Question

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

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

Short answer

he graph of the system of equations $y=\sqrt{4-x^2};y=2x+4$is:

The points of intersection are $(-\frac{6}{5},\frac{8}{5}),(-2,0)$

Step-by-step solution

Step 1. Given

The system of non-linear equation:

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

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.

Equate both the equations,

$\sqrt{4-x^2}=2x+4$

Squaring on both sides, we get,

$$\begin{gathered} 4-x^2={(2x+4)}^2 \\ 4-x^2=4x^2+16x+16 \\ 5x^2+16x+12=0 \end{gathered}$$

Step 4. Solve the equations

$$\begin{gathered} 5x^2+16x+12=0 \\ (5x+6)(x+2)=0 \\ x=-\frac{6}{5},-2 \end{gathered}$$

The value of $x$ is $-\frac{6}{5},-2$

Step 5. Find y

When

$$\begin{gathered} x=-\frac{6}{5}, \\ y=2x+4 \\ =2(-\frac{6}{5})+4 \\ =-\frac{12}{5}+4 \\ =\frac{8}{5} \end{gathered}$$

When $x=-2,$

$$\begin{gathered} y=2x+4 \\ =2(-2)+4 \\ =0 \end{gathered}$$

The point of intersections are $(-\frac{6}{5},\frac{8}{5}),(-2,0)$