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Q. 17

Expert-verified
Found in: Page 772

### Precalculus Enhanced with Graphing Utilities

Book edition 6th
Author(s) Sullivan
Pages 1200 pages
ISBN 9780321795465

# In Problems 5–24, graph each equation of the system. Then solve the system to find the points of intersection.${x}^{2}+{y}^{2}=4;\phantom{\rule{0ex}{0ex}}{y}^{2}-x=4$

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

See the step by step solution

## Step 1. Given

The system of non-linear equation:

${x}^{2}+{y}^{2}=4;\phantom{\rule{0ex}{0ex}}{y}^{2}-x=4$

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,

${\left({y}^{2}-4\right)}^{2}+{y}^{2}=4;\phantom{\rule{0ex}{0ex}}{y}^{4}-8{y}^{2}+16+{y}^{2}=4\phantom{\rule{0ex}{0ex}}{y}^{4}-7{y}^{2}+12=0$

## Step 4. Solve the equation

${y}^{4}-7{y}^{2}+12=0\phantom{\rule{0ex}{0ex}}\left({y}^{2}-4\right)\left({y}^{2}-3\right)=0\phantom{\rule{0ex}{0ex}}{y}^{2}=3,4\phantom{\rule{0ex}{0ex}}y=±\sqrt{3},±2$

## Step 5. Find x

When $y=±\sqrt{3}$,

${x}^{2}+{\left(±\sqrt{3}\right)}^{2}=4\phantom{\rule{0ex}{0ex}}{x}^{2}+3=4\phantom{\rule{0ex}{0ex}}{x}^{2}=1\phantom{\rule{0ex}{0ex}}x=±1$

When $y=±2$,

${x}^{2}+{\left(±2\right)}^{2}=4\phantom{\rule{0ex}{0ex}}{x}^{2}+4=4\phantom{\rule{0ex}{0ex}}{x}^{2}=0\phantom{\rule{0ex}{0ex}}x=0$

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

## Step 6. Plot the point of intersection

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