Graphs of Absolute Value Functions

Absolute value functions turn negative numbers positive while leaving positive numbers unchanged. The most basic absolute value function is y = |x|, which creates a V-shaped graph with its point at the origin.

When graphing absolute value functions, you'll notice they always create this distinctive V-shape. The "point" of the V is an important feature we'll identify as the vertex.

Quick Tip: Remember that absolute value means "distance from zero" on a number line - this explains why the graph forms a V-shape!

Graphing y = |x|

The basic absolute value function y = |x| forms the foundation for all absolute value functions. To graph it, we create a table of values and plot the points.

For each input value of x, we find |x| - the absolute value or distance from zero. For example, |-2| = 2 because -2 is 2 units away from zero. Similarly, |2| = 2 because 2 is also 2 units away from zero.

When we plot these points ((-2,2), (-1,1), (0,0), (1,1), (2,2)), we get a V-shaped graph that opens upward with its point at the origin (0,0).

Remember: The absolute value of any number is always positive or zero, never negative!

Identifying the Vertex

The vertex is the highest or lowest point on the graph of an absolute value function. It's the "point" of the V-shape.

For y = |x|, the vertex is at (0,0) - this is where the function changes direction. For absolute value functions that open upward, the vertex is the lowest point. For those that open downward, it's the highest point.

Finding the vertex helps us understand how the function behaves and is crucial for graphing transformations of absolute value functions.

Math Insight: The vertex is where the function changes from decreasing to increasing (or vice versa) - it's the turning point of the graph!

Axis of Symmetry

The axis of symmetry is a vertical line that divides the graph of an absolute value function into two mirror images. It passes through the vertex.

For the basic function y = |x|, the axis of symmetry is the y-axis $x = 0$. This means if you fold the graph along the y-axis, both sides would match perfectly.

Understanding the axis of symmetry helps us graph absolute value functions more efficiently. We can plot points on one side, then reflect them across the axis to complete the graph.

Visualization Tip: Think of the axis of symmetry as a mirror - whatever happens on one side is reflected exactly on the other side!

Graphing y = |x| + 1 (Vertical Shifts)

Adding a constant to an absolute value function shifts the entire graph vertically. For y = |x| + 1, we're shifting the basic function y = |x| up by 1 unit.

To graph this, we create a table of values. For each x, we find |x| and then add 1. For example, when x = -2, y = |-2| + 1 = 2 + 1 = 3.

The vertex of this function is at (0,1) - shifted up 1 unit from the original (0,0). The V-shape remains the same, but the entire graph moves up, and now the lowest point is at y = 1 instead of y = 0.

Key Pattern: When you add a constant c to |x|, the graph shifts up by c units; when you subtract a constant, it shifts down by that amount.

Graphing y = |x| + 2

Continuing with vertical shifts, y = |x| + 2 shifts the basic absolute value function up by 2 units.

The table of values confirms this: for each x-value, we calculate |x| and add 2. This means every y-coordinate is 2 units higher than in the basic function y = |x|.

The vertex of this function is now at (0,2), shifted up 2 units from the original position. The axis of symmetry remains at x = 0 $the y-axis$, as the vertical shift doesn't affect the left-right positioning.

Try This: Cover up the bottom portion of the graph below y = 2. Can you see how the remaining part looks just like the original y = |x| graph but shifted up?

Graphing y = |x| + 3

With y = |x| + 3, we're shifting the basic absolute value function up by 3 units.

Our table of values shows this clearly - each y-value is 3 more than the corresponding value in y = |x|. For instance, when x = -2, y = |-2| + 3 = 2 + 3 = 5.

The vertex of this function is at (0,3), and the graph maintains its V-shape while sitting 3 units higher than the original function. The axis of symmetry stays at x = 0.

Pattern Alert: Do you see how the vertex moves up by exactly the amount we add to the function? This is true for any vertical shift of the form y = |x| + c.

Graphing y = |x| - 1

When we subtract a constant from an absolute value function, the graph shifts down. For y = |x| - 1, we're shifting the basic function down by 1 unit.

Looking at our table of values, for each x, we find |x| and subtract 1. When x = 0, y = |0| - 1 = 0 - 1 = -1, placing our vertex at (0,-1).

The V-shape is preserved, but the entire graph is now 1 unit lower than the original function. The axis of symmetry remains at x = 0.

Visual Connection: Compare this graph to y = |x| + 1. Do you notice they're the same shape but positioned differently on the y-axis?

Graphing y = |x| - 2

With y = |x| - 2, we're shifting the basic absolute value function down by 2 units.

Our table shows this pattern: when x = 0, y = |0| - 2 = 0 - 2 = -2, making our vertex (0,-2). When x = ±1, y = |±1| - 2 = 1 - 2 = -1.

Notice that some y-values are now negative! This happens because we're subtracting 2 from all values of |x|. The function maintains its V-shape but is positioned 2 units below the original function.

Math Insight: Even though absolute values are always positive (or zero), the function y = |x| - 2 can produce negative outputs because we subtract after taking the absolute value.

Graphing y = |x| - 3

Continuing with vertical shifts downward, y = |x| - 3 shifts the basic absolute value function down by 3 units.

From our table, we can see that the vertex is now at (0,-3), and all points are 3 units lower than in the original function. For example, when x = 2, y = |2| - 3 = 2 - 3 = -1.

The V-shape and axis of symmetry $x = 0$ remain unchanged, but the entire graph now sits 3 units below the basic function y = |x|.

Pattern Challenge: What would the graph of y = |x| - 5 look like? Where would its vertex be? $The vertex would be at (0,-5), 5 units below the origin.$