Functions
The Absolute Value Function and Its Graph
The V-shaped graph of y = |x|, how shifts and reflections move it, and a method for sketching any transformed version.
What this page covers
- Sketch y = |x| and identify its key features
- Locate the vertex of a transformed absolute-value graph
- Find the intercepts of such a graph
- Predict whether the V opens upward or downward
The basic graph
The two-case definition produces two half-lines. For x ≥ 0, |x| = x; for x < 0, |x| = -x. Both pass through the origin, with slopes 1 and -1.
| x | −3 | −2 | −1 | 0 | 1 | 2 | 3 |
|---|---|---|---|---|---|---|---|
| y = |x| | 3 | 2 | 1 | 0 | 1 | 2 | 3 |
- Shape
- A V with its point at the origin
- Vertex
- (0, 0)
- Domain
- All real x
- Range
- y ≥ 0, that is [0, ∞)
- Symmetry
- About the y-axis — the function is even
- Reflecting line
- The x-axis, y = 0
The typed notes call the x-axis the bounce line, then give its proper name: the reflecting line. Everything below it in the graph of y = x is folded up above it.
That folding picture is the most useful way to think about the graph. Draw y = x, then reflect whatever lies below the x-axis upward. The same procedure works for any expression inside the bars.
Transformations
| Form | Effect | Vertex |
|---|---|---|
| y = |x| | The basic V | (0, 0) |
| y = |x - h| | Shift right by h | (h, 0) |
| y = |x| + k | Shift up by k | (0, k) |
| y = -|x| | Reflect in the x-axis; opens downward | (0, 0) |
| y = a|x| | Steeper if |a| > 1, flatter if |a| < 1 | (0, 0) |
| y = k - |x - h| | Shift and reflect; opens downward | (h, k) |
|x + 3| shifts left by 3, not right, because x + 3 = x - (-3). The vertex sits where the inside is zero, which is x = -3.
Sketching a transformed graph
- Find the vertex. Set the expression inside the bars to zero and solve for x; substitute back to get y.
- Decide which way it opens. A positive multiple of the absolute value opens upward; a negative one opens downward.
- Find the y-intercept. Set x = 0.
- Find the x-intercepts, if any. Set y = 0 and solve the absolute-value equation.
- Draw two half-lines from the vertex through the intercepts.
Worked example — y = 2 - |x + 3|
Vertex. x + 3 = 0 gives x = -3, and then y = 2 - 0 = 2. The vertex is (-3, 2).
Direction. The absolute value is subtracted, so the graph opens downward and 2 is the maximum value. The typed notes reason it exactly this way: because we have 2 minus the absolute value, y will never be larger than 2.
x-intercepts. Set y = 0:
y-intercept. At x = 0: y = 2 - |3| = 2 - 3 = -1, giving (0, -1).
The three key x values: intercepts at -5 and -1, vertex directly above -3 at height 2.
The two x-intercepts should be equidistant from the vertex. -5 and -1 are each 2 away from -3. That agreement is a free check on the arithmetic.
Worked example — y = 1 + |x - 3|
Vertex. x - 3 = 0 gives x = 3, and y = 1. The vertex is (3, 1).
Direction. The absolute value is added, so the graph opens upward and 1 is the minimum.
x-intercepts. Setting y = 0 gives |x - 3| = -1, which has no solution. The typed notes ask is this possible? — and it is not. The graph never reaches the x-axis.
y-intercept. At x = 0: y = 1 + |-3| = 1 + 3 = 4.
This is the general pattern: a V opening upward with a vertex above the axis has no x-intercepts, and one opening downward with a vertex below it has none either. Checking the vertex first tells you how many to expect.
How many intercepts to expect
| Opens | Vertex y | x-intercepts |
|---|---|---|
| Upward | Below the axis | Two |
| Upward | On the axis | One (the vertex) |
| Upward | Above the axis | None |
| Downward | Above the axis | Two |
| Downward | On the axis | One |
| Downward | Below the axis | None |
The pattern mirrors that of a parabola, and for the same reason: both graphs have a single turning point and open consistently in one direction.
Common mistakes
| Mistake | Correct | Check |
|---|---|---|
| |x + 3| shifts right | It shifts left | The vertex is where the inside is zero |
| Assuming the vertex is always at the origin | It moves with the transformation | Set the inside to zero |
| Missing that a minus flips the V | 2 - |x + 3| opens downward | Its vertex is a maximum |
| Solving |x - 3| = -1 | No solution | A distance is never negative |
| Giving one x-intercept when there are two | Two cases | They sit symmetrically about the vertex |
| Drawing curved sides | The sides are straight | Each branch is a line of slope ±a |
Frequently asked questions
Why is the graph V-shaped?
For x ≥ 0 the function is y = x, a line of slope 1. For x < 0 it is y = -x, a line of slope -1. Two half-lines meeting at the origin make a V.
Where is the vertex?
Wherever the expression inside the bars is zero. For y = 2 - |x + 3| that is x = -3, and the y value there is 2.
When does the V open downward?
When the absolute value is subtracted, or multiplied by a negative. y = 2 - |x + 3| opens downward, so its vertex is a maximum.
Can such a graph miss the x-axis entirely?
Yes. y = 1 + |x - 3| has minimum value 1, so it never reaches zero. The typed notes pose exactly this question and leave it for the reader.
Source. Handwritten teaching notes, Week 4, page 2, with the typed supplementary fact sheet on absolute value functions.
This page is an original exposition prepared for the KEVOS® knowledge library. It restates, reorganises and verifies the mathematics of the supplied teaching notes; it is not a reproduction of them. Numerical values taken from the notes are identified as source examples and are not presented as engineering standards.