Question

In Problems 5–24, graph each equation of the system. Then solve the system to find the points of intersection.

$$\begin{gathered} x^2+y^2=4; \\ {y}^2-x=4 \end{gathered}$$

Short answer

The graph of the system of equations $x^2+y^2=4;y^2-x=4$is:

Step-by-step solution

Step 1. Given

The system of non-linear equation:

$$\begin{gathered} x^2+y^2=4; \\ {y}^2-x=4 \end{gathered}$$

To graph the equation and to find the point of intersection.

Step 2. Graph the equations

Graph the equations in the same plane.

Step 3. To find the point of intersection.

Substitute the second equation in the first equation,

$$\begin{gathered} {(y^2-4)}^2+y^2=4; \\ {y}^4-8y^2+16+y^2=4 \\ {y}^4-7y^2+12=0 \end{gathered}$$

Step 4. Solve the equation

$$\begin{gathered} y^4-7y^2+12=0 \\ (y^2-4)(y^2-3)=0 \\ {y}^2=3,4 \\ y=\pm \sqrt{3},\pm 2 \end{gathered}$$

Step 5. Find x

When $y=\pm \sqrt{3}$,

$$\begin{gathered} x^2+{(\pm \sqrt{3})}^2=4 \\ {x}^2+3=4 \\ {x}^2=1 \\ x=\pm 1 \end{gathered}$$

When $y=\pm 2$,

$$\begin{gathered} x^2+{(\pm 2)}^2=4 \\ {x}^2+4=4 \\ {x}^2=0 \\ x=0 \end{gathered}$$

The points of intersection are $(-1,-\sqrt{3})(-1,\sqrt{3}),(0,-2),(0,2)$

Step 6. Plot the point of intersection

The graph of the system of non-linear equations with the point of intersection is: