0, 3) ([itex]x^2+ (y-3)^2[/itex] is the square of the distance from (x,y) to (0,3) and 9 is the square of 3). If a point is inside then what can you say about the distance from that point to (0,3)? What can you say about the square of that distance?

And a point is above a parabola if its y coordinate, for give x, is larger then the (x,y) on the parabola for the same x. How would you write that as an inequality?
 
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