Problem 29 - Calculus Early Transcendentals: Pearson New International Edition

Question

, plot the graph of each equation. Begin by checking for symmetries and be sure to find all \(x\) - and \(y\) -intercepts.. $$ |x|+|y|=1 $$

Step-by-Step Solution

Answer

Graph is a diamond shape with vertices (1, 0), (-1, 0), (0, 1), and (0, -1).

Step 1: Identify the equation form

The equation given is \(|x| + |y| = 1\). This is an absolute value equation involving both \(x\) and \(y\).

Step 2: Check for symmetry

This equation is symmetric with respect to both the \(x\)-axis and the \(y\)-axis. Whenever \(x\) is replaced by \(-x\) or \(y\) is replaced by \(-y\), the equation remains the same.

Step 3: Find the x-intercepts

Set \(y = 0\) to find the \(x\)-intercepts. The equation becomes \(|x| + |0| = 1\), which simplifies to \(|x| = 1\). Thus, the \(x\)-intercepts are \((1, 0)\) and \((-1, 0)\).

Step 4: Find the y-intercepts

Set \(x = 0\) to find the \(y\)-intercepts. The equation becomes \(|0| + |y| = 1\), which simplifies to \(|y| = 1\). Thus, the \(y\)-intercepts are \((0, 1)\) and \((0, -1)\).

Step 5: Plotting Points

Use points found from intercepts to sketch the basic structure. The points \((1, 0), (-1, 0), (0, 1), (0, -1)\) form the vertices of a square centered on the origin.

Step 6: Connecting the Points

Join the plotted points with lines. The solution will make a diamond shape with lines connecting \( (1,0) \), \((0, 1)\), \(( -1, 0)\), and \((0, -1)\). Each line segment is straight because the equation \(|x| + |y| = 1\) in each quadrant becomes a line. For example, in the first quadrant, the equation becomes \(x + y = 1\).

Key Concepts

Symmetry in Graphs

When dealing with the equation \( |x| + |y| = 1 \), symmetry plays a crucial role. Understanding symmetry can simplify the graphing process immensely. For this equation, we have symmetrical properties around both the x-axis and y-axis. This means if you flip the graph over the x-axis or y-axis, it will look the same.

• Symmetry with respect to the x-axis: For every point \((x, y)\), the point \((x, -y)\) will also lie on the graph. This happens because replacing \(y\) with \(-y\) still satisfies the equation.
• Symmetry with respect to the y-axis: Similarly, for any point \((x, y)\), the point \((-x, y)\) will lie on the graph. Replacing \(x\) with \(-x\) doesn't change the truth of the equation.

Symmetry checks make it easier to predict the shape of the graph, allowing us to sketch only a section, then mirror it across the axes.

X-Intercepts

Finding the \(x\)-intercepts is a key part of graphing any equation. These are points where the graph meets the x-axis. For our equation, \(|x| + |y| = 1\), we find \(x\)-intercepts by setting \(y = 0\).
The equation simplifies to \(|x| = 1\). Absolute value expressions like this mean \(x\) can equal either \(1\) or \(-1\), resulting in two intercepts: \((1, 0)\) and \((-1, 0)\).
These intercept points help in outlining the boundary or shape of the graph. Always remember, more than one \(x\)-intercept is possible due to absolute values.

Y-Intercepts

To discover where the graph touches the y-axis, we seek the \(y\)-intercepts by setting \(x = 0\). In \(|x| + |y| = 1\), this results in \(|y| = 1\). Absolute value tells us \(y\) can be \(1\) or \(-1\).
Consequently, our y-intercepts are \((0, 1)\) and \((0, -1)\). These points are essential for shaping the graph. Since the path crosses the y-axis in two places, they anchor parts of the graph.
Finding and plotting these intercepts first provides structural guidance when sketching the equation.

Graphing Equations

Graphing involves plotting points and connecting them to visualize their relationships. For \(|x| + |y| = 1\), the intercepts form a useful starting framework.
This equation naturally forms a diamond shape, positioned at coordinates: \((1, 0)\), \((-1, 0)\), \((0, 1)\), and \((0, -1)\). These shape the vertices of a diamond centered at the origin.
To complete the graph, join these plotted intercepts with straight line segments:

• Connect \((1, 0)\) to \((0, 1)\)
• Connect \((0, 1)\) to \((-1, 0)\)
• Connect \((-1, 0)\) to \((0, -1)\)
• Connect \((0, -1)\) back to \((1, 0)\)

These lines represent quadrants where the equation behaves like a linear equation, such as \(x + y = 1\) in one quadrant. This method of connecting intercepts provides a clear, visual representation of absolute value equations.