Question

Graph each equation of the system. Then solve the system to find the points of intersection. $$ \left\\{\begin{array}{l} x^{2}+y^{2}=4 \\ y^{2}-x=4 \end{array}\right. $$

Short answer

The points of intersection are: (0, 2), (0, -2), (-1, \sqrt{5}), (-1, -\sqrt{5}).

Step-by-step solution

Identify the Equations

The system of equations given is: 1. \(x^2 + y^2 = 4\) 2. \(y^2 - x = 4\).

Graph the First Equation

The first equation \(x^2 + y^2 = 4\) represents a circle centered at the origin with a radius of 2. Draw this circle on the coordinate plane.

Solve for y in the Second Equation

Isolate \(y^2\) in the second equation: \(y^2 = x + 4\).

Graph the Parabola

The equation \(y^2 = x + 4\) represents a parabola that opens to the right. Shift the standard parabola \(y^2 = x\) four units left to graph this equation.

Find the Points of Intersection

To find the intersection points, equate \(y^2\) from both equations: \(4 - x^2 = x + 4\). Solve this equation for \(x\).

Solve the Equation

Rearrange the equation to get: \( x^2 + x = 0\), or \(x(x+1) = 0\). The solutions are: \(x = 0\) or \(x = -1\).

Find Corresponding y Values

For \(x = 0\), substitute into the circle equation \(x^2 + y^2 = 4\): \(0 + y^2 = 4\) \(y^2 = 4\), so \(y = \pm 2\). For \(x = -1\), \(-1 + y^2 = 4\) \(y^2 = 5\), so \(y = \pm \sqrt{5}\).

List the Intersection Points

The points of intersection are: \( (0, 2) \) \( (0, -2) \) \( (-1, \sqrt{5}) \) \( (-1, -\sqrt{5}) \) .

Key concepts

Graphing Equations

Graphing equations is a visual way to understand how equations relate to each other. This helps with finding solutions to systems of equations.

When you graph an equation, you plot points that satisfy the equation on a coordinate plane and then connect those points to form a shape, like a line, circle, or parabola.

Let's take the first equation from our exercise, which is a circle: \(x^2 + y^2 = 4\). This represents all points that are exactly 2 units away from the origin (0,0) because it's a circle centered at the origin with a radius of 2.

The second equation is a parabola: \(y^2 = x + 4\). Parabolas usually open either up or down, or left and right. This one opens to the right because the equation has \(y^2\) instead of \(x^2\). When graphing, you'll see how each equation's shape can intersect with another one.

Points of Intersection

Points of intersection are where the graphs of the equations in a system cross each other.
This means these points satisfy all the equations in the system simultaneously.

To find these points, you can graph each equation and look for the coordinates where the shapes overlap.

In our exercise, the circle and the parabola intersect at multiple points:

• Graph the circle \(x^2 + y^2 = 4\).
• Graph the parabola \(y^2 = x + 4\).
• Check where the two graphs overlap to find their intersection points.
Using math, you can also find intersection points by setting the equations equal to each other:
\(4 - x^2 = x + 4\)


Solving this, we get:
\[ x^2 + x = 0 \]

This equation simplifies to find the solutions: \(x = 0\) or \(x = -1\).

Then, substituting each x-value into either equation helps find corresponding y-values, providing the exact intersection points.

Circle and Parabola Intersections

Circle and parabola intersections are special because these shapes are different in how they spread out on the graph.

The circle, represented by \(x^2 + y^2 = 4\), is symmetric around the center point, while the parabola, represented by \(y^2 = x + 4\), is symmetric around its axis.

To find intersections:

• You solve the system by equalizing their y-values or x-values.
• Here, we start with equating y-values because both equations have a y squared term: \(4 - x^2 = x + 4\).

Simplifying this, we get:\
\(x(x + 1) = 0 \).
This gives solutions \(x = 0\) and \(x = -1\).

Now, plug these x-values back into either equation, such as the circle equation, to find y-values:

• For \(x = 0\), \(0 + y^2 = 4\) yields \(y = \pm 2\).
• For \(x = -1\), \(-1 + y^2 = 4\) yields \(y = \pm \sqrt{5}\).

This process shows that the graphs intersect at the points \((0, 2)\), \((0, -2)\), \((-1, \sqrt{5})\), and \((-1, -\sqrt{5})\).

Understanding these intersections helps in visually and algebraically solving systems of equations.

Circle and Parabola Intersections

Circle and parabola intersections are special because these shapes are different in how they spread out on the graph.

The circle, represented by \(x^2 + y^2 = 4\), is symmetric around the center point, while the parabola, represented by \(y^2 = x + 4\), is symmetric around its axis.

To find intersections:

• You solve the system by equalizing their y-values or x-values.
• Here, we start with equating y-values because both equations have a y squared term: \(4 - x^2 = x + 4\).

Simplifying this, we get:\
\(x(x + 1) = 0 \).
This gives solutions \(x = 0\) and \(x = -1\).

Now, plug these x-values back into either equation, such as the circle equation, to find y-values:

• For \(x = 0\), \(0 + y^2 = 4\) yields \(y = \pm 2\).
• For \(x = -1\), \(-1 + y^2 = 4\) yields \(y = \pm \sqrt{5}\).

This process shows that the graphs intersect at the points \((0, 2)\), \((0, -2)\), \((-1, \sqrt{5})\), and \((-1, -\sqrt{5})\).

Understanding these intersections helps in visually and algebraically solving systems of equations.