Question

Show that it is possible to find graphs with equations of the forms y=A-(x-a)^2 and y=B+(x-b)^2 with A>B which do not intersect?

Y=A-(x-a)^2
The graph of Y is a parabola which bends downwards. The simplest equation for this type of parabola is y=-x^2 which is the case when a and A are both set to zero. If a>0, then the axis of symmetry is to the left of the y-axis. If a<0 the axis of symmetry Is to the right of the y-axis. If A>0 the graph of y=-(x-a)^2 moves a distance of A units up in the y-direction. If A<0, the graph of y=-(x-a)^2 moves a distance of A units down in the y-direction.
The possible combinations of signs for B and b
A>0 a>0
A>0 a<0
A<0 a>0
A<0 a<0
Y=B+(x-b)^2
The graph of Y is a parabola which bends upwards. The simplest equation for this type of parabola is y=x^2 which is the case when b and B are both are set to zero. If b>0, then the axis of symmetry is to the left of the y-axis. If b<0 the axis of symmetry Is to the right of the y-axis. If A>0 the graph of y=-(x-a)^2 moves a distance of A units up in the y-direction. If A<0, the graph of y=-(x-a)^2 moves a distance of A units down in the y-direction.
The possible combinations of signs for B and b
B>0 b>0
B>0 b<0
B<0 b>0
B<0 b<0
Provided these conditions
B>0 b>0 or B>0 b<0 and A<0 a>0 or A<0 a<0 the parabolas will not intersect. I can't decide whether or not the parabola will intersect in the case of the other combinations

Answer 1

The parabolas will not intersect for

`2 (A - B)< (a-b)^2`

Explanation:

Supposing that

`A-(x-a)^2=B+(x-b)^2` we have

`A-B = 2x^2-2(a+b)x+a^2+b^2` or

`x^2-(a+b)x+(a^2+b^2+B-A)/2=0`

with solutions

`x = 1/2 (a + b pm sqrt[2 (A - B)-(a-b)^2])`

Those solutions are real if

`2 (A - B)-(a-b)^2 ge 0`

otherwise

`y_1=A-(x-a)^2` and `y_2=B+(x-b)^2` will not intersect.