Question

Graph each equation of the system. Then solve the system to find the points of intersection. $$ \left\\{\begin{array}{l} x^{2}=y \\ x y=1 \end{array}\right. $$

Short answer

The points of intersection are (1, 1) and (-1, 1).

Step-by-step solution

Graph the first equation

The first equation given is \( x^2 = y \). This is a parabolic equation which opens upwards. Plot this parabola on the coordinate system.

Graph the second equation

The second equation given is \( xy = 1 \). This is a hyperbola with asymptotes on the x and y axes. Plot this hyperbola on the coordinate system.

Find the points of intersection

To find the points of intersection, substitute y from the first equation into the second equation: \[ x \times (x^2) = 1 \ x^3 = 1 \ x = 1 \] Now substitute \(x = 1\) back into the first equation: \[ y = x^2 = 1^2 = 1 \] Thus, one point of intersection is \((1, 1)\).

Check for any additional intersections

Also, consider potential negative solutions. Substitute \(y = x^2\) into \(xy = 1\), and solve for \(x\) (negative root): \[x^3 = 1 \ x^3 + (-1)^3 = 0 \ (x + 1)(x^2 - x + 1) = 0 \ x = -1 \] Substituting \(x = -1\) back into the first equation: \[ y = x^2 = (-1)^2 = 1 \] Thus, another point of intersection is \((-1, 1)\).

Key concepts

graphing parabolas

The first equation in our system is \(x^2 = y\). This represents a **parabola**. A parabola is a U-shaped curve that can open either upwards or downwards. In this case, since the coefficient of \(x^2\) is positive, the parabola opens upwards. Here’s how to graph it effectively:
- Start by determining the vertex. For \(x^2 = y\), the vertex is at (0, 0).
- Find several points to plot. For example, if \(x = 1\), then \(y = 1\). If \(x = -1\), then \(y = 1\) as well. Continue this for more points.
- Plot the vertex and these points on a coordinate system.
- Draw a smooth curve through these points to complete the parabola.
Parabolas are symmetric about the y-axis. This symmetry can be helpful in sketching an accurate graph.

graphing hyperbolas

The second equation in our system is \(xy = 1\). This equation represents a **hyperbola**. A hyperbola consists of two separate curves called branches. In this case, the branches are found in diagonally opposite quadrants. Here’s how to graph it:
- Identify the asymptotes. For \(xy = 1\), the x and y axes are asymptotes, meaning the branches will approach these lines but never touch them.
- Find points to plot. For instance, if \(x = 1\), then \(y = 1\). If \(x = -1\), then \(y = -1\). Likewise, if \(x = 0.5\), \(y = 2\), and if \(x = -0.5\), \(y = -2\).
- Plot these points and draw two curves that come close to the asymptotes without touching them.
- Ensure the curves are in opposite quadrants: top right and bottom left for positive values, top left and bottom right for negative values.

The hyperbola's unique shape and asymptotic behavior are key to understanding how it interacts with other curves like parabolas.

points of intersection

To find the **points of intersection** between the parabola \(x^2 = y\) and the hyperbola \(xy = 1\), follow these steps:
- **Substitution**: Start by substituting \(y\) from the first equation into the second. This means replacing \(y\) in \(xy = 1\) with \(x^2\). So, \( x \times x^2 = 1 \rightarrow x^3 = 1 \rightarrow x = 1 \).
- **Evaluate**: Substitute \(x = 1\) back into the first equation to find \(y\). Thus, \(y = 1^2 = 1\). Therefore, one point of intersection is \( (1, 1) \).
- **Check for more solutions**: Consider negative solutions. Solve \( x^3 = 1 \) which involves considering the negative solution: \(-1\). Substituting \(x = -1\) into \(x^2 = y\), we find \(y = (-1)^2 = 1\). Thus, another point of intersection is \((-1, 1)\).

The points of intersection provide solutions to the system and highlight where the curves of the parabola and the hyperbola meet. In this example, the points \( (1, 1) \) and \((-1, 1)\) show where the parabola and hyperbola intersect graphically.