Question

Is (0,0) a solution to this system y>=x^2+x-4; y<=x^2+2x+1

Answer 1

Answer:

It is a solution

Step-by-step explanation:

Check and see if (0,0) satisfies both inequalities.

The first inequality is:

$y \geqslant {x}^{2} + x - 4$

When we put x=0, and y=0, we get:

$0\geqslant {0}^{2} + 0 - 4$

$\implies0\geqslant- 4$

This part is true.

The second inequality is:

$y \leqslant {x}^{2} + 2 x + 1$

We put x=0 and y=0 to get:

$0 \leqslant {0}^{2} + 2 (0) + 1$

$0 \leqslant 1$

This part is also true.

Since the (0,0) not satisfy both, inequalities, it is a solution.

Answer 2

Answer:

yes it is a solution

Step-by-step explanation:

o>-4

0<1

yes both work out to be equal