Question 695832
First let's look at the graph of the two parabolas:
{{{graph(500, 500, -5, 5, -5, 5, x^2+2x+2, -x^2-4x-2)}}}
The red one is {{{y = x^2+2x+2}}} and the green one is {{{y = -x^2-4x-2}}}. The solution to {{{y >= x^2+2x+2}}} would be the red parabola and all the area above/inside the bowl. The solution to {{{y < -x^2-4x-2}}} would be all the area below/inside the green parabola. (Not the green parabola itself because that is where y equals {{{-x^2-4x-2}}} and the inequality does not include "or equal to".)<br>
So the solution to the system is where these two areas overlap each other: The enclosed area between the two parabolas. To express this solution we must first find the two points where the parabolas intersect. Setting
{{{x^2+2x+2 = -x^2-4x-2}}}
and solving for x we should be able to find the points of intersection. Subtracting the entire right side we get:
{{{2x^2+6x+4 = 0}}}
Factor out the GCF of 2:
{{{2(x^2+3x+2) = 0}}}
Factoring the trinomial:
{{{2(x+1)(x+2) = 0}}}
From the Zero Product Property:
x + 1 = 0 or x + 2 = 0
Solving:
x = -1 or x = -2<br>
Using these x values and either equation for the parabola we can find that the y values for each x are 1 and 2, respectively. So the points of intersection are:
(-1, 1) and (-2, 2)
So the x values of the solution are between -1 and -2, inclusive:
{{{-2 < x < -1}}}
and the y values of the solution are:
{{{x^2+2x+2 <= y < -x^2-4x-2}}}