Question

Which of the following nonlinear inequalities is graphed below? Picture included below!! PLEASE HELP ASAP!

Answer 1

I think it is A, cause the horizontal axis is 4 (x), and the vertical axis is 8 (y)
and then the <= means that it is inside the ellipse

Answer 2

The correct option is A.

Explanation:

The general form of an ellipse is

`(x^2)/(a^2)+(y^2)/(b^2)=1`

Where, (0,0) is the center of ellipse, (±a,0) are x-intercepts and (0,±b) are y-intercepts.

From the given graph it is clear that the center of ellipse is (0,0), (±2,0) are x-intercepts and (0,±4) are y-intercepts. So, the equation of ellipse is

`(x^2)/((2)^2)+(y^2)/((4)^2)=1`

`(x^2)/(4)+(y^2)/(16)=1`

The shaded region is inside the ellipse, (0,0) lie in the solution set and the points lie on the ellipse are included in the solution set, therefore the sign of inequality is ≤.

`((0)^2)/(4)+((0)^2)/(16)\leq 1`

`0\leq 1`

This statement is true. The required inequality is

`(x^2)/(4)+(y^2)/(16)\leq 1`

Therefore, the correct option is A.