Problem 30 Graph. $$y \leq|x+2|$$... [FREE SOLUTION] (2024)

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Graphing an absolute value function involves a few key steps:

• First, identify the vertex of the V-shape. As mentioned, for y = |x + 2|, the vertex is at (-2, 0).
• Next, plot this vertex as the central point of the graph.
• Since absolute value functions are symmetric, the arms of the V will mirror each other from the vertex. One side of the V will extend upwards and to the right, and the other side will extend upwards and to the left.

By plotting a couple of points on either side of the vertex, you can draw the V-shaped graph accurately. For example, when x = 0, y = |0 + 2| = 2. Plot the point (0, 2) and similarly plot a point on the left side. These steps help in generating a clear graph.

Once the absolute value function is graphed, the next step is to consider the inequality. Here, we are dealing with the inequality y ≤ |x + 2|. This means we need to shade the region where the y-values are less than or equal to the values on the graph of y = |x + 2|.
In practical terms, shading represents all possible solutions to the inequality. To do this:

• First, ensure that the boundary line (y = |x + 2|) is included in the shaded region because the inequality is less than or equal to (≤).
• Then, shade the entire region below the V-shaped graph. This shaded area shows all the points where the y-values are less than or equal to the corresponding values of |x + 2|.

This shading process visually illustrates the set of all solutions to the inequality.

Boundary conditions are essential in understanding the constraints of the inequality. For the given exercise y ≤ |x + 2|, the boundary condition is represented by the line y = |x + 2|.
Since the inequality includes the 'equal to' part (≤), we need to include the boundary line itself in the solution.
To finalize the graph considering the boundary conditions:

• Plot the line y = |x + 2| precisely.
• Ensure this line is bold or evident to clearly differentiate between the shaded region and the boundary.

By doing this, you capture all points that satisfy the inequality y ≤ |x + 2|, including those that exactly lie on the line y = |x + 2|. This approach completes the graphical solution of the inequality, ensuring no points are missed out.