The question is
Is the point (-3, -2) a solution of the intersection of the following set of quadratic equations:
Y < -X^2
X^2 + Y^2 < 16
I guess I am somewhat confused by the way it's written. Would I be graphing this to find out? I'm so confused with this problem. I have asked for help and alot of people really don't understand it when they look at it. The answer I have for it is 321, but I am not quite sure if that really answers the question. Is the point a solution of the intersection of those equations? Please help me, I feel so lost!
The solution to inequalities is a region, not just a point. In this case, think of the graphs.
y < -x^2 is the whole inside of the downward-opening parabola with vertex at (0,0).
x^2 + y^2 < 16 is the inside of the circle of radius 4 centered at (0,0).
So, is 9+4<16? yes, the point is inside the circle.
Is -2 < -9? No. It is not inside the parabola.
How can you possibly come up with 321 as an answer? Is that x or y?
The point is not a solution of equations, because you don't have any equations; you have inequalities.
visit wolframalpha . com and enter
plot {x^2+y^2 < 16, y < -x^2}
to see the shaded region which satisfies both inequalities.
To determine whether the point (-3, -2) is a solution of the intersection of the given set of quadratic equations, you can follow these steps:
1. Equation 1: Y < -X^2
To check if (-3, -2) satisfies this inequality, substitute the x-coordinate (-3) and the y-coordinate (-2) into the equation:
-2 < -(-3)^2
-2 < -9
Since -2 is not less than -9, the point (-3, -2) does not satisfy Equation 1.
2. Equation 2: X^2 + Y^2 < 16
To check if (-3, -2) satisfies this inequality, substitute the x-coordinate (-3) and the y-coordinate (-2) into the equation:
(-3)^2 + (-2)^2 < 16
9 + 4 < 16
13 < 16
Since 13 is less than 16, the point (-3, -2) does satisfy Equation 2.
To find the intersection of the two equations, we need to find the solutions that satisfy both equations simultaneously. Since (-3, -2) satisfies Equation 2 but not Equation 1, it is not a solution of the intersection of the given equations.
Therefore, the answer is that the point (-3, -2) is not a solution of the intersection of the quadratic equations Y < -X^2 and X^2 + Y^2 < 16.